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Work, Energy and Power AP style. Energy. Energy: the currency of the universe . Everything has to be “paid for” with energy. Energy can’t be created or destroyed, but it can be transformed from one kind to another and it can be transferred from one object to another. - PowerPoint PPT Presentation
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Work Energy and PowerAP style
Energy
Energy the currency of the universe
Everything has to be ldquopaid forrdquo with energy
Energy canrsquot be created or destroyed but it can be transformed from one kind to another and it can be transferred from one object to another
bull Doing WORK is one way to transfer energy from one object to another
Work = Force x displacement
W = F∙dbull Unit for work is Newton x meter One
Newton-meter is also called a Joule J
Work- the transfer of energy
Work = Force middot displacementbull Work is not done unless there is a displacement bull If you hold an object a long time you may get tired
but NO work was done on the objectbull If you push against a solid wall for hours there is
still NO work done on the wall
bull For work to be done the displacement of the object must be along the same direction as the applied force They must be parallel
bull If the force and the displacement are perpendicular to each other NO work is done by the force
bull For example in lifting a book the force exerted by your hands is upward and the displacement is upward- work is done
bull Similarly in lowering a book the force exerted by your hands is still upward and the displacement is downward
bull The force and the displacement are STILL parallel so work is still done
bull But since they are in opposite directions now it is NEGATIVE work
F
F
d
d
bull On the other hand while carrying a book down the hallway the force from your hands is vertical and the displacement of the book is horizontal
bull Therefore NO work is done by your hands
bull Since the book is obviously moving what force IS doing work
The static friction force between your hands and the book is acting parallel to the displacement and IS doing work
F
d
Both Force and displacement are vectors
So using vector multiplication
Work =
This is a DOT product- only parallel components will yield a non-zero solution
In many university texts and sometimes the AP test the displacement which is the change in
position is represented by ldquosrdquo and not ldquodrdquo
W =
The solution for dot products are NOT vectors
Work is not a vector
Letrsquos look an example of using a dot product to calculate workhellip
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement s
s = Dr =
rf ndash ro =(i + 7j) - (-4i + 3j) =
5i + 4j
Then do the dot product W = F middot s(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2 J of work
ExampleHow much work is done to push a 5 kg
cat with a force of 25 N to the top of a ramp that is 7 meters long and 3 meters tall at a constant velocity
W = Force middot displacement
Which measurement is parallel to the force- the length of the ramp or the height of the ramp
W = 25 N x 7 m
W = 175 J 7 m
3 mF = 25 N
ExampleHow much work is done to carry a 5 kg
cat to the top of a ramp that is 7 meters long and 3 meters tall at a constant velocity
W = Force middot displacement
What force is required to carry the cat
Force = weight of the cat
Which is parallel to the Force vector- the length of the ramp or the height
d = height NOT length
W = mg x h
W = 5 kg x 10 ms2 x 3 m
W = 150 J
7 m
3 m
If all four ramps are the same height which ramp would require the greatest amount of work in order to carry an object to the top of the ramp
W = F∙ Dr
For lifting an object the distance d is the height
Because they all have the same height the work for each is the same
bull Andhellipwhile carrying yourself when climbing stairs or walking up an incline at a constant velocity only the height is used to calculate the work you do to get yourself to the top
bull The force required is your weight Horizontal component of d
Ver
tical
com
pone
nt o
f d
Yo
ur
Fo
rce
How much work do you do on a 30 kg cat to carry it from one side of the room to the other if the room is 10 meters long
ZERO because your Force is vertical but the displacement is horizontal
ExampleA boy pushes a
lawnmower 20 meters across the yard If he pushed with a force of 200 N and the angle between the handle and the ground was 50 degrees how much work did he do
q
F
Displacement = 20 m
F cos q
W = (F cos q )dW = (200 N cos 50˚) 20 mW = 2571 J
NOTE If while pushing an object it is moving at a constant velocity
the NET force must be zero
Sohellip Your applied force must be exactly equal to any resistant forces like friction
S F = ma
FA ndash f = ma = 0
A 50 kg box is pulled 6 m across a rough horizontal floor (m = 04) with a force of 80 N at an angle of 35 degrees above the horizontal What is the work done by EACH force exerted on it What is the NET work done
Does the gravitational force do any work NO It is perpendicular to the displacement Does the Normal force do any work No It is perpendicular to the displacement Does the applied Force do any work Yes but ONLY its horizontal component
WF = FAcosq x d = 80cos 35˚ x 6 m = 39319 J Does friction do any work Yes but first what is the normal force Itrsquos NOT mg
Normal = mg ndash FAsinq
Wf = -f x d = -mN∙d = -m(mg ndash FAsin )q ∙d = -747 J What is the NET work done39319 J ndash 747 J = 38572 J
mg
Normal
FA
qf
bull For work to be done the displacement of the object must be in the same direction as the applied force They must be parallel
bull If the force and the displacement are perpendicular to each other NO work is done by the force
So using vector multiplication
W = F bull d (this is a DOT product)
(In many university texts as well as the AP test the displacement is given by d = Dr where r is the
position vector displacement = change in position
W = F bull Dr
If the motion is in only one direction
W = F bull Dx
Or
W = F bull Dy
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement Dr
rf ndash ro =(i + 7j) - (-4i + 3j) =
Dr = 5i + 4j
Then complete the dot product W = F middot Dr(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2J of work
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
If the force varies with displacement-
in other words
Force is a function of displacement
you must integrate to find the work done
Examples of Integration
An object of mass m is subject to a force given by F(x) = 3x3 N What is the work done by the force
W F x dx x dx x 33
43 4
Examples of Definite Integration
To stretch a NON linear spring a distance x requires a force given by F(x) = 4x2 ndash x
What work is done to stretch the spring 2 meters beyond its equilibrium position
W F ds x x ds x x Jmm
m ( ) ( ) ( ) 4
4
3
1
2
4
32
1
22
4
30
1
20 8 6672
0
23
0
22
0
2 3 2 2
Graphing Force vs postion
bull If you graph the applied force vs the position you can find how much work was done by the force
Work = Fmiddotd = ldquoarea under the curverdquoTotal Work = 2 N x 2 m + 3N x 4m = 16 JArea UNDER the x-axis is NEGATIVE work = - 1N x 2m
Force N
Position m
F
dNet work = 16 J ndash 2 J = 14 J
The Integral = the area under the curve
If y = f(x) then the area under the f(x) curve from x = a to x = b is equal to
( )b
a
f x dx
Therefore given a graph of Force vs position the work done is the area under the curve
Work and Energy
Work and EnergyOften some force must do work
to give an object potential or kinetic energy
You push a wagon and it starts moving You do work to stretch a spring and you transform your work energy into spring potential energy
Or you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy
Work = Force x distance = transfer of energy
Kinetic Energy
the energy of motion
K = frac12 mv2
Kinetic Energy
the energy of motion
K = frac12 mv2
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Energy
Energy the currency of the universe
Everything has to be ldquopaid forrdquo with energy
Energy canrsquot be created or destroyed but it can be transformed from one kind to another and it can be transferred from one object to another
bull Doing WORK is one way to transfer energy from one object to another
Work = Force x displacement
W = F∙dbull Unit for work is Newton x meter One
Newton-meter is also called a Joule J
Work- the transfer of energy
Work = Force middot displacementbull Work is not done unless there is a displacement bull If you hold an object a long time you may get tired
but NO work was done on the objectbull If you push against a solid wall for hours there is
still NO work done on the wall
bull For work to be done the displacement of the object must be along the same direction as the applied force They must be parallel
bull If the force and the displacement are perpendicular to each other NO work is done by the force
bull For example in lifting a book the force exerted by your hands is upward and the displacement is upward- work is done
bull Similarly in lowering a book the force exerted by your hands is still upward and the displacement is downward
bull The force and the displacement are STILL parallel so work is still done
bull But since they are in opposite directions now it is NEGATIVE work
F
F
d
d
bull On the other hand while carrying a book down the hallway the force from your hands is vertical and the displacement of the book is horizontal
bull Therefore NO work is done by your hands
bull Since the book is obviously moving what force IS doing work
The static friction force between your hands and the book is acting parallel to the displacement and IS doing work
F
d
Both Force and displacement are vectors
So using vector multiplication
Work =
This is a DOT product- only parallel components will yield a non-zero solution
In many university texts and sometimes the AP test the displacement which is the change in
position is represented by ldquosrdquo and not ldquodrdquo
W =
The solution for dot products are NOT vectors
Work is not a vector
Letrsquos look an example of using a dot product to calculate workhellip
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement s
s = Dr =
rf ndash ro =(i + 7j) - (-4i + 3j) =
5i + 4j
Then do the dot product W = F middot s(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2 J of work
ExampleHow much work is done to push a 5 kg
cat with a force of 25 N to the top of a ramp that is 7 meters long and 3 meters tall at a constant velocity
W = Force middot displacement
Which measurement is parallel to the force- the length of the ramp or the height of the ramp
W = 25 N x 7 m
W = 175 J 7 m
3 mF = 25 N
ExampleHow much work is done to carry a 5 kg
cat to the top of a ramp that is 7 meters long and 3 meters tall at a constant velocity
W = Force middot displacement
What force is required to carry the cat
Force = weight of the cat
Which is parallel to the Force vector- the length of the ramp or the height
d = height NOT length
W = mg x h
W = 5 kg x 10 ms2 x 3 m
W = 150 J
7 m
3 m
If all four ramps are the same height which ramp would require the greatest amount of work in order to carry an object to the top of the ramp
W = F∙ Dr
For lifting an object the distance d is the height
Because they all have the same height the work for each is the same
bull Andhellipwhile carrying yourself when climbing stairs or walking up an incline at a constant velocity only the height is used to calculate the work you do to get yourself to the top
bull The force required is your weight Horizontal component of d
Ver
tical
com
pone
nt o
f d
Yo
ur
Fo
rce
How much work do you do on a 30 kg cat to carry it from one side of the room to the other if the room is 10 meters long
ZERO because your Force is vertical but the displacement is horizontal
ExampleA boy pushes a
lawnmower 20 meters across the yard If he pushed with a force of 200 N and the angle between the handle and the ground was 50 degrees how much work did he do
q
F
Displacement = 20 m
F cos q
W = (F cos q )dW = (200 N cos 50˚) 20 mW = 2571 J
NOTE If while pushing an object it is moving at a constant velocity
the NET force must be zero
Sohellip Your applied force must be exactly equal to any resistant forces like friction
S F = ma
FA ndash f = ma = 0
A 50 kg box is pulled 6 m across a rough horizontal floor (m = 04) with a force of 80 N at an angle of 35 degrees above the horizontal What is the work done by EACH force exerted on it What is the NET work done
Does the gravitational force do any work NO It is perpendicular to the displacement Does the Normal force do any work No It is perpendicular to the displacement Does the applied Force do any work Yes but ONLY its horizontal component
WF = FAcosq x d = 80cos 35˚ x 6 m = 39319 J Does friction do any work Yes but first what is the normal force Itrsquos NOT mg
Normal = mg ndash FAsinq
Wf = -f x d = -mN∙d = -m(mg ndash FAsin )q ∙d = -747 J What is the NET work done39319 J ndash 747 J = 38572 J
mg
Normal
FA
qf
bull For work to be done the displacement of the object must be in the same direction as the applied force They must be parallel
bull If the force and the displacement are perpendicular to each other NO work is done by the force
So using vector multiplication
W = F bull d (this is a DOT product)
(In many university texts as well as the AP test the displacement is given by d = Dr where r is the
position vector displacement = change in position
W = F bull Dr
If the motion is in only one direction
W = F bull Dx
Or
W = F bull Dy
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement Dr
rf ndash ro =(i + 7j) - (-4i + 3j) =
Dr = 5i + 4j
Then complete the dot product W = F middot Dr(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2J of work
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
If the force varies with displacement-
in other words
Force is a function of displacement
you must integrate to find the work done
Examples of Integration
An object of mass m is subject to a force given by F(x) = 3x3 N What is the work done by the force
W F x dx x dx x 33
43 4
Examples of Definite Integration
To stretch a NON linear spring a distance x requires a force given by F(x) = 4x2 ndash x
What work is done to stretch the spring 2 meters beyond its equilibrium position
W F ds x x ds x x Jmm
m ( ) ( ) ( ) 4
4
3
1
2
4
32
1
22
4
30
1
20 8 6672
0
23
0
22
0
2 3 2 2
Graphing Force vs postion
bull If you graph the applied force vs the position you can find how much work was done by the force
Work = Fmiddotd = ldquoarea under the curverdquoTotal Work = 2 N x 2 m + 3N x 4m = 16 JArea UNDER the x-axis is NEGATIVE work = - 1N x 2m
Force N
Position m
F
dNet work = 16 J ndash 2 J = 14 J
The Integral = the area under the curve
If y = f(x) then the area under the f(x) curve from x = a to x = b is equal to
( )b
a
f x dx
Therefore given a graph of Force vs position the work done is the area under the curve
Work and Energy
Work and EnergyOften some force must do work
to give an object potential or kinetic energy
You push a wagon and it starts moving You do work to stretch a spring and you transform your work energy into spring potential energy
Or you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy
Work = Force x distance = transfer of energy
Kinetic Energy
the energy of motion
K = frac12 mv2
Kinetic Energy
the energy of motion
K = frac12 mv2
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
bull Doing WORK is one way to transfer energy from one object to another
Work = Force x displacement
W = F∙dbull Unit for work is Newton x meter One
Newton-meter is also called a Joule J
Work- the transfer of energy
Work = Force middot displacementbull Work is not done unless there is a displacement bull If you hold an object a long time you may get tired
but NO work was done on the objectbull If you push against a solid wall for hours there is
still NO work done on the wall
bull For work to be done the displacement of the object must be along the same direction as the applied force They must be parallel
bull If the force and the displacement are perpendicular to each other NO work is done by the force
bull For example in lifting a book the force exerted by your hands is upward and the displacement is upward- work is done
bull Similarly in lowering a book the force exerted by your hands is still upward and the displacement is downward
bull The force and the displacement are STILL parallel so work is still done
bull But since they are in opposite directions now it is NEGATIVE work
F
F
d
d
bull On the other hand while carrying a book down the hallway the force from your hands is vertical and the displacement of the book is horizontal
bull Therefore NO work is done by your hands
bull Since the book is obviously moving what force IS doing work
The static friction force between your hands and the book is acting parallel to the displacement and IS doing work
F
d
Both Force and displacement are vectors
So using vector multiplication
Work =
This is a DOT product- only parallel components will yield a non-zero solution
In many university texts and sometimes the AP test the displacement which is the change in
position is represented by ldquosrdquo and not ldquodrdquo
W =
The solution for dot products are NOT vectors
Work is not a vector
Letrsquos look an example of using a dot product to calculate workhellip
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement s
s = Dr =
rf ndash ro =(i + 7j) - (-4i + 3j) =
5i + 4j
Then do the dot product W = F middot s(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2 J of work
ExampleHow much work is done to push a 5 kg
cat with a force of 25 N to the top of a ramp that is 7 meters long and 3 meters tall at a constant velocity
W = Force middot displacement
Which measurement is parallel to the force- the length of the ramp or the height of the ramp
W = 25 N x 7 m
W = 175 J 7 m
3 mF = 25 N
ExampleHow much work is done to carry a 5 kg
cat to the top of a ramp that is 7 meters long and 3 meters tall at a constant velocity
W = Force middot displacement
What force is required to carry the cat
Force = weight of the cat
Which is parallel to the Force vector- the length of the ramp or the height
d = height NOT length
W = mg x h
W = 5 kg x 10 ms2 x 3 m
W = 150 J
7 m
3 m
If all four ramps are the same height which ramp would require the greatest amount of work in order to carry an object to the top of the ramp
W = F∙ Dr
For lifting an object the distance d is the height
Because they all have the same height the work for each is the same
bull Andhellipwhile carrying yourself when climbing stairs or walking up an incline at a constant velocity only the height is used to calculate the work you do to get yourself to the top
bull The force required is your weight Horizontal component of d
Ver
tical
com
pone
nt o
f d
Yo
ur
Fo
rce
How much work do you do on a 30 kg cat to carry it from one side of the room to the other if the room is 10 meters long
ZERO because your Force is vertical but the displacement is horizontal
ExampleA boy pushes a
lawnmower 20 meters across the yard If he pushed with a force of 200 N and the angle between the handle and the ground was 50 degrees how much work did he do
q
F
Displacement = 20 m
F cos q
W = (F cos q )dW = (200 N cos 50˚) 20 mW = 2571 J
NOTE If while pushing an object it is moving at a constant velocity
the NET force must be zero
Sohellip Your applied force must be exactly equal to any resistant forces like friction
S F = ma
FA ndash f = ma = 0
A 50 kg box is pulled 6 m across a rough horizontal floor (m = 04) with a force of 80 N at an angle of 35 degrees above the horizontal What is the work done by EACH force exerted on it What is the NET work done
Does the gravitational force do any work NO It is perpendicular to the displacement Does the Normal force do any work No It is perpendicular to the displacement Does the applied Force do any work Yes but ONLY its horizontal component
WF = FAcosq x d = 80cos 35˚ x 6 m = 39319 J Does friction do any work Yes but first what is the normal force Itrsquos NOT mg
Normal = mg ndash FAsinq
Wf = -f x d = -mN∙d = -m(mg ndash FAsin )q ∙d = -747 J What is the NET work done39319 J ndash 747 J = 38572 J
mg
Normal
FA
qf
bull For work to be done the displacement of the object must be in the same direction as the applied force They must be parallel
bull If the force and the displacement are perpendicular to each other NO work is done by the force
So using vector multiplication
W = F bull d (this is a DOT product)
(In many university texts as well as the AP test the displacement is given by d = Dr where r is the
position vector displacement = change in position
W = F bull Dr
If the motion is in only one direction
W = F bull Dx
Or
W = F bull Dy
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement Dr
rf ndash ro =(i + 7j) - (-4i + 3j) =
Dr = 5i + 4j
Then complete the dot product W = F middot Dr(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2J of work
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
If the force varies with displacement-
in other words
Force is a function of displacement
you must integrate to find the work done
Examples of Integration
An object of mass m is subject to a force given by F(x) = 3x3 N What is the work done by the force
W F x dx x dx x 33
43 4
Examples of Definite Integration
To stretch a NON linear spring a distance x requires a force given by F(x) = 4x2 ndash x
What work is done to stretch the spring 2 meters beyond its equilibrium position
W F ds x x ds x x Jmm
m ( ) ( ) ( ) 4
4
3
1
2
4
32
1
22
4
30
1
20 8 6672
0
23
0
22
0
2 3 2 2
Graphing Force vs postion
bull If you graph the applied force vs the position you can find how much work was done by the force
Work = Fmiddotd = ldquoarea under the curverdquoTotal Work = 2 N x 2 m + 3N x 4m = 16 JArea UNDER the x-axis is NEGATIVE work = - 1N x 2m
Force N
Position m
F
dNet work = 16 J ndash 2 J = 14 J
The Integral = the area under the curve
If y = f(x) then the area under the f(x) curve from x = a to x = b is equal to
( )b
a
f x dx
Therefore given a graph of Force vs position the work done is the area under the curve
Work and Energy
Work and EnergyOften some force must do work
to give an object potential or kinetic energy
You push a wagon and it starts moving You do work to stretch a spring and you transform your work energy into spring potential energy
Or you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy
Work = Force x distance = transfer of energy
Kinetic Energy
the energy of motion
K = frac12 mv2
Kinetic Energy
the energy of motion
K = frac12 mv2
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Work- the transfer of energy
Work = Force middot displacementbull Work is not done unless there is a displacement bull If you hold an object a long time you may get tired
but NO work was done on the objectbull If you push against a solid wall for hours there is
still NO work done on the wall
bull For work to be done the displacement of the object must be along the same direction as the applied force They must be parallel
bull If the force and the displacement are perpendicular to each other NO work is done by the force
bull For example in lifting a book the force exerted by your hands is upward and the displacement is upward- work is done
bull Similarly in lowering a book the force exerted by your hands is still upward and the displacement is downward
bull The force and the displacement are STILL parallel so work is still done
bull But since they are in opposite directions now it is NEGATIVE work
F
F
d
d
bull On the other hand while carrying a book down the hallway the force from your hands is vertical and the displacement of the book is horizontal
bull Therefore NO work is done by your hands
bull Since the book is obviously moving what force IS doing work
The static friction force between your hands and the book is acting parallel to the displacement and IS doing work
F
d
Both Force and displacement are vectors
So using vector multiplication
Work =
This is a DOT product- only parallel components will yield a non-zero solution
In many university texts and sometimes the AP test the displacement which is the change in
position is represented by ldquosrdquo and not ldquodrdquo
W =
The solution for dot products are NOT vectors
Work is not a vector
Letrsquos look an example of using a dot product to calculate workhellip
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement s
s = Dr =
rf ndash ro =(i + 7j) - (-4i + 3j) =
5i + 4j
Then do the dot product W = F middot s(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2 J of work
ExampleHow much work is done to push a 5 kg
cat with a force of 25 N to the top of a ramp that is 7 meters long and 3 meters tall at a constant velocity
W = Force middot displacement
Which measurement is parallel to the force- the length of the ramp or the height of the ramp
W = 25 N x 7 m
W = 175 J 7 m
3 mF = 25 N
ExampleHow much work is done to carry a 5 kg
cat to the top of a ramp that is 7 meters long and 3 meters tall at a constant velocity
W = Force middot displacement
What force is required to carry the cat
Force = weight of the cat
Which is parallel to the Force vector- the length of the ramp or the height
d = height NOT length
W = mg x h
W = 5 kg x 10 ms2 x 3 m
W = 150 J
7 m
3 m
If all four ramps are the same height which ramp would require the greatest amount of work in order to carry an object to the top of the ramp
W = F∙ Dr
For lifting an object the distance d is the height
Because they all have the same height the work for each is the same
bull Andhellipwhile carrying yourself when climbing stairs or walking up an incline at a constant velocity only the height is used to calculate the work you do to get yourself to the top
bull The force required is your weight Horizontal component of d
Ver
tical
com
pone
nt o
f d
Yo
ur
Fo
rce
How much work do you do on a 30 kg cat to carry it from one side of the room to the other if the room is 10 meters long
ZERO because your Force is vertical but the displacement is horizontal
ExampleA boy pushes a
lawnmower 20 meters across the yard If he pushed with a force of 200 N and the angle between the handle and the ground was 50 degrees how much work did he do
q
F
Displacement = 20 m
F cos q
W = (F cos q )dW = (200 N cos 50˚) 20 mW = 2571 J
NOTE If while pushing an object it is moving at a constant velocity
the NET force must be zero
Sohellip Your applied force must be exactly equal to any resistant forces like friction
S F = ma
FA ndash f = ma = 0
A 50 kg box is pulled 6 m across a rough horizontal floor (m = 04) with a force of 80 N at an angle of 35 degrees above the horizontal What is the work done by EACH force exerted on it What is the NET work done
Does the gravitational force do any work NO It is perpendicular to the displacement Does the Normal force do any work No It is perpendicular to the displacement Does the applied Force do any work Yes but ONLY its horizontal component
WF = FAcosq x d = 80cos 35˚ x 6 m = 39319 J Does friction do any work Yes but first what is the normal force Itrsquos NOT mg
Normal = mg ndash FAsinq
Wf = -f x d = -mN∙d = -m(mg ndash FAsin )q ∙d = -747 J What is the NET work done39319 J ndash 747 J = 38572 J
mg
Normal
FA
qf
bull For work to be done the displacement of the object must be in the same direction as the applied force They must be parallel
bull If the force and the displacement are perpendicular to each other NO work is done by the force
So using vector multiplication
W = F bull d (this is a DOT product)
(In many university texts as well as the AP test the displacement is given by d = Dr where r is the
position vector displacement = change in position
W = F bull Dr
If the motion is in only one direction
W = F bull Dx
Or
W = F bull Dy
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement Dr
rf ndash ro =(i + 7j) - (-4i + 3j) =
Dr = 5i + 4j
Then complete the dot product W = F middot Dr(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2J of work
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
If the force varies with displacement-
in other words
Force is a function of displacement
you must integrate to find the work done
Examples of Integration
An object of mass m is subject to a force given by F(x) = 3x3 N What is the work done by the force
W F x dx x dx x 33
43 4
Examples of Definite Integration
To stretch a NON linear spring a distance x requires a force given by F(x) = 4x2 ndash x
What work is done to stretch the spring 2 meters beyond its equilibrium position
W F ds x x ds x x Jmm
m ( ) ( ) ( ) 4
4
3
1
2
4
32
1
22
4
30
1
20 8 6672
0
23
0
22
0
2 3 2 2
Graphing Force vs postion
bull If you graph the applied force vs the position you can find how much work was done by the force
Work = Fmiddotd = ldquoarea under the curverdquoTotal Work = 2 N x 2 m + 3N x 4m = 16 JArea UNDER the x-axis is NEGATIVE work = - 1N x 2m
Force N
Position m
F
dNet work = 16 J ndash 2 J = 14 J
The Integral = the area under the curve
If y = f(x) then the area under the f(x) curve from x = a to x = b is equal to
( )b
a
f x dx
Therefore given a graph of Force vs position the work done is the area under the curve
Work and Energy
Work and EnergyOften some force must do work
to give an object potential or kinetic energy
You push a wagon and it starts moving You do work to stretch a spring and you transform your work energy into spring potential energy
Or you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy
Work = Force x distance = transfer of energy
Kinetic Energy
the energy of motion
K = frac12 mv2
Kinetic Energy
the energy of motion
K = frac12 mv2
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Work = Force middot displacementbull Work is not done unless there is a displacement bull If you hold an object a long time you may get tired
but NO work was done on the objectbull If you push against a solid wall for hours there is
still NO work done on the wall
bull For work to be done the displacement of the object must be along the same direction as the applied force They must be parallel
bull If the force and the displacement are perpendicular to each other NO work is done by the force
bull For example in lifting a book the force exerted by your hands is upward and the displacement is upward- work is done
bull Similarly in lowering a book the force exerted by your hands is still upward and the displacement is downward
bull The force and the displacement are STILL parallel so work is still done
bull But since they are in opposite directions now it is NEGATIVE work
F
F
d
d
bull On the other hand while carrying a book down the hallway the force from your hands is vertical and the displacement of the book is horizontal
bull Therefore NO work is done by your hands
bull Since the book is obviously moving what force IS doing work
The static friction force between your hands and the book is acting parallel to the displacement and IS doing work
F
d
Both Force and displacement are vectors
So using vector multiplication
Work =
This is a DOT product- only parallel components will yield a non-zero solution
In many university texts and sometimes the AP test the displacement which is the change in
position is represented by ldquosrdquo and not ldquodrdquo
W =
The solution for dot products are NOT vectors
Work is not a vector
Letrsquos look an example of using a dot product to calculate workhellip
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement s
s = Dr =
rf ndash ro =(i + 7j) - (-4i + 3j) =
5i + 4j
Then do the dot product W = F middot s(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2 J of work
ExampleHow much work is done to push a 5 kg
cat with a force of 25 N to the top of a ramp that is 7 meters long and 3 meters tall at a constant velocity
W = Force middot displacement
Which measurement is parallel to the force- the length of the ramp or the height of the ramp
W = 25 N x 7 m
W = 175 J 7 m
3 mF = 25 N
ExampleHow much work is done to carry a 5 kg
cat to the top of a ramp that is 7 meters long and 3 meters tall at a constant velocity
W = Force middot displacement
What force is required to carry the cat
Force = weight of the cat
Which is parallel to the Force vector- the length of the ramp or the height
d = height NOT length
W = mg x h
W = 5 kg x 10 ms2 x 3 m
W = 150 J
7 m
3 m
If all four ramps are the same height which ramp would require the greatest amount of work in order to carry an object to the top of the ramp
W = F∙ Dr
For lifting an object the distance d is the height
Because they all have the same height the work for each is the same
bull Andhellipwhile carrying yourself when climbing stairs or walking up an incline at a constant velocity only the height is used to calculate the work you do to get yourself to the top
bull The force required is your weight Horizontal component of d
Ver
tical
com
pone
nt o
f d
Yo
ur
Fo
rce
How much work do you do on a 30 kg cat to carry it from one side of the room to the other if the room is 10 meters long
ZERO because your Force is vertical but the displacement is horizontal
ExampleA boy pushes a
lawnmower 20 meters across the yard If he pushed with a force of 200 N and the angle between the handle and the ground was 50 degrees how much work did he do
q
F
Displacement = 20 m
F cos q
W = (F cos q )dW = (200 N cos 50˚) 20 mW = 2571 J
NOTE If while pushing an object it is moving at a constant velocity
the NET force must be zero
Sohellip Your applied force must be exactly equal to any resistant forces like friction
S F = ma
FA ndash f = ma = 0
A 50 kg box is pulled 6 m across a rough horizontal floor (m = 04) with a force of 80 N at an angle of 35 degrees above the horizontal What is the work done by EACH force exerted on it What is the NET work done
Does the gravitational force do any work NO It is perpendicular to the displacement Does the Normal force do any work No It is perpendicular to the displacement Does the applied Force do any work Yes but ONLY its horizontal component
WF = FAcosq x d = 80cos 35˚ x 6 m = 39319 J Does friction do any work Yes but first what is the normal force Itrsquos NOT mg
Normal = mg ndash FAsinq
Wf = -f x d = -mN∙d = -m(mg ndash FAsin )q ∙d = -747 J What is the NET work done39319 J ndash 747 J = 38572 J
mg
Normal
FA
qf
bull For work to be done the displacement of the object must be in the same direction as the applied force They must be parallel
bull If the force and the displacement are perpendicular to each other NO work is done by the force
So using vector multiplication
W = F bull d (this is a DOT product)
(In many university texts as well as the AP test the displacement is given by d = Dr where r is the
position vector displacement = change in position
W = F bull Dr
If the motion is in only one direction
W = F bull Dx
Or
W = F bull Dy
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement Dr
rf ndash ro =(i + 7j) - (-4i + 3j) =
Dr = 5i + 4j
Then complete the dot product W = F middot Dr(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2J of work
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
If the force varies with displacement-
in other words
Force is a function of displacement
you must integrate to find the work done
Examples of Integration
An object of mass m is subject to a force given by F(x) = 3x3 N What is the work done by the force
W F x dx x dx x 33
43 4
Examples of Definite Integration
To stretch a NON linear spring a distance x requires a force given by F(x) = 4x2 ndash x
What work is done to stretch the spring 2 meters beyond its equilibrium position
W F ds x x ds x x Jmm
m ( ) ( ) ( ) 4
4
3
1
2
4
32
1
22
4
30
1
20 8 6672
0
23
0
22
0
2 3 2 2
Graphing Force vs postion
bull If you graph the applied force vs the position you can find how much work was done by the force
Work = Fmiddotd = ldquoarea under the curverdquoTotal Work = 2 N x 2 m + 3N x 4m = 16 JArea UNDER the x-axis is NEGATIVE work = - 1N x 2m
Force N
Position m
F
dNet work = 16 J ndash 2 J = 14 J
The Integral = the area under the curve
If y = f(x) then the area under the f(x) curve from x = a to x = b is equal to
( )b
a
f x dx
Therefore given a graph of Force vs position the work done is the area under the curve
Work and Energy
Work and EnergyOften some force must do work
to give an object potential or kinetic energy
You push a wagon and it starts moving You do work to stretch a spring and you transform your work energy into spring potential energy
Or you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy
Work = Force x distance = transfer of energy
Kinetic Energy
the energy of motion
K = frac12 mv2
Kinetic Energy
the energy of motion
K = frac12 mv2
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
bull For work to be done the displacement of the object must be along the same direction as the applied force They must be parallel
bull If the force and the displacement are perpendicular to each other NO work is done by the force
bull For example in lifting a book the force exerted by your hands is upward and the displacement is upward- work is done
bull Similarly in lowering a book the force exerted by your hands is still upward and the displacement is downward
bull The force and the displacement are STILL parallel so work is still done
bull But since they are in opposite directions now it is NEGATIVE work
F
F
d
d
bull On the other hand while carrying a book down the hallway the force from your hands is vertical and the displacement of the book is horizontal
bull Therefore NO work is done by your hands
bull Since the book is obviously moving what force IS doing work
The static friction force between your hands and the book is acting parallel to the displacement and IS doing work
F
d
Both Force and displacement are vectors
So using vector multiplication
Work =
This is a DOT product- only parallel components will yield a non-zero solution
In many university texts and sometimes the AP test the displacement which is the change in
position is represented by ldquosrdquo and not ldquodrdquo
W =
The solution for dot products are NOT vectors
Work is not a vector
Letrsquos look an example of using a dot product to calculate workhellip
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement s
s = Dr =
rf ndash ro =(i + 7j) - (-4i + 3j) =
5i + 4j
Then do the dot product W = F middot s(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2 J of work
ExampleHow much work is done to push a 5 kg
cat with a force of 25 N to the top of a ramp that is 7 meters long and 3 meters tall at a constant velocity
W = Force middot displacement
Which measurement is parallel to the force- the length of the ramp or the height of the ramp
W = 25 N x 7 m
W = 175 J 7 m
3 mF = 25 N
ExampleHow much work is done to carry a 5 kg
cat to the top of a ramp that is 7 meters long and 3 meters tall at a constant velocity
W = Force middot displacement
What force is required to carry the cat
Force = weight of the cat
Which is parallel to the Force vector- the length of the ramp or the height
d = height NOT length
W = mg x h
W = 5 kg x 10 ms2 x 3 m
W = 150 J
7 m
3 m
If all four ramps are the same height which ramp would require the greatest amount of work in order to carry an object to the top of the ramp
W = F∙ Dr
For lifting an object the distance d is the height
Because they all have the same height the work for each is the same
bull Andhellipwhile carrying yourself when climbing stairs or walking up an incline at a constant velocity only the height is used to calculate the work you do to get yourself to the top
bull The force required is your weight Horizontal component of d
Ver
tical
com
pone
nt o
f d
Yo
ur
Fo
rce
How much work do you do on a 30 kg cat to carry it from one side of the room to the other if the room is 10 meters long
ZERO because your Force is vertical but the displacement is horizontal
ExampleA boy pushes a
lawnmower 20 meters across the yard If he pushed with a force of 200 N and the angle between the handle and the ground was 50 degrees how much work did he do
q
F
Displacement = 20 m
F cos q
W = (F cos q )dW = (200 N cos 50˚) 20 mW = 2571 J
NOTE If while pushing an object it is moving at a constant velocity
the NET force must be zero
Sohellip Your applied force must be exactly equal to any resistant forces like friction
S F = ma
FA ndash f = ma = 0
A 50 kg box is pulled 6 m across a rough horizontal floor (m = 04) with a force of 80 N at an angle of 35 degrees above the horizontal What is the work done by EACH force exerted on it What is the NET work done
Does the gravitational force do any work NO It is perpendicular to the displacement Does the Normal force do any work No It is perpendicular to the displacement Does the applied Force do any work Yes but ONLY its horizontal component
WF = FAcosq x d = 80cos 35˚ x 6 m = 39319 J Does friction do any work Yes but first what is the normal force Itrsquos NOT mg
Normal = mg ndash FAsinq
Wf = -f x d = -mN∙d = -m(mg ndash FAsin )q ∙d = -747 J What is the NET work done39319 J ndash 747 J = 38572 J
mg
Normal
FA
qf
bull For work to be done the displacement of the object must be in the same direction as the applied force They must be parallel
bull If the force and the displacement are perpendicular to each other NO work is done by the force
So using vector multiplication
W = F bull d (this is a DOT product)
(In many university texts as well as the AP test the displacement is given by d = Dr where r is the
position vector displacement = change in position
W = F bull Dr
If the motion is in only one direction
W = F bull Dx
Or
W = F bull Dy
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement Dr
rf ndash ro =(i + 7j) - (-4i + 3j) =
Dr = 5i + 4j
Then complete the dot product W = F middot Dr(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2J of work
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
If the force varies with displacement-
in other words
Force is a function of displacement
you must integrate to find the work done
Examples of Integration
An object of mass m is subject to a force given by F(x) = 3x3 N What is the work done by the force
W F x dx x dx x 33
43 4
Examples of Definite Integration
To stretch a NON linear spring a distance x requires a force given by F(x) = 4x2 ndash x
What work is done to stretch the spring 2 meters beyond its equilibrium position
W F ds x x ds x x Jmm
m ( ) ( ) ( ) 4
4
3
1
2
4
32
1
22
4
30
1
20 8 6672
0
23
0
22
0
2 3 2 2
Graphing Force vs postion
bull If you graph the applied force vs the position you can find how much work was done by the force
Work = Fmiddotd = ldquoarea under the curverdquoTotal Work = 2 N x 2 m + 3N x 4m = 16 JArea UNDER the x-axis is NEGATIVE work = - 1N x 2m
Force N
Position m
F
dNet work = 16 J ndash 2 J = 14 J
The Integral = the area under the curve
If y = f(x) then the area under the f(x) curve from x = a to x = b is equal to
( )b
a
f x dx
Therefore given a graph of Force vs position the work done is the area under the curve
Work and Energy
Work and EnergyOften some force must do work
to give an object potential or kinetic energy
You push a wagon and it starts moving You do work to stretch a spring and you transform your work energy into spring potential energy
Or you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy
Work = Force x distance = transfer of energy
Kinetic Energy
the energy of motion
K = frac12 mv2
Kinetic Energy
the energy of motion
K = frac12 mv2
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
bull For example in lifting a book the force exerted by your hands is upward and the displacement is upward- work is done
bull Similarly in lowering a book the force exerted by your hands is still upward and the displacement is downward
bull The force and the displacement are STILL parallel so work is still done
bull But since they are in opposite directions now it is NEGATIVE work
F
F
d
d
bull On the other hand while carrying a book down the hallway the force from your hands is vertical and the displacement of the book is horizontal
bull Therefore NO work is done by your hands
bull Since the book is obviously moving what force IS doing work
The static friction force between your hands and the book is acting parallel to the displacement and IS doing work
F
d
Both Force and displacement are vectors
So using vector multiplication
Work =
This is a DOT product- only parallel components will yield a non-zero solution
In many university texts and sometimes the AP test the displacement which is the change in
position is represented by ldquosrdquo and not ldquodrdquo
W =
The solution for dot products are NOT vectors
Work is not a vector
Letrsquos look an example of using a dot product to calculate workhellip
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement s
s = Dr =
rf ndash ro =(i + 7j) - (-4i + 3j) =
5i + 4j
Then do the dot product W = F middot s(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2 J of work
ExampleHow much work is done to push a 5 kg
cat with a force of 25 N to the top of a ramp that is 7 meters long and 3 meters tall at a constant velocity
W = Force middot displacement
Which measurement is parallel to the force- the length of the ramp or the height of the ramp
W = 25 N x 7 m
W = 175 J 7 m
3 mF = 25 N
ExampleHow much work is done to carry a 5 kg
cat to the top of a ramp that is 7 meters long and 3 meters tall at a constant velocity
W = Force middot displacement
What force is required to carry the cat
Force = weight of the cat
Which is parallel to the Force vector- the length of the ramp or the height
d = height NOT length
W = mg x h
W = 5 kg x 10 ms2 x 3 m
W = 150 J
7 m
3 m
If all four ramps are the same height which ramp would require the greatest amount of work in order to carry an object to the top of the ramp
W = F∙ Dr
For lifting an object the distance d is the height
Because they all have the same height the work for each is the same
bull Andhellipwhile carrying yourself when climbing stairs or walking up an incline at a constant velocity only the height is used to calculate the work you do to get yourself to the top
bull The force required is your weight Horizontal component of d
Ver
tical
com
pone
nt o
f d
Yo
ur
Fo
rce
How much work do you do on a 30 kg cat to carry it from one side of the room to the other if the room is 10 meters long
ZERO because your Force is vertical but the displacement is horizontal
ExampleA boy pushes a
lawnmower 20 meters across the yard If he pushed with a force of 200 N and the angle between the handle and the ground was 50 degrees how much work did he do
q
F
Displacement = 20 m
F cos q
W = (F cos q )dW = (200 N cos 50˚) 20 mW = 2571 J
NOTE If while pushing an object it is moving at a constant velocity
the NET force must be zero
Sohellip Your applied force must be exactly equal to any resistant forces like friction
S F = ma
FA ndash f = ma = 0
A 50 kg box is pulled 6 m across a rough horizontal floor (m = 04) with a force of 80 N at an angle of 35 degrees above the horizontal What is the work done by EACH force exerted on it What is the NET work done
Does the gravitational force do any work NO It is perpendicular to the displacement Does the Normal force do any work No It is perpendicular to the displacement Does the applied Force do any work Yes but ONLY its horizontal component
WF = FAcosq x d = 80cos 35˚ x 6 m = 39319 J Does friction do any work Yes but first what is the normal force Itrsquos NOT mg
Normal = mg ndash FAsinq
Wf = -f x d = -mN∙d = -m(mg ndash FAsin )q ∙d = -747 J What is the NET work done39319 J ndash 747 J = 38572 J
mg
Normal
FA
qf
bull For work to be done the displacement of the object must be in the same direction as the applied force They must be parallel
bull If the force and the displacement are perpendicular to each other NO work is done by the force
So using vector multiplication
W = F bull d (this is a DOT product)
(In many university texts as well as the AP test the displacement is given by d = Dr where r is the
position vector displacement = change in position
W = F bull Dr
If the motion is in only one direction
W = F bull Dx
Or
W = F bull Dy
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement Dr
rf ndash ro =(i + 7j) - (-4i + 3j) =
Dr = 5i + 4j
Then complete the dot product W = F middot Dr(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2J of work
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
If the force varies with displacement-
in other words
Force is a function of displacement
you must integrate to find the work done
Examples of Integration
An object of mass m is subject to a force given by F(x) = 3x3 N What is the work done by the force
W F x dx x dx x 33
43 4
Examples of Definite Integration
To stretch a NON linear spring a distance x requires a force given by F(x) = 4x2 ndash x
What work is done to stretch the spring 2 meters beyond its equilibrium position
W F ds x x ds x x Jmm
m ( ) ( ) ( ) 4
4
3
1
2
4
32
1
22
4
30
1
20 8 6672
0
23
0
22
0
2 3 2 2
Graphing Force vs postion
bull If you graph the applied force vs the position you can find how much work was done by the force
Work = Fmiddotd = ldquoarea under the curverdquoTotal Work = 2 N x 2 m + 3N x 4m = 16 JArea UNDER the x-axis is NEGATIVE work = - 1N x 2m
Force N
Position m
F
dNet work = 16 J ndash 2 J = 14 J
The Integral = the area under the curve
If y = f(x) then the area under the f(x) curve from x = a to x = b is equal to
( )b
a
f x dx
Therefore given a graph of Force vs position the work done is the area under the curve
Work and Energy
Work and EnergyOften some force must do work
to give an object potential or kinetic energy
You push a wagon and it starts moving You do work to stretch a spring and you transform your work energy into spring potential energy
Or you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy
Work = Force x distance = transfer of energy
Kinetic Energy
the energy of motion
K = frac12 mv2
Kinetic Energy
the energy of motion
K = frac12 mv2
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
bull On the other hand while carrying a book down the hallway the force from your hands is vertical and the displacement of the book is horizontal
bull Therefore NO work is done by your hands
bull Since the book is obviously moving what force IS doing work
The static friction force between your hands and the book is acting parallel to the displacement and IS doing work
F
d
Both Force and displacement are vectors
So using vector multiplication
Work =
This is a DOT product- only parallel components will yield a non-zero solution
In many university texts and sometimes the AP test the displacement which is the change in
position is represented by ldquosrdquo and not ldquodrdquo
W =
The solution for dot products are NOT vectors
Work is not a vector
Letrsquos look an example of using a dot product to calculate workhellip
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement s
s = Dr =
rf ndash ro =(i + 7j) - (-4i + 3j) =
5i + 4j
Then do the dot product W = F middot s(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2 J of work
ExampleHow much work is done to push a 5 kg
cat with a force of 25 N to the top of a ramp that is 7 meters long and 3 meters tall at a constant velocity
W = Force middot displacement
Which measurement is parallel to the force- the length of the ramp or the height of the ramp
W = 25 N x 7 m
W = 175 J 7 m
3 mF = 25 N
ExampleHow much work is done to carry a 5 kg
cat to the top of a ramp that is 7 meters long and 3 meters tall at a constant velocity
W = Force middot displacement
What force is required to carry the cat
Force = weight of the cat
Which is parallel to the Force vector- the length of the ramp or the height
d = height NOT length
W = mg x h
W = 5 kg x 10 ms2 x 3 m
W = 150 J
7 m
3 m
If all four ramps are the same height which ramp would require the greatest amount of work in order to carry an object to the top of the ramp
W = F∙ Dr
For lifting an object the distance d is the height
Because they all have the same height the work for each is the same
bull Andhellipwhile carrying yourself when climbing stairs or walking up an incline at a constant velocity only the height is used to calculate the work you do to get yourself to the top
bull The force required is your weight Horizontal component of d
Ver
tical
com
pone
nt o
f d
Yo
ur
Fo
rce
How much work do you do on a 30 kg cat to carry it from one side of the room to the other if the room is 10 meters long
ZERO because your Force is vertical but the displacement is horizontal
ExampleA boy pushes a
lawnmower 20 meters across the yard If he pushed with a force of 200 N and the angle between the handle and the ground was 50 degrees how much work did he do
q
F
Displacement = 20 m
F cos q
W = (F cos q )dW = (200 N cos 50˚) 20 mW = 2571 J
NOTE If while pushing an object it is moving at a constant velocity
the NET force must be zero
Sohellip Your applied force must be exactly equal to any resistant forces like friction
S F = ma
FA ndash f = ma = 0
A 50 kg box is pulled 6 m across a rough horizontal floor (m = 04) with a force of 80 N at an angle of 35 degrees above the horizontal What is the work done by EACH force exerted on it What is the NET work done
Does the gravitational force do any work NO It is perpendicular to the displacement Does the Normal force do any work No It is perpendicular to the displacement Does the applied Force do any work Yes but ONLY its horizontal component
WF = FAcosq x d = 80cos 35˚ x 6 m = 39319 J Does friction do any work Yes but first what is the normal force Itrsquos NOT mg
Normal = mg ndash FAsinq
Wf = -f x d = -mN∙d = -m(mg ndash FAsin )q ∙d = -747 J What is the NET work done39319 J ndash 747 J = 38572 J
mg
Normal
FA
qf
bull For work to be done the displacement of the object must be in the same direction as the applied force They must be parallel
bull If the force and the displacement are perpendicular to each other NO work is done by the force
So using vector multiplication
W = F bull d (this is a DOT product)
(In many university texts as well as the AP test the displacement is given by d = Dr where r is the
position vector displacement = change in position
W = F bull Dr
If the motion is in only one direction
W = F bull Dx
Or
W = F bull Dy
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement Dr
rf ndash ro =(i + 7j) - (-4i + 3j) =
Dr = 5i + 4j
Then complete the dot product W = F middot Dr(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2J of work
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
If the force varies with displacement-
in other words
Force is a function of displacement
you must integrate to find the work done
Examples of Integration
An object of mass m is subject to a force given by F(x) = 3x3 N What is the work done by the force
W F x dx x dx x 33
43 4
Examples of Definite Integration
To stretch a NON linear spring a distance x requires a force given by F(x) = 4x2 ndash x
What work is done to stretch the spring 2 meters beyond its equilibrium position
W F ds x x ds x x Jmm
m ( ) ( ) ( ) 4
4
3
1
2
4
32
1
22
4
30
1
20 8 6672
0
23
0
22
0
2 3 2 2
Graphing Force vs postion
bull If you graph the applied force vs the position you can find how much work was done by the force
Work = Fmiddotd = ldquoarea under the curverdquoTotal Work = 2 N x 2 m + 3N x 4m = 16 JArea UNDER the x-axis is NEGATIVE work = - 1N x 2m
Force N
Position m
F
dNet work = 16 J ndash 2 J = 14 J
The Integral = the area under the curve
If y = f(x) then the area under the f(x) curve from x = a to x = b is equal to
( )b
a
f x dx
Therefore given a graph of Force vs position the work done is the area under the curve
Work and Energy
Work and EnergyOften some force must do work
to give an object potential or kinetic energy
You push a wagon and it starts moving You do work to stretch a spring and you transform your work energy into spring potential energy
Or you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy
Work = Force x distance = transfer of energy
Kinetic Energy
the energy of motion
K = frac12 mv2
Kinetic Energy
the energy of motion
K = frac12 mv2
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Both Force and displacement are vectors
So using vector multiplication
Work =
This is a DOT product- only parallel components will yield a non-zero solution
In many university texts and sometimes the AP test the displacement which is the change in
position is represented by ldquosrdquo and not ldquodrdquo
W =
The solution for dot products are NOT vectors
Work is not a vector
Letrsquos look an example of using a dot product to calculate workhellip
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement s
s = Dr =
rf ndash ro =(i + 7j) - (-4i + 3j) =
5i + 4j
Then do the dot product W = F middot s(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2 J of work
ExampleHow much work is done to push a 5 kg
cat with a force of 25 N to the top of a ramp that is 7 meters long and 3 meters tall at a constant velocity
W = Force middot displacement
Which measurement is parallel to the force- the length of the ramp or the height of the ramp
W = 25 N x 7 m
W = 175 J 7 m
3 mF = 25 N
ExampleHow much work is done to carry a 5 kg
cat to the top of a ramp that is 7 meters long and 3 meters tall at a constant velocity
W = Force middot displacement
What force is required to carry the cat
Force = weight of the cat
Which is parallel to the Force vector- the length of the ramp or the height
d = height NOT length
W = mg x h
W = 5 kg x 10 ms2 x 3 m
W = 150 J
7 m
3 m
If all four ramps are the same height which ramp would require the greatest amount of work in order to carry an object to the top of the ramp
W = F∙ Dr
For lifting an object the distance d is the height
Because they all have the same height the work for each is the same
bull Andhellipwhile carrying yourself when climbing stairs or walking up an incline at a constant velocity only the height is used to calculate the work you do to get yourself to the top
bull The force required is your weight Horizontal component of d
Ver
tical
com
pone
nt o
f d
Yo
ur
Fo
rce
How much work do you do on a 30 kg cat to carry it from one side of the room to the other if the room is 10 meters long
ZERO because your Force is vertical but the displacement is horizontal
ExampleA boy pushes a
lawnmower 20 meters across the yard If he pushed with a force of 200 N and the angle between the handle and the ground was 50 degrees how much work did he do
q
F
Displacement = 20 m
F cos q
W = (F cos q )dW = (200 N cos 50˚) 20 mW = 2571 J
NOTE If while pushing an object it is moving at a constant velocity
the NET force must be zero
Sohellip Your applied force must be exactly equal to any resistant forces like friction
S F = ma
FA ndash f = ma = 0
A 50 kg box is pulled 6 m across a rough horizontal floor (m = 04) with a force of 80 N at an angle of 35 degrees above the horizontal What is the work done by EACH force exerted on it What is the NET work done
Does the gravitational force do any work NO It is perpendicular to the displacement Does the Normal force do any work No It is perpendicular to the displacement Does the applied Force do any work Yes but ONLY its horizontal component
WF = FAcosq x d = 80cos 35˚ x 6 m = 39319 J Does friction do any work Yes but first what is the normal force Itrsquos NOT mg
Normal = mg ndash FAsinq
Wf = -f x d = -mN∙d = -m(mg ndash FAsin )q ∙d = -747 J What is the NET work done39319 J ndash 747 J = 38572 J
mg
Normal
FA
qf
bull For work to be done the displacement of the object must be in the same direction as the applied force They must be parallel
bull If the force and the displacement are perpendicular to each other NO work is done by the force
So using vector multiplication
W = F bull d (this is a DOT product)
(In many university texts as well as the AP test the displacement is given by d = Dr where r is the
position vector displacement = change in position
W = F bull Dr
If the motion is in only one direction
W = F bull Dx
Or
W = F bull Dy
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement Dr
rf ndash ro =(i + 7j) - (-4i + 3j) =
Dr = 5i + 4j
Then complete the dot product W = F middot Dr(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2J of work
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
If the force varies with displacement-
in other words
Force is a function of displacement
you must integrate to find the work done
Examples of Integration
An object of mass m is subject to a force given by F(x) = 3x3 N What is the work done by the force
W F x dx x dx x 33
43 4
Examples of Definite Integration
To stretch a NON linear spring a distance x requires a force given by F(x) = 4x2 ndash x
What work is done to stretch the spring 2 meters beyond its equilibrium position
W F ds x x ds x x Jmm
m ( ) ( ) ( ) 4
4
3
1
2
4
32
1
22
4
30
1
20 8 6672
0
23
0
22
0
2 3 2 2
Graphing Force vs postion
bull If you graph the applied force vs the position you can find how much work was done by the force
Work = Fmiddotd = ldquoarea under the curverdquoTotal Work = 2 N x 2 m + 3N x 4m = 16 JArea UNDER the x-axis is NEGATIVE work = - 1N x 2m
Force N
Position m
F
dNet work = 16 J ndash 2 J = 14 J
The Integral = the area under the curve
If y = f(x) then the area under the f(x) curve from x = a to x = b is equal to
( )b
a
f x dx
Therefore given a graph of Force vs position the work done is the area under the curve
Work and Energy
Work and EnergyOften some force must do work
to give an object potential or kinetic energy
You push a wagon and it starts moving You do work to stretch a spring and you transform your work energy into spring potential energy
Or you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy
Work = Force x distance = transfer of energy
Kinetic Energy
the energy of motion
K = frac12 mv2
Kinetic Energy
the energy of motion
K = frac12 mv2
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement s
s = Dr =
rf ndash ro =(i + 7j) - (-4i + 3j) =
5i + 4j
Then do the dot product W = F middot s(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2 J of work
ExampleHow much work is done to push a 5 kg
cat with a force of 25 N to the top of a ramp that is 7 meters long and 3 meters tall at a constant velocity
W = Force middot displacement
Which measurement is parallel to the force- the length of the ramp or the height of the ramp
W = 25 N x 7 m
W = 175 J 7 m
3 mF = 25 N
ExampleHow much work is done to carry a 5 kg
cat to the top of a ramp that is 7 meters long and 3 meters tall at a constant velocity
W = Force middot displacement
What force is required to carry the cat
Force = weight of the cat
Which is parallel to the Force vector- the length of the ramp or the height
d = height NOT length
W = mg x h
W = 5 kg x 10 ms2 x 3 m
W = 150 J
7 m
3 m
If all four ramps are the same height which ramp would require the greatest amount of work in order to carry an object to the top of the ramp
W = F∙ Dr
For lifting an object the distance d is the height
Because they all have the same height the work for each is the same
bull Andhellipwhile carrying yourself when climbing stairs or walking up an incline at a constant velocity only the height is used to calculate the work you do to get yourself to the top
bull The force required is your weight Horizontal component of d
Ver
tical
com
pone
nt o
f d
Yo
ur
Fo
rce
How much work do you do on a 30 kg cat to carry it from one side of the room to the other if the room is 10 meters long
ZERO because your Force is vertical but the displacement is horizontal
ExampleA boy pushes a
lawnmower 20 meters across the yard If he pushed with a force of 200 N and the angle between the handle and the ground was 50 degrees how much work did he do
q
F
Displacement = 20 m
F cos q
W = (F cos q )dW = (200 N cos 50˚) 20 mW = 2571 J
NOTE If while pushing an object it is moving at a constant velocity
the NET force must be zero
Sohellip Your applied force must be exactly equal to any resistant forces like friction
S F = ma
FA ndash f = ma = 0
A 50 kg box is pulled 6 m across a rough horizontal floor (m = 04) with a force of 80 N at an angle of 35 degrees above the horizontal What is the work done by EACH force exerted on it What is the NET work done
Does the gravitational force do any work NO It is perpendicular to the displacement Does the Normal force do any work No It is perpendicular to the displacement Does the applied Force do any work Yes but ONLY its horizontal component
WF = FAcosq x d = 80cos 35˚ x 6 m = 39319 J Does friction do any work Yes but first what is the normal force Itrsquos NOT mg
Normal = mg ndash FAsinq
Wf = -f x d = -mN∙d = -m(mg ndash FAsin )q ∙d = -747 J What is the NET work done39319 J ndash 747 J = 38572 J
mg
Normal
FA
qf
bull For work to be done the displacement of the object must be in the same direction as the applied force They must be parallel
bull If the force and the displacement are perpendicular to each other NO work is done by the force
So using vector multiplication
W = F bull d (this is a DOT product)
(In many university texts as well as the AP test the displacement is given by d = Dr where r is the
position vector displacement = change in position
W = F bull Dr
If the motion is in only one direction
W = F bull Dx
Or
W = F bull Dy
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement Dr
rf ndash ro =(i + 7j) - (-4i + 3j) =
Dr = 5i + 4j
Then complete the dot product W = F middot Dr(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2J of work
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
If the force varies with displacement-
in other words
Force is a function of displacement
you must integrate to find the work done
Examples of Integration
An object of mass m is subject to a force given by F(x) = 3x3 N What is the work done by the force
W F x dx x dx x 33
43 4
Examples of Definite Integration
To stretch a NON linear spring a distance x requires a force given by F(x) = 4x2 ndash x
What work is done to stretch the spring 2 meters beyond its equilibrium position
W F ds x x ds x x Jmm
m ( ) ( ) ( ) 4
4
3
1
2
4
32
1
22
4
30
1
20 8 6672
0
23
0
22
0
2 3 2 2
Graphing Force vs postion
bull If you graph the applied force vs the position you can find how much work was done by the force
Work = Fmiddotd = ldquoarea under the curverdquoTotal Work = 2 N x 2 m + 3N x 4m = 16 JArea UNDER the x-axis is NEGATIVE work = - 1N x 2m
Force N
Position m
F
dNet work = 16 J ndash 2 J = 14 J
The Integral = the area under the curve
If y = f(x) then the area under the f(x) curve from x = a to x = b is equal to
( )b
a
f x dx
Therefore given a graph of Force vs position the work done is the area under the curve
Work and Energy
Work and EnergyOften some force must do work
to give an object potential or kinetic energy
You push a wagon and it starts moving You do work to stretch a spring and you transform your work energy into spring potential energy
Or you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy
Work = Force x distance = transfer of energy
Kinetic Energy
the energy of motion
K = frac12 mv2
Kinetic Energy
the energy of motion
K = frac12 mv2
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
ExampleHow much work is done to push a 5 kg
cat with a force of 25 N to the top of a ramp that is 7 meters long and 3 meters tall at a constant velocity
W = Force middot displacement
Which measurement is parallel to the force- the length of the ramp or the height of the ramp
W = 25 N x 7 m
W = 175 J 7 m
3 mF = 25 N
ExampleHow much work is done to carry a 5 kg
cat to the top of a ramp that is 7 meters long and 3 meters tall at a constant velocity
W = Force middot displacement
What force is required to carry the cat
Force = weight of the cat
Which is parallel to the Force vector- the length of the ramp or the height
d = height NOT length
W = mg x h
W = 5 kg x 10 ms2 x 3 m
W = 150 J
7 m
3 m
If all four ramps are the same height which ramp would require the greatest amount of work in order to carry an object to the top of the ramp
W = F∙ Dr
For lifting an object the distance d is the height
Because they all have the same height the work for each is the same
bull Andhellipwhile carrying yourself when climbing stairs or walking up an incline at a constant velocity only the height is used to calculate the work you do to get yourself to the top
bull The force required is your weight Horizontal component of d
Ver
tical
com
pone
nt o
f d
Yo
ur
Fo
rce
How much work do you do on a 30 kg cat to carry it from one side of the room to the other if the room is 10 meters long
ZERO because your Force is vertical but the displacement is horizontal
ExampleA boy pushes a
lawnmower 20 meters across the yard If he pushed with a force of 200 N and the angle between the handle and the ground was 50 degrees how much work did he do
q
F
Displacement = 20 m
F cos q
W = (F cos q )dW = (200 N cos 50˚) 20 mW = 2571 J
NOTE If while pushing an object it is moving at a constant velocity
the NET force must be zero
Sohellip Your applied force must be exactly equal to any resistant forces like friction
S F = ma
FA ndash f = ma = 0
A 50 kg box is pulled 6 m across a rough horizontal floor (m = 04) with a force of 80 N at an angle of 35 degrees above the horizontal What is the work done by EACH force exerted on it What is the NET work done
Does the gravitational force do any work NO It is perpendicular to the displacement Does the Normal force do any work No It is perpendicular to the displacement Does the applied Force do any work Yes but ONLY its horizontal component
WF = FAcosq x d = 80cos 35˚ x 6 m = 39319 J Does friction do any work Yes but first what is the normal force Itrsquos NOT mg
Normal = mg ndash FAsinq
Wf = -f x d = -mN∙d = -m(mg ndash FAsin )q ∙d = -747 J What is the NET work done39319 J ndash 747 J = 38572 J
mg
Normal
FA
qf
bull For work to be done the displacement of the object must be in the same direction as the applied force They must be parallel
bull If the force and the displacement are perpendicular to each other NO work is done by the force
So using vector multiplication
W = F bull d (this is a DOT product)
(In many university texts as well as the AP test the displacement is given by d = Dr where r is the
position vector displacement = change in position
W = F bull Dr
If the motion is in only one direction
W = F bull Dx
Or
W = F bull Dy
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement Dr
rf ndash ro =(i + 7j) - (-4i + 3j) =
Dr = 5i + 4j
Then complete the dot product W = F middot Dr(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2J of work
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
If the force varies with displacement-
in other words
Force is a function of displacement
you must integrate to find the work done
Examples of Integration
An object of mass m is subject to a force given by F(x) = 3x3 N What is the work done by the force
W F x dx x dx x 33
43 4
Examples of Definite Integration
To stretch a NON linear spring a distance x requires a force given by F(x) = 4x2 ndash x
What work is done to stretch the spring 2 meters beyond its equilibrium position
W F ds x x ds x x Jmm
m ( ) ( ) ( ) 4
4
3
1
2
4
32
1
22
4
30
1
20 8 6672
0
23
0
22
0
2 3 2 2
Graphing Force vs postion
bull If you graph the applied force vs the position you can find how much work was done by the force
Work = Fmiddotd = ldquoarea under the curverdquoTotal Work = 2 N x 2 m + 3N x 4m = 16 JArea UNDER the x-axis is NEGATIVE work = - 1N x 2m
Force N
Position m
F
dNet work = 16 J ndash 2 J = 14 J
The Integral = the area under the curve
If y = f(x) then the area under the f(x) curve from x = a to x = b is equal to
( )b
a
f x dx
Therefore given a graph of Force vs position the work done is the area under the curve
Work and Energy
Work and EnergyOften some force must do work
to give an object potential or kinetic energy
You push a wagon and it starts moving You do work to stretch a spring and you transform your work energy into spring potential energy
Or you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy
Work = Force x distance = transfer of energy
Kinetic Energy
the energy of motion
K = frac12 mv2
Kinetic Energy
the energy of motion
K = frac12 mv2
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
ExampleHow much work is done to carry a 5 kg
cat to the top of a ramp that is 7 meters long and 3 meters tall at a constant velocity
W = Force middot displacement
What force is required to carry the cat
Force = weight of the cat
Which is parallel to the Force vector- the length of the ramp or the height
d = height NOT length
W = mg x h
W = 5 kg x 10 ms2 x 3 m
W = 150 J
7 m
3 m
If all four ramps are the same height which ramp would require the greatest amount of work in order to carry an object to the top of the ramp
W = F∙ Dr
For lifting an object the distance d is the height
Because they all have the same height the work for each is the same
bull Andhellipwhile carrying yourself when climbing stairs or walking up an incline at a constant velocity only the height is used to calculate the work you do to get yourself to the top
bull The force required is your weight Horizontal component of d
Ver
tical
com
pone
nt o
f d
Yo
ur
Fo
rce
How much work do you do on a 30 kg cat to carry it from one side of the room to the other if the room is 10 meters long
ZERO because your Force is vertical but the displacement is horizontal
ExampleA boy pushes a
lawnmower 20 meters across the yard If he pushed with a force of 200 N and the angle between the handle and the ground was 50 degrees how much work did he do
q
F
Displacement = 20 m
F cos q
W = (F cos q )dW = (200 N cos 50˚) 20 mW = 2571 J
NOTE If while pushing an object it is moving at a constant velocity
the NET force must be zero
Sohellip Your applied force must be exactly equal to any resistant forces like friction
S F = ma
FA ndash f = ma = 0
A 50 kg box is pulled 6 m across a rough horizontal floor (m = 04) with a force of 80 N at an angle of 35 degrees above the horizontal What is the work done by EACH force exerted on it What is the NET work done
Does the gravitational force do any work NO It is perpendicular to the displacement Does the Normal force do any work No It is perpendicular to the displacement Does the applied Force do any work Yes but ONLY its horizontal component
WF = FAcosq x d = 80cos 35˚ x 6 m = 39319 J Does friction do any work Yes but first what is the normal force Itrsquos NOT mg
Normal = mg ndash FAsinq
Wf = -f x d = -mN∙d = -m(mg ndash FAsin )q ∙d = -747 J What is the NET work done39319 J ndash 747 J = 38572 J
mg
Normal
FA
qf
bull For work to be done the displacement of the object must be in the same direction as the applied force They must be parallel
bull If the force and the displacement are perpendicular to each other NO work is done by the force
So using vector multiplication
W = F bull d (this is a DOT product)
(In many university texts as well as the AP test the displacement is given by d = Dr where r is the
position vector displacement = change in position
W = F bull Dr
If the motion is in only one direction
W = F bull Dx
Or
W = F bull Dy
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement Dr
rf ndash ro =(i + 7j) - (-4i + 3j) =
Dr = 5i + 4j
Then complete the dot product W = F middot Dr(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2J of work
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
If the force varies with displacement-
in other words
Force is a function of displacement
you must integrate to find the work done
Examples of Integration
An object of mass m is subject to a force given by F(x) = 3x3 N What is the work done by the force
W F x dx x dx x 33
43 4
Examples of Definite Integration
To stretch a NON linear spring a distance x requires a force given by F(x) = 4x2 ndash x
What work is done to stretch the spring 2 meters beyond its equilibrium position
W F ds x x ds x x Jmm
m ( ) ( ) ( ) 4
4
3
1
2
4
32
1
22
4
30
1
20 8 6672
0
23
0
22
0
2 3 2 2
Graphing Force vs postion
bull If you graph the applied force vs the position you can find how much work was done by the force
Work = Fmiddotd = ldquoarea under the curverdquoTotal Work = 2 N x 2 m + 3N x 4m = 16 JArea UNDER the x-axis is NEGATIVE work = - 1N x 2m
Force N
Position m
F
dNet work = 16 J ndash 2 J = 14 J
The Integral = the area under the curve
If y = f(x) then the area under the f(x) curve from x = a to x = b is equal to
( )b
a
f x dx
Therefore given a graph of Force vs position the work done is the area under the curve
Work and Energy
Work and EnergyOften some force must do work
to give an object potential or kinetic energy
You push a wagon and it starts moving You do work to stretch a spring and you transform your work energy into spring potential energy
Or you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy
Work = Force x distance = transfer of energy
Kinetic Energy
the energy of motion
K = frac12 mv2
Kinetic Energy
the energy of motion
K = frac12 mv2
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
If all four ramps are the same height which ramp would require the greatest amount of work in order to carry an object to the top of the ramp
W = F∙ Dr
For lifting an object the distance d is the height
Because they all have the same height the work for each is the same
bull Andhellipwhile carrying yourself when climbing stairs or walking up an incline at a constant velocity only the height is used to calculate the work you do to get yourself to the top
bull The force required is your weight Horizontal component of d
Ver
tical
com
pone
nt o
f d
Yo
ur
Fo
rce
How much work do you do on a 30 kg cat to carry it from one side of the room to the other if the room is 10 meters long
ZERO because your Force is vertical but the displacement is horizontal
ExampleA boy pushes a
lawnmower 20 meters across the yard If he pushed with a force of 200 N and the angle between the handle and the ground was 50 degrees how much work did he do
q
F
Displacement = 20 m
F cos q
W = (F cos q )dW = (200 N cos 50˚) 20 mW = 2571 J
NOTE If while pushing an object it is moving at a constant velocity
the NET force must be zero
Sohellip Your applied force must be exactly equal to any resistant forces like friction
S F = ma
FA ndash f = ma = 0
A 50 kg box is pulled 6 m across a rough horizontal floor (m = 04) with a force of 80 N at an angle of 35 degrees above the horizontal What is the work done by EACH force exerted on it What is the NET work done
Does the gravitational force do any work NO It is perpendicular to the displacement Does the Normal force do any work No It is perpendicular to the displacement Does the applied Force do any work Yes but ONLY its horizontal component
WF = FAcosq x d = 80cos 35˚ x 6 m = 39319 J Does friction do any work Yes but first what is the normal force Itrsquos NOT mg
Normal = mg ndash FAsinq
Wf = -f x d = -mN∙d = -m(mg ndash FAsin )q ∙d = -747 J What is the NET work done39319 J ndash 747 J = 38572 J
mg
Normal
FA
qf
bull For work to be done the displacement of the object must be in the same direction as the applied force They must be parallel
bull If the force and the displacement are perpendicular to each other NO work is done by the force
So using vector multiplication
W = F bull d (this is a DOT product)
(In many university texts as well as the AP test the displacement is given by d = Dr where r is the
position vector displacement = change in position
W = F bull Dr
If the motion is in only one direction
W = F bull Dx
Or
W = F bull Dy
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement Dr
rf ndash ro =(i + 7j) - (-4i + 3j) =
Dr = 5i + 4j
Then complete the dot product W = F middot Dr(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2J of work
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
If the force varies with displacement-
in other words
Force is a function of displacement
you must integrate to find the work done
Examples of Integration
An object of mass m is subject to a force given by F(x) = 3x3 N What is the work done by the force
W F x dx x dx x 33
43 4
Examples of Definite Integration
To stretch a NON linear spring a distance x requires a force given by F(x) = 4x2 ndash x
What work is done to stretch the spring 2 meters beyond its equilibrium position
W F ds x x ds x x Jmm
m ( ) ( ) ( ) 4
4
3
1
2
4
32
1
22
4
30
1
20 8 6672
0
23
0
22
0
2 3 2 2
Graphing Force vs postion
bull If you graph the applied force vs the position you can find how much work was done by the force
Work = Fmiddotd = ldquoarea under the curverdquoTotal Work = 2 N x 2 m + 3N x 4m = 16 JArea UNDER the x-axis is NEGATIVE work = - 1N x 2m
Force N
Position m
F
dNet work = 16 J ndash 2 J = 14 J
The Integral = the area under the curve
If y = f(x) then the area under the f(x) curve from x = a to x = b is equal to
( )b
a
f x dx
Therefore given a graph of Force vs position the work done is the area under the curve
Work and Energy
Work and EnergyOften some force must do work
to give an object potential or kinetic energy
You push a wagon and it starts moving You do work to stretch a spring and you transform your work energy into spring potential energy
Or you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy
Work = Force x distance = transfer of energy
Kinetic Energy
the energy of motion
K = frac12 mv2
Kinetic Energy
the energy of motion
K = frac12 mv2
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
bull Andhellipwhile carrying yourself when climbing stairs or walking up an incline at a constant velocity only the height is used to calculate the work you do to get yourself to the top
bull The force required is your weight Horizontal component of d
Ver
tical
com
pone
nt o
f d
Yo
ur
Fo
rce
How much work do you do on a 30 kg cat to carry it from one side of the room to the other if the room is 10 meters long
ZERO because your Force is vertical but the displacement is horizontal
ExampleA boy pushes a
lawnmower 20 meters across the yard If he pushed with a force of 200 N and the angle between the handle and the ground was 50 degrees how much work did he do
q
F
Displacement = 20 m
F cos q
W = (F cos q )dW = (200 N cos 50˚) 20 mW = 2571 J
NOTE If while pushing an object it is moving at a constant velocity
the NET force must be zero
Sohellip Your applied force must be exactly equal to any resistant forces like friction
S F = ma
FA ndash f = ma = 0
A 50 kg box is pulled 6 m across a rough horizontal floor (m = 04) with a force of 80 N at an angle of 35 degrees above the horizontal What is the work done by EACH force exerted on it What is the NET work done
Does the gravitational force do any work NO It is perpendicular to the displacement Does the Normal force do any work No It is perpendicular to the displacement Does the applied Force do any work Yes but ONLY its horizontal component
WF = FAcosq x d = 80cos 35˚ x 6 m = 39319 J Does friction do any work Yes but first what is the normal force Itrsquos NOT mg
Normal = mg ndash FAsinq
Wf = -f x d = -mN∙d = -m(mg ndash FAsin )q ∙d = -747 J What is the NET work done39319 J ndash 747 J = 38572 J
mg
Normal
FA
qf
bull For work to be done the displacement of the object must be in the same direction as the applied force They must be parallel
bull If the force and the displacement are perpendicular to each other NO work is done by the force
So using vector multiplication
W = F bull d (this is a DOT product)
(In many university texts as well as the AP test the displacement is given by d = Dr where r is the
position vector displacement = change in position
W = F bull Dr
If the motion is in only one direction
W = F bull Dx
Or
W = F bull Dy
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement Dr
rf ndash ro =(i + 7j) - (-4i + 3j) =
Dr = 5i + 4j
Then complete the dot product W = F middot Dr(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2J of work
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
If the force varies with displacement-
in other words
Force is a function of displacement
you must integrate to find the work done
Examples of Integration
An object of mass m is subject to a force given by F(x) = 3x3 N What is the work done by the force
W F x dx x dx x 33
43 4
Examples of Definite Integration
To stretch a NON linear spring a distance x requires a force given by F(x) = 4x2 ndash x
What work is done to stretch the spring 2 meters beyond its equilibrium position
W F ds x x ds x x Jmm
m ( ) ( ) ( ) 4
4
3
1
2
4
32
1
22
4
30
1
20 8 6672
0
23
0
22
0
2 3 2 2
Graphing Force vs postion
bull If you graph the applied force vs the position you can find how much work was done by the force
Work = Fmiddotd = ldquoarea under the curverdquoTotal Work = 2 N x 2 m + 3N x 4m = 16 JArea UNDER the x-axis is NEGATIVE work = - 1N x 2m
Force N
Position m
F
dNet work = 16 J ndash 2 J = 14 J
The Integral = the area under the curve
If y = f(x) then the area under the f(x) curve from x = a to x = b is equal to
( )b
a
f x dx
Therefore given a graph of Force vs position the work done is the area under the curve
Work and Energy
Work and EnergyOften some force must do work
to give an object potential or kinetic energy
You push a wagon and it starts moving You do work to stretch a spring and you transform your work energy into spring potential energy
Or you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy
Work = Force x distance = transfer of energy
Kinetic Energy
the energy of motion
K = frac12 mv2
Kinetic Energy
the energy of motion
K = frac12 mv2
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
How much work do you do on a 30 kg cat to carry it from one side of the room to the other if the room is 10 meters long
ZERO because your Force is vertical but the displacement is horizontal
ExampleA boy pushes a
lawnmower 20 meters across the yard If he pushed with a force of 200 N and the angle between the handle and the ground was 50 degrees how much work did he do
q
F
Displacement = 20 m
F cos q
W = (F cos q )dW = (200 N cos 50˚) 20 mW = 2571 J
NOTE If while pushing an object it is moving at a constant velocity
the NET force must be zero
Sohellip Your applied force must be exactly equal to any resistant forces like friction
S F = ma
FA ndash f = ma = 0
A 50 kg box is pulled 6 m across a rough horizontal floor (m = 04) with a force of 80 N at an angle of 35 degrees above the horizontal What is the work done by EACH force exerted on it What is the NET work done
Does the gravitational force do any work NO It is perpendicular to the displacement Does the Normal force do any work No It is perpendicular to the displacement Does the applied Force do any work Yes but ONLY its horizontal component
WF = FAcosq x d = 80cos 35˚ x 6 m = 39319 J Does friction do any work Yes but first what is the normal force Itrsquos NOT mg
Normal = mg ndash FAsinq
Wf = -f x d = -mN∙d = -m(mg ndash FAsin )q ∙d = -747 J What is the NET work done39319 J ndash 747 J = 38572 J
mg
Normal
FA
qf
bull For work to be done the displacement of the object must be in the same direction as the applied force They must be parallel
bull If the force and the displacement are perpendicular to each other NO work is done by the force
So using vector multiplication
W = F bull d (this is a DOT product)
(In many university texts as well as the AP test the displacement is given by d = Dr where r is the
position vector displacement = change in position
W = F bull Dr
If the motion is in only one direction
W = F bull Dx
Or
W = F bull Dy
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement Dr
rf ndash ro =(i + 7j) - (-4i + 3j) =
Dr = 5i + 4j
Then complete the dot product W = F middot Dr(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2J of work
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
If the force varies with displacement-
in other words
Force is a function of displacement
you must integrate to find the work done
Examples of Integration
An object of mass m is subject to a force given by F(x) = 3x3 N What is the work done by the force
W F x dx x dx x 33
43 4
Examples of Definite Integration
To stretch a NON linear spring a distance x requires a force given by F(x) = 4x2 ndash x
What work is done to stretch the spring 2 meters beyond its equilibrium position
W F ds x x ds x x Jmm
m ( ) ( ) ( ) 4
4
3
1
2
4
32
1
22
4
30
1
20 8 6672
0
23
0
22
0
2 3 2 2
Graphing Force vs postion
bull If you graph the applied force vs the position you can find how much work was done by the force
Work = Fmiddotd = ldquoarea under the curverdquoTotal Work = 2 N x 2 m + 3N x 4m = 16 JArea UNDER the x-axis is NEGATIVE work = - 1N x 2m
Force N
Position m
F
dNet work = 16 J ndash 2 J = 14 J
The Integral = the area under the curve
If y = f(x) then the area under the f(x) curve from x = a to x = b is equal to
( )b
a
f x dx
Therefore given a graph of Force vs position the work done is the area under the curve
Work and Energy
Work and EnergyOften some force must do work
to give an object potential or kinetic energy
You push a wagon and it starts moving You do work to stretch a spring and you transform your work energy into spring potential energy
Or you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy
Work = Force x distance = transfer of energy
Kinetic Energy
the energy of motion
K = frac12 mv2
Kinetic Energy
the energy of motion
K = frac12 mv2
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
ExampleA boy pushes a
lawnmower 20 meters across the yard If he pushed with a force of 200 N and the angle between the handle and the ground was 50 degrees how much work did he do
q
F
Displacement = 20 m
F cos q
W = (F cos q )dW = (200 N cos 50˚) 20 mW = 2571 J
NOTE If while pushing an object it is moving at a constant velocity
the NET force must be zero
Sohellip Your applied force must be exactly equal to any resistant forces like friction
S F = ma
FA ndash f = ma = 0
A 50 kg box is pulled 6 m across a rough horizontal floor (m = 04) with a force of 80 N at an angle of 35 degrees above the horizontal What is the work done by EACH force exerted on it What is the NET work done
Does the gravitational force do any work NO It is perpendicular to the displacement Does the Normal force do any work No It is perpendicular to the displacement Does the applied Force do any work Yes but ONLY its horizontal component
WF = FAcosq x d = 80cos 35˚ x 6 m = 39319 J Does friction do any work Yes but first what is the normal force Itrsquos NOT mg
Normal = mg ndash FAsinq
Wf = -f x d = -mN∙d = -m(mg ndash FAsin )q ∙d = -747 J What is the NET work done39319 J ndash 747 J = 38572 J
mg
Normal
FA
qf
bull For work to be done the displacement of the object must be in the same direction as the applied force They must be parallel
bull If the force and the displacement are perpendicular to each other NO work is done by the force
So using vector multiplication
W = F bull d (this is a DOT product)
(In many university texts as well as the AP test the displacement is given by d = Dr where r is the
position vector displacement = change in position
W = F bull Dr
If the motion is in only one direction
W = F bull Dx
Or
W = F bull Dy
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement Dr
rf ndash ro =(i + 7j) - (-4i + 3j) =
Dr = 5i + 4j
Then complete the dot product W = F middot Dr(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2J of work
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
If the force varies with displacement-
in other words
Force is a function of displacement
you must integrate to find the work done
Examples of Integration
An object of mass m is subject to a force given by F(x) = 3x3 N What is the work done by the force
W F x dx x dx x 33
43 4
Examples of Definite Integration
To stretch a NON linear spring a distance x requires a force given by F(x) = 4x2 ndash x
What work is done to stretch the spring 2 meters beyond its equilibrium position
W F ds x x ds x x Jmm
m ( ) ( ) ( ) 4
4
3
1
2
4
32
1
22
4
30
1
20 8 6672
0
23
0
22
0
2 3 2 2
Graphing Force vs postion
bull If you graph the applied force vs the position you can find how much work was done by the force
Work = Fmiddotd = ldquoarea under the curverdquoTotal Work = 2 N x 2 m + 3N x 4m = 16 JArea UNDER the x-axis is NEGATIVE work = - 1N x 2m
Force N
Position m
F
dNet work = 16 J ndash 2 J = 14 J
The Integral = the area under the curve
If y = f(x) then the area under the f(x) curve from x = a to x = b is equal to
( )b
a
f x dx
Therefore given a graph of Force vs position the work done is the area under the curve
Work and Energy
Work and EnergyOften some force must do work
to give an object potential or kinetic energy
You push a wagon and it starts moving You do work to stretch a spring and you transform your work energy into spring potential energy
Or you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy
Work = Force x distance = transfer of energy
Kinetic Energy
the energy of motion
K = frac12 mv2
Kinetic Energy
the energy of motion
K = frac12 mv2
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
NOTE If while pushing an object it is moving at a constant velocity
the NET force must be zero
Sohellip Your applied force must be exactly equal to any resistant forces like friction
S F = ma
FA ndash f = ma = 0
A 50 kg box is pulled 6 m across a rough horizontal floor (m = 04) with a force of 80 N at an angle of 35 degrees above the horizontal What is the work done by EACH force exerted on it What is the NET work done
Does the gravitational force do any work NO It is perpendicular to the displacement Does the Normal force do any work No It is perpendicular to the displacement Does the applied Force do any work Yes but ONLY its horizontal component
WF = FAcosq x d = 80cos 35˚ x 6 m = 39319 J Does friction do any work Yes but first what is the normal force Itrsquos NOT mg
Normal = mg ndash FAsinq
Wf = -f x d = -mN∙d = -m(mg ndash FAsin )q ∙d = -747 J What is the NET work done39319 J ndash 747 J = 38572 J
mg
Normal
FA
qf
bull For work to be done the displacement of the object must be in the same direction as the applied force They must be parallel
bull If the force and the displacement are perpendicular to each other NO work is done by the force
So using vector multiplication
W = F bull d (this is a DOT product)
(In many university texts as well as the AP test the displacement is given by d = Dr where r is the
position vector displacement = change in position
W = F bull Dr
If the motion is in only one direction
W = F bull Dx
Or
W = F bull Dy
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement Dr
rf ndash ro =(i + 7j) - (-4i + 3j) =
Dr = 5i + 4j
Then complete the dot product W = F middot Dr(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2J of work
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
If the force varies with displacement-
in other words
Force is a function of displacement
you must integrate to find the work done
Examples of Integration
An object of mass m is subject to a force given by F(x) = 3x3 N What is the work done by the force
W F x dx x dx x 33
43 4
Examples of Definite Integration
To stretch a NON linear spring a distance x requires a force given by F(x) = 4x2 ndash x
What work is done to stretch the spring 2 meters beyond its equilibrium position
W F ds x x ds x x Jmm
m ( ) ( ) ( ) 4
4
3
1
2
4
32
1
22
4
30
1
20 8 6672
0
23
0
22
0
2 3 2 2
Graphing Force vs postion
bull If you graph the applied force vs the position you can find how much work was done by the force
Work = Fmiddotd = ldquoarea under the curverdquoTotal Work = 2 N x 2 m + 3N x 4m = 16 JArea UNDER the x-axis is NEGATIVE work = - 1N x 2m
Force N
Position m
F
dNet work = 16 J ndash 2 J = 14 J
The Integral = the area under the curve
If y = f(x) then the area under the f(x) curve from x = a to x = b is equal to
( )b
a
f x dx
Therefore given a graph of Force vs position the work done is the area under the curve
Work and Energy
Work and EnergyOften some force must do work
to give an object potential or kinetic energy
You push a wagon and it starts moving You do work to stretch a spring and you transform your work energy into spring potential energy
Or you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy
Work = Force x distance = transfer of energy
Kinetic Energy
the energy of motion
K = frac12 mv2
Kinetic Energy
the energy of motion
K = frac12 mv2
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
A 50 kg box is pulled 6 m across a rough horizontal floor (m = 04) with a force of 80 N at an angle of 35 degrees above the horizontal What is the work done by EACH force exerted on it What is the NET work done
Does the gravitational force do any work NO It is perpendicular to the displacement Does the Normal force do any work No It is perpendicular to the displacement Does the applied Force do any work Yes but ONLY its horizontal component
WF = FAcosq x d = 80cos 35˚ x 6 m = 39319 J Does friction do any work Yes but first what is the normal force Itrsquos NOT mg
Normal = mg ndash FAsinq
Wf = -f x d = -mN∙d = -m(mg ndash FAsin )q ∙d = -747 J What is the NET work done39319 J ndash 747 J = 38572 J
mg
Normal
FA
qf
bull For work to be done the displacement of the object must be in the same direction as the applied force They must be parallel
bull If the force and the displacement are perpendicular to each other NO work is done by the force
So using vector multiplication
W = F bull d (this is a DOT product)
(In many university texts as well as the AP test the displacement is given by d = Dr where r is the
position vector displacement = change in position
W = F bull Dr
If the motion is in only one direction
W = F bull Dx
Or
W = F bull Dy
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement Dr
rf ndash ro =(i + 7j) - (-4i + 3j) =
Dr = 5i + 4j
Then complete the dot product W = F middot Dr(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2J of work
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
If the force varies with displacement-
in other words
Force is a function of displacement
you must integrate to find the work done
Examples of Integration
An object of mass m is subject to a force given by F(x) = 3x3 N What is the work done by the force
W F x dx x dx x 33
43 4
Examples of Definite Integration
To stretch a NON linear spring a distance x requires a force given by F(x) = 4x2 ndash x
What work is done to stretch the spring 2 meters beyond its equilibrium position
W F ds x x ds x x Jmm
m ( ) ( ) ( ) 4
4
3
1
2
4
32
1
22
4
30
1
20 8 6672
0
23
0
22
0
2 3 2 2
Graphing Force vs postion
bull If you graph the applied force vs the position you can find how much work was done by the force
Work = Fmiddotd = ldquoarea under the curverdquoTotal Work = 2 N x 2 m + 3N x 4m = 16 JArea UNDER the x-axis is NEGATIVE work = - 1N x 2m
Force N
Position m
F
dNet work = 16 J ndash 2 J = 14 J
The Integral = the area under the curve
If y = f(x) then the area under the f(x) curve from x = a to x = b is equal to
( )b
a
f x dx
Therefore given a graph of Force vs position the work done is the area under the curve
Work and Energy
Work and EnergyOften some force must do work
to give an object potential or kinetic energy
You push a wagon and it starts moving You do work to stretch a spring and you transform your work energy into spring potential energy
Or you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy
Work = Force x distance = transfer of energy
Kinetic Energy
the energy of motion
K = frac12 mv2
Kinetic Energy
the energy of motion
K = frac12 mv2
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
bull For work to be done the displacement of the object must be in the same direction as the applied force They must be parallel
bull If the force and the displacement are perpendicular to each other NO work is done by the force
So using vector multiplication
W = F bull d (this is a DOT product)
(In many university texts as well as the AP test the displacement is given by d = Dr where r is the
position vector displacement = change in position
W = F bull Dr
If the motion is in only one direction
W = F bull Dx
Or
W = F bull Dy
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement Dr
rf ndash ro =(i + 7j) - (-4i + 3j) =
Dr = 5i + 4j
Then complete the dot product W = F middot Dr(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2J of work
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
If the force varies with displacement-
in other words
Force is a function of displacement
you must integrate to find the work done
Examples of Integration
An object of mass m is subject to a force given by F(x) = 3x3 N What is the work done by the force
W F x dx x dx x 33
43 4
Examples of Definite Integration
To stretch a NON linear spring a distance x requires a force given by F(x) = 4x2 ndash x
What work is done to stretch the spring 2 meters beyond its equilibrium position
W F ds x x ds x x Jmm
m ( ) ( ) ( ) 4
4
3
1
2
4
32
1
22
4
30
1
20 8 6672
0
23
0
22
0
2 3 2 2
Graphing Force vs postion
bull If you graph the applied force vs the position you can find how much work was done by the force
Work = Fmiddotd = ldquoarea under the curverdquoTotal Work = 2 N x 2 m + 3N x 4m = 16 JArea UNDER the x-axis is NEGATIVE work = - 1N x 2m
Force N
Position m
F
dNet work = 16 J ndash 2 J = 14 J
The Integral = the area under the curve
If y = f(x) then the area under the f(x) curve from x = a to x = b is equal to
( )b
a
f x dx
Therefore given a graph of Force vs position the work done is the area under the curve
Work and Energy
Work and EnergyOften some force must do work
to give an object potential or kinetic energy
You push a wagon and it starts moving You do work to stretch a spring and you transform your work energy into spring potential energy
Or you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy
Work = Force x distance = transfer of energy
Kinetic Energy
the energy of motion
K = frac12 mv2
Kinetic Energy
the energy of motion
K = frac12 mv2
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
An object is subject to a force given byF = 6i ndash 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j (ldquorrdquo is a position vector)
What work was done by this forceFirst find the displacement Dr
rf ndash ro =(i + 7j) - (-4i + 3j) =
Dr = 5i + 4j
Then complete the dot product W = F middot Dr(6i ndash 8j) middot (5i + 4j) =
30 ndash 32 = -2J of work
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
If the force varies with displacement-
in other words
Force is a function of displacement
you must integrate to find the work done
Examples of Integration
An object of mass m is subject to a force given by F(x) = 3x3 N What is the work done by the force
W F x dx x dx x 33
43 4
Examples of Definite Integration
To stretch a NON linear spring a distance x requires a force given by F(x) = 4x2 ndash x
What work is done to stretch the spring 2 meters beyond its equilibrium position
W F ds x x ds x x Jmm
m ( ) ( ) ( ) 4
4
3
1
2
4
32
1
22
4
30
1
20 8 6672
0
23
0
22
0
2 3 2 2
Graphing Force vs postion
bull If you graph the applied force vs the position you can find how much work was done by the force
Work = Fmiddotd = ldquoarea under the curverdquoTotal Work = 2 N x 2 m + 3N x 4m = 16 JArea UNDER the x-axis is NEGATIVE work = - 1N x 2m
Force N
Position m
F
dNet work = 16 J ndash 2 J = 14 J
The Integral = the area under the curve
If y = f(x) then the area under the f(x) curve from x = a to x = b is equal to
( )b
a
f x dx
Therefore given a graph of Force vs position the work done is the area under the curve
Work and Energy
Work and EnergyOften some force must do work
to give an object potential or kinetic energy
You push a wagon and it starts moving You do work to stretch a spring and you transform your work energy into spring potential energy
Or you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy
Work = Force x distance = transfer of energy
Kinetic Energy
the energy of motion
K = frac12 mv2
Kinetic Energy
the energy of motion
K = frac12 mv2
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
If the force varies with displacement-
in other words
Force is a function of displacement
you must integrate to find the work done
Examples of Integration
An object of mass m is subject to a force given by F(x) = 3x3 N What is the work done by the force
W F x dx x dx x 33
43 4
Examples of Definite Integration
To stretch a NON linear spring a distance x requires a force given by F(x) = 4x2 ndash x
What work is done to stretch the spring 2 meters beyond its equilibrium position
W F ds x x ds x x Jmm
m ( ) ( ) ( ) 4
4
3
1
2
4
32
1
22
4
30
1
20 8 6672
0
23
0
22
0
2 3 2 2
Graphing Force vs postion
bull If you graph the applied force vs the position you can find how much work was done by the force
Work = Fmiddotd = ldquoarea under the curverdquoTotal Work = 2 N x 2 m + 3N x 4m = 16 JArea UNDER the x-axis is NEGATIVE work = - 1N x 2m
Force N
Position m
F
dNet work = 16 J ndash 2 J = 14 J
The Integral = the area under the curve
If y = f(x) then the area under the f(x) curve from x = a to x = b is equal to
( )b
a
f x dx
Therefore given a graph of Force vs position the work done is the area under the curve
Work and Energy
Work and EnergyOften some force must do work
to give an object potential or kinetic energy
You push a wagon and it starts moving You do work to stretch a spring and you transform your work energy into spring potential energy
Or you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy
Work = Force x distance = transfer of energy
Kinetic Energy
the energy of motion
K = frac12 mv2
Kinetic Energy
the energy of motion
K = frac12 mv2
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Examples of Integration
An object of mass m is subject to a force given by F(x) = 3x3 N What is the work done by the force
W F x dx x dx x 33
43 4
Examples of Definite Integration
To stretch a NON linear spring a distance x requires a force given by F(x) = 4x2 ndash x
What work is done to stretch the spring 2 meters beyond its equilibrium position
W F ds x x ds x x Jmm
m ( ) ( ) ( ) 4
4
3
1
2
4
32
1
22
4
30
1
20 8 6672
0
23
0
22
0
2 3 2 2
Graphing Force vs postion
bull If you graph the applied force vs the position you can find how much work was done by the force
Work = Fmiddotd = ldquoarea under the curverdquoTotal Work = 2 N x 2 m + 3N x 4m = 16 JArea UNDER the x-axis is NEGATIVE work = - 1N x 2m
Force N
Position m
F
dNet work = 16 J ndash 2 J = 14 J
The Integral = the area under the curve
If y = f(x) then the area under the f(x) curve from x = a to x = b is equal to
( )b
a
f x dx
Therefore given a graph of Force vs position the work done is the area under the curve
Work and Energy
Work and EnergyOften some force must do work
to give an object potential or kinetic energy
You push a wagon and it starts moving You do work to stretch a spring and you transform your work energy into spring potential energy
Or you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy
Work = Force x distance = transfer of energy
Kinetic Energy
the energy of motion
K = frac12 mv2
Kinetic Energy
the energy of motion
K = frac12 mv2
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Examples of Definite Integration
To stretch a NON linear spring a distance x requires a force given by F(x) = 4x2 ndash x
What work is done to stretch the spring 2 meters beyond its equilibrium position
W F ds x x ds x x Jmm
m ( ) ( ) ( ) 4
4
3
1
2
4
32
1
22
4
30
1
20 8 6672
0
23
0
22
0
2 3 2 2
Graphing Force vs postion
bull If you graph the applied force vs the position you can find how much work was done by the force
Work = Fmiddotd = ldquoarea under the curverdquoTotal Work = 2 N x 2 m + 3N x 4m = 16 JArea UNDER the x-axis is NEGATIVE work = - 1N x 2m
Force N
Position m
F
dNet work = 16 J ndash 2 J = 14 J
The Integral = the area under the curve
If y = f(x) then the area under the f(x) curve from x = a to x = b is equal to
( )b
a
f x dx
Therefore given a graph of Force vs position the work done is the area under the curve
Work and Energy
Work and EnergyOften some force must do work
to give an object potential or kinetic energy
You push a wagon and it starts moving You do work to stretch a spring and you transform your work energy into spring potential energy
Or you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy
Work = Force x distance = transfer of energy
Kinetic Energy
the energy of motion
K = frac12 mv2
Kinetic Energy
the energy of motion
K = frac12 mv2
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Graphing Force vs postion
bull If you graph the applied force vs the position you can find how much work was done by the force
Work = Fmiddotd = ldquoarea under the curverdquoTotal Work = 2 N x 2 m + 3N x 4m = 16 JArea UNDER the x-axis is NEGATIVE work = - 1N x 2m
Force N
Position m
F
dNet work = 16 J ndash 2 J = 14 J
The Integral = the area under the curve
If y = f(x) then the area under the f(x) curve from x = a to x = b is equal to
( )b
a
f x dx
Therefore given a graph of Force vs position the work done is the area under the curve
Work and Energy
Work and EnergyOften some force must do work
to give an object potential or kinetic energy
You push a wagon and it starts moving You do work to stretch a spring and you transform your work energy into spring potential energy
Or you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy
Work = Force x distance = transfer of energy
Kinetic Energy
the energy of motion
K = frac12 mv2
Kinetic Energy
the energy of motion
K = frac12 mv2
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
The Integral = the area under the curve
If y = f(x) then the area under the f(x) curve from x = a to x = b is equal to
( )b
a
f x dx
Therefore given a graph of Force vs position the work done is the area under the curve
Work and Energy
Work and EnergyOften some force must do work
to give an object potential or kinetic energy
You push a wagon and it starts moving You do work to stretch a spring and you transform your work energy into spring potential energy
Or you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy
Work = Force x distance = transfer of energy
Kinetic Energy
the energy of motion
K = frac12 mv2
Kinetic Energy
the energy of motion
K = frac12 mv2
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Therefore given a graph of Force vs position the work done is the area under the curve
Work and Energy
Work and EnergyOften some force must do work
to give an object potential or kinetic energy
You push a wagon and it starts moving You do work to stretch a spring and you transform your work energy into spring potential energy
Or you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy
Work = Force x distance = transfer of energy
Kinetic Energy
the energy of motion
K = frac12 mv2
Kinetic Energy
the energy of motion
K = frac12 mv2
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Work and Energy
Work and EnergyOften some force must do work
to give an object potential or kinetic energy
You push a wagon and it starts moving You do work to stretch a spring and you transform your work energy into spring potential energy
Or you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy
Work = Force x distance = transfer of energy
Kinetic Energy
the energy of motion
K = frac12 mv2
Kinetic Energy
the energy of motion
K = frac12 mv2
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Work and EnergyOften some force must do work
to give an object potential or kinetic energy
You push a wagon and it starts moving You do work to stretch a spring and you transform your work energy into spring potential energy
Or you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy
Work = Force x distance = transfer of energy
Kinetic Energy
the energy of motion
K = frac12 mv2
Kinetic Energy
the energy of motion
K = frac12 mv2
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Kinetic Energy
the energy of motion
K = frac12 mv2
Kinetic Energy
the energy of motion
K = frac12 mv2
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Where does K = frac12 mv2 come from
Did your teacher just amazingly miraculously make that equation up
Hmmmhellip
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
The ldquoWork- Kinetic Energy TheoremrdquoS Work = DK
S F bull d = DK = frac12 mvf2 - frac12 mvo
2
College Board AP Objective You are supposed to be able to derive the work-kinetic energy theoremhellip (this does not mean to take a
derivative- it means to start with basic principles- S S F= ma S W = S F bulld and kinematics equations
and develop the equation for kinetic energy)
Sohellip do it Then yoursquoll see where the equation we call ldquokinetic energyrdquo comes from
(Hint start with S F = ma and use the kinematics equation that doesnrsquot involve time)
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
NET Work by all forces = D Kinetic EnergySFmiddotd = Dfrac12 mv2
How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)
The work done by the forces stopping the car = the change in the kinetic energy
SFmiddotd = Dfrac12 mv2
With TWICE the speed the car hasFOUR times the kinetic energy
Therefore it takes FOUR times the stopping distance
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
A car going 20 kmh will skid to a stop over a distance of 7 meters
If the same car was moving at 50 kmh how many meters would be required for it to come to a stop
The velocity changed by a factor of 25 therefore the stopping distance is 252 times the original distance
7 meters x 625 = 4375 meters
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Example Wnet = SF∙d = DKA 500 kg car moving at 15 ms skids 20 m to a stop
How much kinetic energy did the car lose
DK = frac12 mvf2
DK = -frac12 (500 kg)(15 ms) 2
DK = -56250 J
What is the magnitude of the net force applied to stop the car
SFmiddotd = DK
SF = DK d
SF = 56250 J 20 m
F = 28125 N
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
SF∙d = D frac12 mv2
A 002 kg bullet moving at 90 ms strikes a block of wood If the bullet comes to a stop at a depth of 25 cm inside the wood how much force did the wood exert on the bullet
F = 3240 N
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Example Wnet = SF∙d = DKA 500 kg car moving on a flat road at 15 ms skids to
a stopHow much kinetic energy did the car loseDK = Dfrac12 mvf
2 DK = -frac12 (500 kg)(15 ms)2
DK = -56250 JHow far did the car skid if the effective coefficient of
friction was = m 06Stopping force = friction = mN = mmgSFmiddotd = DK-(mmg)middotd = DKd = DK (-mmg) be careful to group in the denominatord = 56250 J (06 middot 500 kg middot 98 ms2) = 1913 m
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
First a review of Conservative Forces and Potential Energy
from the noteshellipWhat we have covered so far- the College Board Objectives state that the student should know
The two definitions of conservative forces and why they are equivalentTwo examples each of conservative and non-conservative forcesThe general relationship between a conservative force and potential energyCalculate the potential energy if given the forceCalculate the force if given the potential energy
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Conservation of Mechanical Energy
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Mechanical Energy
Mechanical Energy = Kinetic Energy + Potential Energy
E = frac12 mv2 + U
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Potential EnergyStored energy
It is called potential energy because it has the potential to do work
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
bull Example 1 Spring potential energy in the stretched string of a bow or spring or rubber band Us= frac12 kx2
bull Example 2 Chemical potential energy in fuels- gasoline propane batteries food
bull Example 3 Gravitational potential energy- stored in an object due to its position from a chosen reference point
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Gravitational potential energy
Ug = weight x height
Ug = mgh
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
bull The Ug may be negative For example if your reference point is the top of a cliff and the object is at its base its ldquoheightrdquo would be negative so mgh would also be negative
bull The Ug only depends on the weight and the height not on the path that it took to get to that height
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
ldquoConservativerdquo forces - mechanical energy is conserved if these are the only forces acting on an object
The two main conservative forces are Gravity spring forces
ldquoNon-conservativerdquo forces - mechanical energy is NOT conserved if these forces are acting on an object
Forces like kinetic friction air resistance (which is really friction)
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Conservation of Mechanical EnergyIf there is no kinetic friction or air resistance then the
total mechanical energy of an object remains the same
If the object loses kinetic energy it gains potential energy
If it loses potential energy it gains kinetic energy
For example tossing a ball upward
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Conservation of Mechanical Energy
The ball starts with kinetic energyhellip
Which changes to potential energyhellip
Which changes back to kinetic energy
K = frac12 mv2
PE = mgh
K = frac12 mv2
Energybottom = Energytop
frac12 mvb2 = mght
What about the energy when it is not at the top or bottom
E = frac12 mv2 + mgh
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Examples
bull dropping an objectbull box sliding down an inclinebull tossing a ball upwardsbull a pendulum swinging back and forthbull A block attached to a spring oscillating
back and forth
First letrsquos look at examples where there is NO friction and NO air resistancehellip
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Conservation of Mechanical Energy
If there is no friction or air resistance set the mechanical energies at each location equal Remember there may be BOTH kinds of energy at any location And there may be more than one form of potential energy U
E1 = E2
U1 + frac12mv12 = U2 + frac12 mv2
2
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Some EASY examples with kinetic and gravitational potential energyhellip
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
A 30 kg Egg sits on top of a 12 m tall wall
What is its potential energy
U = mgh
U = 3 kg x 10 ms2 x 12 m = 360 J
If the Egg falls what is its kinetic energy just before it hits the ground
Potential energy = Kinetic energy = 360 J
What is its speed just before it strikes the ground
360 J = K = frac12 mv2
2(360 J) 3 kg = v2
v = 1549 ms
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Example of Conservation of Mechanical Energy
Rapunzel dropped her hairbrush from the top of the castle where she was held captive If the window was 80 m high how fast was the brush moving just before it hit the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
gh = frac12 v2
2gh = v2
Donrsquot forget to take the square root
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Nowhellip do one on your own
An apple falls from a tree that is 18 m tall How fast is it moving just before it hits the ground (g = 10 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh = frac12 mv2
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
And another onehellipA woman throws a ball straight up with an initial velocity of 12 ms How high above the release point will the ball rise g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
frac12 mv2 = mgh
h = frac12 v2 g
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
And another onehellip
Mario the pizza man tosses the dough upward at 8 ms How high above the release point will the dough rise
g = 10 ms2
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Conservation of Mechanical Energy- another lookA skater has a kinetic energy of 57 J at position 1 the bottom of the ramp (and NO potential energy)At his very highest position 3 he comes to a stop for just a moment so that he has 57 J of potential energy (and NO kinetic energy)Mechanical energy = K + UWhat is his kinetic energy at position 2 if his potential energy at position 2 is 257 J
E = 57 J
E = 57 J
U = 257 JK =
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Conservation of Mechanical Energyhellip more difficult
A stork at a height of 80 m flying at 18 ms releases his ldquopackagerdquo How fast will the baby be moving just before he hits the ground
Energyoriginal = Energyfinal
mgh + frac12 mvo2 = frac12 mvf
2
Vf = 435 ms
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Now you do one hellip
The car on a roller coaster starts from rest at the top of a hill that is 60 m high How fast will the car be moving at a height of 10 m (use g = 98 ms2)
mgh1 + frac12 mv12 = mgh2 + frac12 mv2
2
mgh1 = mgh2 + frac12 mv22
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Conservation of Energy with Springs
Example One
A 6 x 105 kg subway train is brought to a stop from a speed of 05 ms in 04 m by a large spring bumper at the end of its track What is the force constant k of the spring Assume a linear spring unless told otherwise
SolutionAll the initial kinetic energy of the subway car is converted into elastic potential
energy
frac12 mv2 = frac12 kx2
Here x is the distance the spring bumper is compressed x = 04 m
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
A spring with spring constant k = 500 Nm is compressed a distance x = 010 m A block of mass m = 0250 kg is placed
next to the spring sitting on a frictionless horizontal plane When the spring and block are released how fast is
the block moving as it leaves the spring
When the spring is compressed work is required and the spring gains potential energy
Us = frac12 k x2
Us = frac12 (500 Nm) (010 m)2
Us = 25 J
As the spring and mass are released this amount of work done to the mass to change its kinetic energy from zero to a final
value of K = frac12 m v2
K = frac12 (025 kg) v2 = 25 J = Us = Ws
v2 = 20 m2 s2
v = 447 ms
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
How fast is the block moving when the spring is compressed 005 m
E = 25 J25 J = frac12 mv2 + frac12 kx2
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
E1 = E2
mgh1 + frac12 mv12 + frac12 kx1
2 = mgh2 + frac12 mv22 + frac12 kx2
2
Be careful about measuring height
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
What about objects suspended from a springThere are 3 types of energy involved
Kinetic gravitational and spring potential
However since potential energy is determined using a reference point you may choose the reference point for potential energy to be at the equilibrium location (with the mass attached) rather than the position where the spring is unstretched By doing so the total potential energy including both Us and Ug can be shown to be
Us + Ug = U = frac12 ky2
where y is the displacement measured from the equilibrium point
equilibrium
y
E = K + U
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
If there is kinetic friction or air resistance mechanical energy will not be conserved
Mechanical energy will be lost in the form of heat energy
The DIFFERENCE between the
original energy and the final energy
is the amount of mechanical energy lost due to heat
Final energy ndash original energy = energy loss
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Letrsquos try onehellipA 2 kg cannonball is shot straight up from the ground at 18 ms It reaches a highest point of 14 m How much mechanical energy was lost due to the air resistance g = 10 ms2
Final energy ndash original energy = Energy loss
mgh ndash frac12 mv2 = Heat loss
2 kg(10 ms2)(14 m) ndash frac12 (2 kg)(18 ms)2 =
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
And one morehellip
A 1 kg flying squirrel drops from the top of a tree 5 m tall Just before he hits the ground his speed is 9 ms How much mechanical energy was lost due to the air resistance
g = 10 ms2
Final energy ndash original energy = Energy loss
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Sometimes mechanical energy is actually INCREASED
For example A bomb sitting on the floor explodes
InitiallyKinetic energy = 0 Gravitational Potential energy = 0 Mechanical Energy = 0After the explosion therersquos lots of kinetic
and gravitational potential energyDid we break the laws of the universe and
create energyOf course not NO ONE NO ONE NO ONE
can break the lawsThe mechanical energy that now appears
came from the chemical potential energy stored within the bomb itself
Donrsquot even think about ithellip
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
According to the Law of Conservation of Energy
energy cannot be created or destroyed
But one form of energy may be transformed into another form as conditions change
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Power
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Power
The rate at which work is done
1 Power = Work divide time
Unit for power = J divide s
= Watt W
What is a Watt in ldquofundamental unitsrdquo
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Example
A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 05 seconds How powerful was he
P = W t
W = Fd
W = mg x h
W = 80 kg x 10 ms2 x 2 m = 1600 J
P = 1600 J 05 s
P = 3200 W
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Since work is also the energy transferred or transformed ldquopowerrdquo is the rate at which energy is transferred or transformed
2 Power = Energy divide time
This energy could be in ANY form heat light potential chemical nuclear
Example How long does it take a 60 W light bulb to give off 1000 J of energy
Time = Energy divide Power
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Since NET work = D K
3 P = DK divide t
Example How much power is generated when a 1500 kg car is brought from rest up to a speed of 20 ms in 50 s
Power = frac12 mv2 divide t
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
And yet another approach
P = W divide t = (Fd) divide t = F middot (d divide t)
4 P = F middot v This is a DOT product
Example What power is generated when a force given by F = 3i - 8j produces a velocity given by v = 6i + 2j
P = (3i x 6i) + (-8j x 2j) = 2 Watts
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Calculus ApplicationsA particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t where a and b are constants due to an applied force Fy(t) What is the power P(t) delivered by the force
P = Fmiddotv
We need to find both v and F v t at bt a t at b F ma m at b 4 2 2 12 2 12 23 2 2( ) ( )
P F v m at b at bty y ( )( )12 2 4 2 22 3
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Varying Forces
The rule ishellip
ldquoIf the Force varies you must integraterdquo
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Work = Fmiddotd Power = FmiddotvIf the force varies with displacement in other words
force is a function of displacement F(x)
you must integrate to find the work done
If the force is a function of velocity F(v) you must integrate to find the power output
(AND to determine velocity as a function of time v(t) - all that natural log
stuff we did for air resistance)
If the force varies with time F(t)hellip well thatrsquos coming up next unit
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Examples
If F(x) = 5x3 what work is done by the force as the object moves from x = 2 to x = 5
If F(v) = 4v2 what power was developed as the velocity changed from 3 ms to 7 ms
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
Another tricky example using definite integration hellipA 288 kg particle starts from rest at x = 0 and moves under
the influence of a single force F = 6 + 4x -3x2 where F is in Newtons and x is in meters Find the power delivered to the particle when it is at x = 3 m
P = Fmiddotv and finding the force at x = 3 is easy but what is the velocity at x = 3 m F = ma and integrating a(t) would yield velocity but the variable here is NOT t - itrsquos x
HmmmhellipW F x dx and W K so ( )
F x dx K mv mvx
x
f o
o
f
( ) 12
12
2 2 ( )6 4 3 12
2
0
3
32 x x dx K mv x
( ) ( )6 4 3 64
2
3
36 3
4
23
3
33 0 9 1
22
0
3
2 3
0
3 2 33
2 x x dx x x x J mv x
v m s and F N so P F v Wx x 3 322 5 6 4 3 3 3 9 9 2 5 22 5
