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Conservation of EnergyConservation of Energy
1-D Variable Force Example: Spring1-D Variable Force Example: Spring
For a spring we recall that For a spring we recall that FFxx = -kx = -kx..
F(x) x2
x
x1
-kxrelaxed position
F = - k x1
F = - k x2
the mass
Review: SpringsReview: Springs Hooke’s Law: Hooke’s Law: The force exerted by a spring is The force exerted by a spring is
proportional to the distance the spring is stretched or proportional to the distance the spring is stretched or compressed from its relaxed position.compressed from its relaxed position.
FFXX = -kx = -kx
Where Where xx is the displacement from the equilibrium and is the displacement from the equilibrium and kk is the constant of proportionality. is the constant of proportionality.
relaxed position
FX = 0
x
More Spring ReviewMore Spring Review The work done The work done by the spring Wby the spring Wss during a displacement from during a displacement from xx11 to to xx22 is the area under the is the area under the F(x) F(x) vs vs xx plot between plot between xx11 and and xx22..
2
1
2
1
2
1
2
2 2s 2 1
( )
( )
1
1W
2
2
x
s
x
x
x
x
x
W F x dx
kx dx
k
k x
x
x
F(x) x2
Ws
x
x1
-kx
In this example it is a negative number. The spring does negative work on the mass
Problem: Spring pulls on mass.Problem: Spring pulls on mass. A spring (constant A spring (constant kk) is stretched a distance ) is stretched a distance dd, and a mass , and a mass mm is hooked to its end. The mass is released (from rest). What is the is hooked to its end. The mass is released (from rest). What is the
speed of the mass when it returns to the relaxed position if it slides without friction?speed of the mass when it returns to the relaxed position if it slides without friction?
relaxed position
stretched position (at rest)
dafter release
back at relaxed position
vr
v
m
m
m
m
Solution: Spring pulls on mass.Solution: Spring pulls on mass. First find the net work done on the mass during the motion from First find the net work done on the mass during the motion from x = d x = d to to x = 0 x = 0 (only due to the spring):(only due to the spring):
stretched position (at rest)
d
relaxed position
vr
m
m
2 2 2 2 22 1
1 1 10
2 2 2sW k x x k d kd
x1x2
Problem: Spring pulls on mass.Problem: Spring pulls on mass. Now find the change in Now find the change in kinetic energy kinetic energy of the mass:of the mass:
2 22 1
2 2
2r
1 1ΔK mv mv
2 21 1
mv m02 21
mv2
r
stretched position (at rest)
d
vr
m
m
x1x2
relaxed position
Solution: Spring pulls on mass.Solution: Spring pulls on mass. Now use work kinetic-energy theorem: Now use work kinetic-energy theorem: WWnetnet = W = WSS = = KK..
21
2kd 2
r
1mv
2 r
kv d
m
stretched position (at rest)
d
vr
m
m
x1x2
relaxed position
Springs : UnderstandingSprings : Understanding
A spring with spring constant A spring with spring constant 40 N/m40 N/m has a relaxed length of has a relaxed length of 1 m1 m. When the spring is stretched so . When the spring is stretched so that it is that it is 1.5 m1.5 m long, what force is exerted on a block attached to the end of the spring? long, what force is exerted on a block attached to the end of the spring?
x = 0
M
k k
M
x = 0 x = 1.5
(a) -20 N (b) 60 N (c) -60 N
x = 1
FX = - (40N/m) ( .5m)
FX = - 20 N
FX = -kx
UnderstandingUnderstandingForces and Motion Forces and Motion
A block of mass A block of mass M = 5.1 kgM = 5.1 kg is supported on a frictionless ramp by a spring having constant is supported on a frictionless ramp by a spring having constant k = 125 N/mk = 125 N/m. When the ramp is horizontal the equilibrium position of the mass is at . When the ramp is horizontal the equilibrium position of the mass is at x = 0x = 0. When the angle of the ramp is . When the angle of the ramp is changed to changed to 3030oo what is the new equilibrium position of the block what is the new equilibrium position of the block xx11??
(a(a) ) xx11 = 20cm = 20cm (b) (b) xx11 = 25cm = 25cm (c) (c) xx11 = 30cm = 30cm
x = 0
M
k
x 1 = ?
Mk
= 30o
Solution Solution
x 1
Mk
x
y
Choose the x-axis to be along downward direction of ramp.
Mg
FBD: The total force on the block is zero since it’s at rest.
N
Fx,g = Mg sin
Force of gravity on block is Fx,g = Mg sin(
Consider x-direction:
Force of spring on block is Fx,s = -kx1
F x,s = -kx 1
Solution Solution
x 1
Mk
x
y
Since the total force in the x-direction must be 0:
F x,g = Mg sin
F x,s = -kx 1
Mg sinkx1 1
Μg sin θx
k
2
1
5.1 9.80.2
1 0.5
125
kg m sx
N mm
Work by variable force in 3-D: Nice Work by variable force in 3-D: Nice to know explanationto know explanation
Work dWF of a force FF acting
through an infinitesimal
displacement r r is:
dW = FF.rr
The work of a big displacement through a variable force will be the integral of a set of infinitesimal displacements:
WTOT = FF.rr
FF
rr
Work/Kinetic Energy Theorem for a Work/Kinetic Energy Theorem for a Variable Force in 3DVariable Force in 3D
2
1
2 21 12 12 2
r
r
W F dr mv mv kinetic energy
Sum up F.dr along path
That’s the work integral
That equals change in KE
For conservative forces, the work is path independent and depends only on starting point and end point
Work by variable force in 3-D:Work by variable force in 3-D:Newton’s Gravitational ForceNewton’s Gravitational Force Integrate dWg to find the total work done by gravity in a “big”
displacement:
Wg = dWg = (-GMm / R2) dR = GMm (1/R2 - 1/R1)
FFg(R1)
R1
R2
FFg(R2)
R1
R2
R1
R2
m
M
Work by variable force in 3-D:Work by variable force in 3-D:Newton’s Gravitational ForceNewton’s Gravitational Force Work done depends only on R1 and R2, not on the path takennot on the path taken.
R1
R2
m
M
2 1
1 1gW GMm
R R
Potential EnergyPotential Energy For any conservative force For any conservative force FF we can define a we can define a
potential energy function potential energy function UU in the following way: in the following way:
The work done by a conservative force is equal and opposite to The work done by a conservative force is equal and opposite to the change in the potential energy function.the change in the potential energy function.
This can be written as:This can be written as:
W = FF.dr r = -U
U = U2 - U1 = -W = - FF.drrr1
r2
r1
r2 U2
U1
Gravitational Potential EnergyGravitational Potential Energy
So we see that the change in So we see that the change in UU near the Earth’s surface is:near the Earth’s surface is: U = -WU = -Wgg = mg = mg yy = mg= mg((yy2 2 -y-y11).).
SoSo U = mg y + UU = mg y + U0 0 wherewhere UU00 is an is an arbitrary constantarbitrary constant..
Having an arbitrary constant Having an arbitrary constant UU00 is equivalent to saying that we can is equivalent to saying that we can choose the choose the y y location where location where U = 0 U = 0 to be anywhere we want to.to be anywhere we want to. Floor level of 400Floor level of 400 Hazel StHazel St Science Office (potential is zero here, for sure!)Science Office (potential is zero here, for sure!)
y1
m
Wg = -mg yy2
Conservative Forces:Conservative Forces:
We have seen that the work done by gravity does not We have seen that the work done by gravity does not depend on the path taken.depend on the path taken.
R1
R2
M
m
hm
Wg = -mgh
2 1
1 1gW GMm
R R
UnderstandingUnderstandingWork & EnergyWork & Energy
A rock is dropped from a distance RE above the surface of the earth, and is observed to have kinetic energy K1 when it hits the ground. An identical rock is dropped from twice the height (2RE) above the earth’s surface and has kinetic energy K2 when it hits. RE is the radius of the earth.
What is K2 / K1?
(a)(a)
(b)(b)
(c)(c)
3
2
4
3
2
RE
RE
2RE
The easiest way to solve this problem is to use the W=K property.
Be careful!
SolutionSolution
Since energy is conserved, Since energy is conserved, K = WK = WGG..
G2 1
1 1W =GMm
R R
2 1
1 1ΔK=c
R R
Where c = GMm is the same for both rocks
RE
RE
2RE
Do not use mgh formula as this only works when h is very small.
1E E E
1 1 1 1K =c c
R 2R 2 R
2E E E
1 1 2 1K =c c
R 3R 3 R
2
1
2K 3=
1K2
4
3
For the second rock:
For the first rock:
Conservative Forces:Conservative Forces: In general, if the work done does not depend on the path taken (only In general, if the work done does not depend on the path taken (only
depends the initial and final distances between objects), the force depends the initial and final distances between objects), the force involved is said to be involved is said to be conservativeconservative..
Gravity is a Gravity is a conservativeconservative force:force:
Gravity near the Earth’s surface:Gravity near the Earth’s surface:
A spring produces a A spring produces a conservativeconservative force:force: 2 22 1
1
2sW k x x
gW mg y
2 1
1 1gW GMm
R R
Conservative Forces:Conservative Forces:
A force that offers the opportunity of two-way conversion A force that offers the opportunity of two-way conversion between kinetic and potential energies is called a between kinetic and potential energies is called a conservative force.conservative force.
The work done by a conservative force always has these The work done by a conservative force always has these properties:properties: It can always be expressed as the difference between It can always be expressed as the difference between
thethe initial initial andand final final values of potential energy function.values of potential energy function. It is reversible.It is reversible. It is independent of the path of the body and depends It is independent of the path of the body and depends
only on the starting and ending points.only on the starting and ending points. When the starting and ending points are the same, the When the starting and ending points are the same, the
total work is zero.total work is zero.
Conservative Forces:Conservative Forces: When the only forces that do work are conservative forces, then the When the only forces that do work are conservative forces, then the
total mechanical energy is total mechanical energy is E = K + UE = K + U Conservative forces have the nice property of being able to be Conservative forces have the nice property of being able to be
defined in terms of a defined in terms of a potential energypotential energy. The usual definition of . The usual definition of potential energy is through the work-energy theorem as for kinetic potential energy is through the work-energy theorem as for kinetic energy, i.e. energy, i.e. W = UW = Uii - U - Uff. .
1 2totalW U U
2 1totalW K K
1 2 2 1
1 1 2 2
U U K K
U K U K
NonConservative Forces:NonConservative Forces: Not all forces are conservative.Not all forces are conservative. Consider the friction force applied to a crate, the total Consider the friction force applied to a crate, the total
work done by friction force when sliding the crate up a work done by friction force when sliding the crate up a ramp and back down is not zero. (ramp and back down is not zero. (when the direction of when the direction of the motion reverses so does the friction force, and the the motion reverses so does the friction force, and the friction does negative work in both directionsfriction does negative work in both directions.).)
Conservative Forces:Conservative Forces: We have seen that the work done by a conservative force We have seen that the work done by a conservative force
does not depend on the path taken.does not depend on the path taken.
W1
W2
W1
W2
W1 = W2
WNET = W1 - W2
= W1 - W1 = 0
Therefore the work done by a conservative force in a closed path is 0.
Potential energy change from one point to another does not depend on path
UnderstandingUnderstandingConservative ForcesConservative Forces
The pictures below show force vectors at different points in space for The pictures below show force vectors at different points in space for two forces. Which one is conservative ? two forces. Which one is conservative ?
(a)(a) 1 (b)(b) 2 (c)(c) both
(1) (2)
y
x
y
x
SolutionSolution Consider the work done by force when moving along different paths in Consider the work done by force when moving along different paths in
each case: each case:
(1) (2)
WA = WB WA > WB
No work is done when going perpendicular to force.
SolutionSolution In fact, you could make money on type (2) if it ever existed:In fact, you could make money on type (2) if it ever existed:
Work done by this force in a “round trip” is Work done by this force in a “round trip” is > 0> 0!! Free kinetic energy!! Free kinetic energy!!
W = 15 J
W = 0
W = -5 J
W = 0WNET = 10 J = K
Potential Energy Recap:Potential Energy Recap: For any conservative force we can define a potential energy function For any conservative force we can define a potential energy function
UU such that: such that:
The potential energy function The potential energy function UU is always defined only up to an is always defined only up to an additive constant.additive constant.
You can choose the location where You can choose the location where U = 0 U = 0 to be anywhere to be anywhere convenient.convenient.
U = U2 - U1 = -W = - FF.drrS1
S2
Conservative Forces & Potential Conservative Forces & Potential Energies Energies
(stuff you should know):(stuff you should know):Force Force
FF
Work Work (done by force)(done by force)
W
Change in P.EChange in P.E
U = U2 - U1
P.E. functionP.E. function
U
FFg = -mg -mg(y2-y1) mg(y2-y1) mgy + C
FFg =2 1
1 1GMm
R R
2 1
1 1GMm
R R
2
GMm
R
GMm
RC
Fs = -kx 2 22 1
1
2k x x 2 2
2 1
1
2k x x 1
22kx C
(R is the center-to-center distance, x is the spring (R is the center-to-center distance, x is the spring stretch)stretch)
UnderstandingUnderstandingPotential EnergyPotential Energy
All springs and masses are identical. (Gravity acts down). All springs and masses are identical. (Gravity acts down). Which of the systems below has the most potential energy stored in its spring(s), relative to the Which of the systems below has the most potential energy stored in its spring(s), relative to the
relaxed position? relaxed position?
(a)(a) 1
(b)(b) 2
(c)(c) same
(1) (2)
SolutionSolution
The displacement of The displacement of (1)(1) from equilibrium will be half of that of from equilibrium will be half of that of (2)(2) (each spring exerts (each spring exerts half of the force needed to balance half of the force needed to balance mgmg))
(1) (2)
d 2d
0
The potential energy stored in (1) is: 2 212 kd kd
2
The potential energy stored in (2) is: 2 21k 2d 2kd
2
The spring P.E. is twice as big in (2) !
Conservation of Mechanical EnergyConservation of Mechanical Energy If only conservative forces are presentIf only conservative forces are present, , the total kinetic plus potential the total kinetic plus potential
energy of is conserved, i.e. the total “energy of is conserved, i.e. the total “mechanical energymechanical energy” is ” is conserved (def. of conserved (def. of MEME). ).
(note: (note: E=EE=Emechanicalmechanical throughout this discussion)throughout this discussion)
Both Both KK and and UU can change, but can change, but E = K + U E = K + U remainsremains constant.constant. But we’ll see that if dissipative forces act, then energy can be “lost” to But we’ll see that if dissipative forces act, then energy can be “lost” to
other modes (thermal, sound, etc) changing other modes (thermal, sound, etc) changing EEmechanicalmechanical and external forces and external forces can change can change EEmechanicalmechanical
E = K + UE = K + U = W + U = W + (-W) = 0
using K = W using U = -W
E = K + U is constantconstant!!!!!!
Example: The simple pendulumExample: The simple pendulum
Suppose we release a mass Suppose we release a mass mm from rest a distance from rest a distance hh11 above its lowest possible point.above its lowest possible point. What is the maximum speed of the mass and whereWhat is the maximum speed of the mass and where
does this happen?does this happen? To what height To what height hh22 does it rise on the other side? does it rise on the other side?
v
h1 h2
m
Example: The simple pendulumExample: The simple pendulum
Kinetic+potential energy is conserved since gravity is a Kinetic+potential energy is conserved since gravity is a conservative force (conservative force (E = K + U E = K + U is constant)is constant)
Choose Choose y = 0 y = 0 at the bottom of the swing, at the bottom of the swing, and and U = 0 U = 0 at at y = 0 y = 0 ((arbitrary choicearbitrary choice))
E = E = 11//22mvmv22 + mgy + mgy
v
h1 h2
y
y = 0
Example: The simple pendulumExample: The simple pendulum
E = E = 11//22mvmv22 + mgy + mgy.. Initially, Initially, y = hy = h11 and and v = 0v = 0, so , so E = mghE = mgh11.. Since Since E = mghE = mgh11 initially, initially, E = mghE = mgh11 always since energyalways since energy isis conserved.conserved.
y
y = 0
Example: The simple pendulumExample: The simple pendulum
11//22mvmv22 will be maximum at the bottom of the swing. will be maximum at the bottom of the swing. So at So at y = 0 y = 0 11//22mvmv22 = mgh = mgh11 vv22 = 2gh = 2gh11
v
h1
y
y = h1
v gh 2 1
y = 0
Example: The simple pendulumExample: The simple pendulum
Since Since E = mghE = mgh1 1 == 11//22mvmv22 + mgy + mgy it is clear that the maximum it is clear that the maximum
height on the other side will be at height on the other side will be at y = hy = h1 1 = h= h22 and and v = 0v = 0.. The ball returns to its original height.The ball returns to its original height.
y
y = h1 = h2
y = 0
Example: The simple pendulumExample: The simple pendulum
The ball will oscillate back and forth. The limits on its The ball will oscillate back and forth. The limits on its height and speed are a consequence of the sharing of height and speed are a consequence of the sharing of energy between energy between KK and and UU. .
E = E = 11//22mvmv22 + mgy + mgy = K + U == K + U = constant constant..
y
Example: Airtrack & GliderExample: Airtrack & Glider
A glider of mass A glider of mass MM is initially at rest on a horizontal frictionless is initially at rest on a horizontal frictionless track. A mass track. A mass mm is attached to it with a massless string hung is attached to it with a massless string hung over a massless pulley as shown. What is the speed over a massless pulley as shown. What is the speed vv of of MM after after mm has fallen a distance has fallen a distance dd ? ?
d
M
m
v
v
Example: Airtrack & GliderExample: Airtrack & Glider Kinetic+potential energy is conserved since all forces are Kinetic+potential energy is conserved since all forces are
conservative.conservative. Choose initial configuration to have Choose initial configuration to have U=0U=0..
K = -K = -UU
2mgdv
m M
d
M
m
v
1
22m M v mgd
Problem: HotwheelProblem: Hotwheel
A toy car slides on the frictionless track shown below. It A toy car slides on the frictionless track shown below. It starts at rest, drops a distance starts at rest, drops a distance dd,, moves horizontally at moves horizontally at speed speed vv11, rises a distance , rises a distance hh, and ends up moving , and ends up moving horizontally with speed horizontally with speed vv22.. Find Find vv11 and and vv22..
hd
v1
v2
Problem: Hotwheel...Problem: Hotwheel...
K+U K+U energy is conserved, so energy is conserved, so E = 0 E = 0 K = - K = - UU Moving down a distance Moving down a distance dd, , U = -mgdU = -mgd, , K = K = 11//22mvmv11
22
Solving for the speed: Solving for the speed:
hd
v1
v gd1 2
Problem: Hotwheel...Problem: Hotwheel...
At the end, we are a distance At the end, we are a distance d - hd - h below our starting point.below our starting point. U = -mg(d - hU = -mg(d - h)), , K = K = 11//22mvmv22
22
Solving for the speed:Solving for the speed:
hd
v2
v g d h2 2
d - h
Hooke’s Law (review)Hooke’s Law (review)The magnitude of the force exerted by the spring is directly proportional to the distance the spring has moved from its equilibrium.
xF kx Force is opposite to the direction spring is moved
xF kx This is the Force applied to the spring
ExampleExampleA 0.085 kg mass is hung from a vertical spring that is allowed to stretch slowly from its unstretched equilibrium position until it comes to its new equilibrium position 0.20 m below its initial one.
a)Determine the force constant of the spring?b)If the ball is returned to the spring’s initial unstreched equilibrium position and then allowed to fall, what is the Net Force on the mass when it has dropped 0.082 m?c)Determine the acceleration of the mass at position b)
M
F=-kx
F=mg
SolutionSolution
a) Determine the force constant of the spring?
0
0
0.085 9.
4.1 5
8
0.20
6
yF
mg kx
kx mg
mgk
x
Nkg
kg
m
mN
Therefore k= 4.2 N/m
F=-kx
F=mg
SolutionSolution
b) If the ball is returned to the spring’s initial unstreched equilibrium position and then allowed to fall, what is the Net Force on the mass when it has dropped 0.082 m?
0.4915
0.085 9.8 4.165 0.082
NF mg kx
N Nkg m
kg m
N
Therefore F= 0.49 N
F=-kx
F=mg
SolutionSolution
c) Determine the acceleration of the mass at position b)
2
0.4915
0.4915
0.
5.7
8
82
0 5
y y
y
y
F ma
N ma
N
g
m
ak
s
Therefore a= 5.8 m/s2 down
Elastic Potential Energy Elastic Potential Energy (review)(review)
The energy stored in objects that are stretched, compressed, bent, or twisted.
21
2sU kx
UnderstandingUnderstandingA 0.10 kg mass is hung from a vertical spring (k=9.6 N/m). The mass is held so that the spring is at its unstretched equilibrium position. The mass is then allowed to fall. Neglect the mass of the spring.
a)How much elastic potential energy is stored in the spring when the mass has fallen 11 cm?b)What is the speed of the mass when it has fallen 11cm?
M
SolutionSolution
a) How much elastic potential energy is stored in the spring when the mass has fallen 11 cm?
2
2
2
1
21
9.
5
6 0.11
.8 0
2
1
sU kx
Nm
m
J
x=0 cm
x=11 cm
M
SolutionSolution
b) What is the speed of the mass when it has fallen 11cm?
2 2
2 2
2
2
2
1 1
2 2
2
2
2 0.1
1
0 9.8 0.11 9.6 0.11
0.
.0
10
i f
f
f
f
E E
mgh mv kx
mgh kx mv
mgh kxv
m
m Nkg m
s
ms
m
mkg
x=0 cm
x=11 cm
FlashFlash
