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Old assignments and midterm exams
(solutions have been posted on
the web)
can be picked up in my office
(LB-212)
AssignmentAssignment 66
Textbook (Giancoli, 6th edition), Chapter 6:
Due on Thursday, October 30, 2008
- Problem 19 - page 162 of the textbook
- Problems 22 and 31 - page 163 of the textbook
- Problem 89 - page 166 of the textbook
Chapter 6 (Today)
• Work Done by a Constant Force
• Kinetic Energy, and the Work-Energy Principle
• Potential Energy• Potential Energy
The work done by a constant force is defined as the distance moved multiplied by
the component of the force in the direction of displacement:
Work Done by a Constant ForceWork Done by a Constant Force
(6-1)
Work Done by a Constant ForceWork Done by a Constant Force
Problem 6-10 (textbook): What is the minimum work needed to push a 950-kg car
810 m up along a 9.0º incline?
(a) Ignore friction.
(b) Assume the effective coefficient of friction retarding the car is 0.25.
Work Done by a Constant ForceWork Done by a Constant Force
Problem 6-10:
Draw a free-body diagram of the car on the incline. Include a frictional force, but
ignore it in part (a) of the problem.
The minimum work will occur when the car is moved at a constant velocity.
y
θ
θ
y
x
m gr
NFr
f rFr
PFr
Work Done by a Constant ForceWork Done by a Constant Force
Problem 6-10:
(a)
Write Newton’s 2nd law in both the x and y directions, noting that the car is
unaccelerated
N Ncos 0 cos
yF F mg F mgθ θ= − = → =∑
The work done by in moving the car a distance d along the plane (parallel
to ) is given by
P Psin 0 sin
xF F mg F mgθ θ= − = → =
∑
∑
PFr
PFr
( ) ( )( )o 2 o 6
P Pcos0 sin 950 kg 9.80m s 810 m sin9.0 1.2 10 JW F d mgd θ= = = = ×
Work Done by a Constant ForceWork Done by a Constant Force
Problem 6-10:
(b)
Now include the frictional force, given by:
We still assume that the car is not accelerated. We again write Newton’s 2nd
law for each direction:
� The y-forces are unchanged by the addition of friction, and so we still have
fr NkF Fµ=
� The y-forces are unchanged by the addition of friction, and so we still have
� In the x-direction we have
The work done by in moving the car a distance d along the plane (parallel to )
is given by
Nc o sF m g θ=
P fr P frsin 0 sin cos sin
x kF F F mg F F mg mg mgθ θ µ θ θ= − − = → = + = +∑
PFr
PFr
( )
( )( )( )( )
o
P P
2 o o 6
cos 0 sin cos
950 kg 9.80 m s 810 m sin 9.0 0.25 cos 9.0 3.0 10 J
kW F d mgd θ µ θ= = +
= + = ×
Kinetic Energy, and the Work Energy PrincipleKinetic Energy, and the Work Energy Principle
I mentioned, in the begin of this lecture, that the concept of energy is very important
in physics. Yet, I started discussing about work done by a force.
� What is the connection between these two quantities?
A good definition of energy was introduced by the Einstein’s theory of relativity:
E = mc2
But this is beyond the scope of this course.But this is beyond the scope of this course.
In mechanics, we can use a less precise definition of energy as:
“The ability to do work”
In this course we will be defining translational kinetic energy and some forms of
potential energies, though other forms the energy exist such as nuclear energy,
heat energy, etc.
Kinetic Energy, and the Work Energy PrincipleKinetic Energy, and the Work Energy Principle
Kinetic Energy
A moving object can do work on a second object. Example:
A car hits another car � the first car applies a force on the second (say, at rest)
which consequently undergoes a displacement.
This has the implication that the first car has the ability to do work, consequently it
has (or carries) energy.
The energy of motion is called
kinetic energy (or KE for short).
� But… how do we measure kinetic energy?
Kinetic Energy, and the Work Energy PrincipleKinetic Energy, and the Work Energy Principle
How do we measure kinetic energy?
Consider a rigid body of mass m, say the bus in the figure, moving along a line.
A net force acts on the bus changing its velocity from to over a
displacement of magnitude d.
The work done on the bus by the force is given by:
Using Newton’s 2nd law, , we can rewrite the above equation as:Using Newton’s 2nd law, , we can rewrite the above equation as:
Kinetic Energy, and the Work Energy PrincipleKinetic Energy, and the Work Energy Principle
How do we measure kinetic energy?
We can now use the equation of motion
with
x – x0
= d
v = v2
v0
= v1
Previous slide:
(6-2)
Kinetic Energy, and the Work Energy PrincipleKinetic Energy, and the Work Energy Principle
How do we measure kinetic energy?
We define translational kinetic energy as:
With this definition, we can rewrite eq. 6.2:
(6-3)
Previous slide:
(6-4)
Where,
We can also write
where
Note: Eq. 6.3 is valid in more general cases even when the force varies.
(6-6)
(6-7)
(6-4)
(6-5)
Kinetic Energy, and the Work Energy PrincipleKinetic Energy, and the Work Energy Principle
How do we measure kinetic energy?
Equation 6.6 represents the work-energy
principle which can be stated as
“ The net work done on a object is equal to the change in the
object’s kinetic energy ”
Previous slide:
object’s kinetic energy ”
The work-energy principle is ONLY valid for the net work done on an object:
� in other words, the work done by the net force applied on the object.
Kinetic Energy, and the Work Energy PrincipleKinetic Energy, and the Work Energy Principle
Notes:
Observing the previous equations:
We can conclude the following
1) If the net work done on an object is positive, then ∆KE > 0 implying that there 1) If the net work done on an object is positive, then ∆KE > 0 implying that there
was an increase in the object’s velocity ( v2
> v1
).
2) If the net work done on an object is negative, then ∆KE < 0 implying that there
was an decrease in the object’s velocity ( v2
< v1
).
Notes:
Because work and kinetic energy can be equated, they must have the same units:
kinetic energy is measured in joules. It is also a scalar quantity.
Kinetic Energy, and the Work Energy PrincipleKinetic Energy, and the Work Energy Principle
The change (decrease) in kinetic energy of the
hammer is equal to the energy gained by the
nail (it starts moving)
or, more generally: The decrease in kinetic energy of
the hammer is equal to the work the hammer can do
on other objects.
Kinetic Energy, and the Work Energy PrincipleKinetic Energy, and the Work Energy Principle
Problem 6-21 (textbook): If the speed of a car is increased by 50%, by what factor
will its minimum braking distance be increased, assuming all else is the same? Ignore
the driver’s reaction time.
Assume:
1) Only friction acts along the direction of the car’s displacement
2) The driver locks the brakes.
Kinetic Energy, and the Work Energy PrincipleKinetic Energy, and the Work Energy Principle
Problem 6-21:
The work needed to stop the car is equal to the change in the car’s kinetic energy.
That work comes from the force of friction on the car. Assume the maximum possible
frictional force, which results in the minimum braking distance. Thus
Where the static friction is used since the driver locks the
brakes. The normal force is equal to the car’s weight if
it is on a level surface, and so
fr NsF Fµ=
d = stopping distance
it is on a level surface, and so
In the diagram, the car is traveling to the right.
Since , if the velocity is increases by 50%, or is multiplied by 1.5, then d
will be multiplied by a factor of (1.5)2 , or 2.25.
f r sF m gµ=
2
o 2 2 2 11 1 1
fr 2 1 12 2 2 cos180
2s
s
vW KE F d mv mv mgd mv d
gµ
µ= ∆ → = − → − = − → =
mgr
NFr
frFr
2
1d v∝
Work Done by a Constant ForceWork Done by a Constant Force
Problem 6-25 (textbook): A 285-kg load is lifted 22.0 m vertically with an
acceleration a = 0.160g by a single cable. Determine
(a) the tension in the cable,
(b) the net work done on the load,
(c) the work done by the cable on the load,
(d) the work done by gravity on the load, and (d) the work done by gravity on the load, and
(e) the final speed of the load assuming it started from rest.
Work Done by a Constant ForceWork Done by a Constant Force
Problem 6-25:
(a)
From the free-body diagram for the load being lifted, write Newton’s 2nd law for
the vertical direction, with up being positive.
( )( )
T
2 3
T
0.160
1.16 1.16 285 kg 9.80 m s 3.24 10 N
F F mg ma mg
F mg
= − = = →
= = = ×
∑Fnet =
( )( )T
TFr
m gr
Kinetic Energy, and the Work Energy PrincipleKinetic Energy, and the Work Energy Principle
Problem 6-25:
(b)
The net work done on the load is found from the net force.
( ) ( )( )( )o 2
net net
3
cos 0 0.160 0.160 285 kg 9.80 m s 22.0 m
9.83 10 J
W F d mg d= = =
= ×
TFr
m gr
Kinetic Energy, and the Work Energy PrincipleKinetic Energy, and the Work Energy Principle
Problem 6-25:
(c)
The work done by the cable on the load is
( ) ( )( )( )o 2 4
cable Tcos 0 1.160 1.16 285 kg 9.80m s 22.0 m 7.13 10 JW F d mg d= = = = ×
TFr
m gr
Kinetic Energy, and the Work Energy PrincipleKinetic Energy, and the Work Energy Principle
Problem 6-25:
(d)
The work done by gravity on the load is
( ) ( )( )o 2 4
Gcos180 285 kg 9.80 m s 22.0 m 6.14 10 JW mgd mgd= = − = − = − ×
TFr
m gr
Kinetic Energy, and the Work Energy PrincipleKinetic Energy, and the Work Energy Principle
Problem 6-25:
(e)
Use the work-energy theory to find the final speed, with an initial speed of 0.
( )
2 21 1
n e t 2 1 2 12 2
3
2
2 1
2 9 .8 3 1 0 J20 8 .3 1 m s
2 8 5 k g
n e t
W K E K E m v m v
Wv v
m
= − = − →
×= + = + =
TFr
m gr
2 8 5 k gm