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Chapter 6
Work and Kinetic
Energy
PowerPoint® Lectures for
University Physics, Twelfth Edition
– Hugh D. Young and Roger A. Freedman
Lectures by James Pazun
Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley
Goals for Chapter 6
• To understand and calculate work done by a
force
• To study and apply kinetic energy
• To learn and use the work-energy theorem
• To see the difference from the first bullet and
then calculate work done by a varying force
along a curved path
• To add time to the calculation and determine
the power in a physical situation
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Introduction
• We’ve studied how Newton’s
Second Law allows us to
calculate an acceleration from
a force but what if the force
changes during its
application? We must be able
to account for things like an
archer’s bow.
• We must look at action–
reaction pairs that are not
immediately obvious (like the
shotgun expelling the pellets
with expanding gas but having
the expanding gas do work on
the shotgun at the same
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6.1 Work
Caution: Work = W, weight = w
• Work is done when a force exerted on a body caused the body to
undergo a displacement.
 
• Work W is defined as a dot product: W  F  s  F s cos 
The SI unit of work is the joule
1 joule = (1 newton)·(1 meter) or 1 J = 1 N∙m
1 J = 1 kg∙m2/s2
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Work is a scalar quantity
Even though work is calculated by using two
vector quantities (force and displacement), it is a
dot product. Dot product is a scalar quantity. It has
only a number value. It has no direction!
F∙s = Fxsx + Fysy + Fzsz
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Example 6.1 - Use the parallel component if the force acts at an angle
a. Steve exerts a steady force of magnitude 210 N on the
stalled car as he pushes it a distance of 18 m. The car also
has a flat tire, so to make the car track straight Steve must
push at and angle of 30o to the direction of motion. How
much work does Steve do?
W = 3300 J
b. in a helpful mood,
Steve pushes a second stalled car with a

steady force F  (160 N )iˆ  (40 N ) ˆj. The displacement of the car

is s  (14m)iˆ  (11m) ˆj. How much work does Steve do in this
case?
W = 1800 J
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Positive Work, Negative Work and No Work
When 0 ≤ Φ < 90o
work is positive
When 90o < Φ ≤ 180o
work is negative
When Φ = 90o
work is zero
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How can it be such a great “workout” with no work?
• When positive and negative work cancel, the net
work is zero even though muscles are exercising.
CAUTION keep track of who’s doing the work
The total work Wtot done on the body is the algebraic
sum of the quantities of work done by the individual
forces.
Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley
Example 6.2- solution of work done by several forces
• A farmer hitches her tractor to a
sled loaded with firewood and pulls
it a distance of 20 m along level
ground. The total weight of sled and
load is 14,700 N. the tractor exerts a
constant 5000-N force at an angle of
36.9o above the horizontal. There is
a 3500 N friction force opposing the
sled’s motion. Find the work done
by each force acting on the sled and
the total work done by all the forces.
WT = 80 kJ
Fx = 500 N
Wf = -70 kJ

Wtot  ( Fx )s
Wtot = 10 kJ
Wtot = (500)(20m)=10 kJ
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Test Your Understanding 6.1
• An electron moves in a straight line toward the east with a
constant speed of 8 x 107 m/s. It has electric, magnetic, and
gravitational forces acting on it. During a 1-m displacement,
the total work done on the electron is
1.Positive
2.Negative
3.Zero
4.Not enough information given to decide.
Answer: iii
Since the electron has constant velocity, its acceleration is zero, and the
net force on the electron is also zero. Therefore the total work done by
all the forces must be zero as well.
Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley
6.2 kinetic energy and work-energy theorem
• The work done by the net force on a particle equals the
change in the particle’s kinetic energy:
v f  vi
2
a
1
1
2
2
Fd  W  mv f  mvi
2
2
2
2d
v f  vi
2
F  ma  m
2
2d
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We can compare the kinetic energy of different bodies
• Kinetic energy and work
have the same units:
• 1 J = 1 N∙m = 1 kg∙m2/s2
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Example 6.3
• Let’s look again at Example 6.2. Suppose the initial speed v1 is
2.0 m/s. What is the speed of the sled after it moves 20 m?
v2 = 4.2 m/s
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Example 6.4 - Forces on a hammerhead
• In a pile driver, a steel hammerhead with mass 200 kg is lifted 3.00 m above the
top of a vertical I-beam being driven into the ground. The hammer is then
dropped, driving the I-beam 7.4 cm farther into the ground. The vertical rails that
guide the hammerhead exert a constant 50 N friction force on the hammerhead.
Use the work-energy theorem to find
a. The speed of the hammerhead just as it hits the I-beam
b. The average force the hammerhead exerts on the I-beam.
• Ignore the effects of the air.
a. v = 7.55 m/s
b. F = 78900 N
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Different objects, different kinetic energies
• Two iceboats hold a race on a
frictionless horizontal lake. The
two iceboats have masses m and
2m. Each iceboat has an identical
sail, so the wind exerts the same
constant force F on each iceboat.
The two iceboats start from rest
and cross the finish line a distance
s away. Which iceboat crosses the
finish line with greater kinetic
energy?
same
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Test Your Understanding 6.2
• Rank the following bodies in order of their kinetic energy,
from least to greatest.
i. A 20.0 kg body moving at 5.0 m/s
ii. A 1.0 kg body that initially was at rest and then had 30 J of
work don on it;
iii. A 1.0 kg body that initially was moving at 4.0 m/s and then
had 30 J of work done on it;
iv. A 20.0 kg body that initially was moving at 10 m/s and then
did 80 J of work on another body.
iv > i > iii > ii
Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley
6.3 Work and energy with varying forces
• Perhaps the best
example is driving
a car, alternating
your attention
between the gas
and the brake.
• The effect is a
variable positive or
negative force of
various magnitude
along a straight
line.
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Area = work
Area = work
The integral represents the area under the curve between x1
and x2.
On a graph of force as a function of position, the total work
done by the force is represented by the area under the curve
between the initial and final positions.
The stretch of a spring and the force that caused it
• The force applied to an
ideal spring will be
proportional to its
stretch.
• The graph of force on
the y axis versus stretch
on the x axis will yield a
slope of k, the spring
constant.
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When spring is compressed, Hooke’s law still holds.
Both force and displacement are negative, so the total
work is still positive.
Example 6.6 Stepping on a scale
• A woman weighing 600 N steps on a bathroom scale containing a
stiff spring. In equilibrium the spring is compressed 1.0 cm under
her weight. Find the force constant of the spring and the total work
done on it during the compression.,
k = 6.0x104 N/m
W = 3.0 J
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Work-Energy Theorem for Straight-Line Motion, Varying Forces
The work-energy theorem, Wtot = K2 – K1, is true even when
the force varies with position.
Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley
Example 6.7 Motion with a varying force
• An air-track glider of mass
0.100 kg is attached to the
end of a horizontal air track
by a spring with force
constant 20.0 N/m. Initially
the spring is unstretched
and the glider is moving at
1.50 m/s to the right. Find
the maximum distance d
that the glider moves to the
right
a) If the air track is turned on
so that there is no friction.
a. 10.6 cm
b) If the air is turned off so that
there is kinetic friction with
coefficient µk = 0.47 b. 8.6 cm
Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley
Work-Energy Theorem for Motion Along a Curve
Wtot = ∆K = K2 – K1 is true in general, no matter what the path
and no matter what the character of the forces.
Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley
Example 6.8 Motion on a curved path
• At a family picnic you are appointed to push your cousin Joe in a
swing. His weight is w, the length of the chains is R, and you push
Joe until the chains make an angle θo with the vertical. To do this,
you exert a varying horizontal force F that starts at zero and
gradually increases just enough so that Joe and the swing move
very slowly and remain very nearly in equilibrium.
a. What is the total work done on Joe by all forces?
b. What is the work done by the tension T in the chains?
c. What is the work you do by exerting the force F?
a. 0
b. 0
c. wR(1-cosθo)
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Test Your Understanding 6.3
• In a conical pendulum, if the speed of the penulum bob remains
constant as it travels around the circle,
a. Over one complete circle, how much work doens the tension force F
do on the bob?
i. A positive amount
ii. A negative amount
iii. Zero
b. Over once complete circle, how much work does the weight do on the
bob?
i. A positive amount
ii. A negative amount
iii. Zero.
a. 0 (iii)
b. 0 (iii)
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6.4 power
Power is the time rate at which work is done.
Like work and energy, power is a scalar
quantity.
Pavg 
W
t
W dW
P  lim

t 0 t
dt
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6.4 Watt about power?
• Once work is calculated, dividing by the
time that passed determines power.
Pavg
W

t
W dW
P  lim

t 0 t
dt
• The pun is credit to James Watt, the
inventor of first modern steam engine.
• Another power unit – horsepower
1 hp = 746 watts
• The energy you use may be noted from
the meter the electric company probably
installed to measure your consumption of
energy in kilowatt-hours.
Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley
• In mechanics we can also express power in terms of force
and velocity.
• Suppose that a force F acts on a body while it undergoes a
vector displacement s. If F// is the component of F tangent
to the path (parallel to ∆s), then the work done by the force
is ∆W = F// ∆s. the average power is
Instantaneous power P is the limit of this expression as
t approaches to zero
Example 6.10
• Each of the two jet engines in a Boeing 767 airliner
develops a thrust (a forward force) of 197,000 N.
When the airplane is flying at 250 m/s, what
horsepower does each engine develop?
66.000 hp
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Example 6.11
• A 50.0 kg marathon runner runs up the stairs to the
top of Chicago’s 443 m tall Sears Tower, the tallest
building the United States. To lift herself to the top
in 15.0 minutes, what must be her average power
output in watts? In kilowatts? In horsepower?
241 W
0.241 kW
0.323 hp
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