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Chapter 6: Work and Energy
Answers and Solutions
1.
Picture the Problem: This is a follow-up question to Guided Example 6.1. An intern pushes a 72-kg patient on a 15-kg
gurney, producing an acceleration of 0.60 m/s2.
Strategy: Use the definition of work to write a ratio that will predict how the amount of work will change if the force is
doubled and the distance is halved.
Solution: 1. Write a ratio for the work:
Wnew Fnew d new (2 Fold ) ( 12 d old )
=
=
= 1 ⇒ Wnew = Wold
Wold
Fold d old
Fold d old
2. We conclude from the ratio that if the force exerted by the intern is doubled, and the distance is halved, the work
done by the intern will remain the same.
Insight: If the intern were to push with twice the force over the same distance, the work would double, and so would
the acceleration.
2.
Picture the Problem: A finch cracks open a seed case by exerting a force over a distance.
Strategy: Use the definition of work to find the work done by the finch.
W = F d = (205 N )(0.0040 m ) = 0.82 J
Solution: Apply the definition of work:
Insight: Work transfers energy into a system. In this case the work done by the finch goes into deformation of the seed
case and some into thermal energy (both the finch’s beak and the seed case get a little warmer).
3.
Picture the Problem: A pumpkin is lifted vertically from the ground.
Strategy: Use the definition of work to find how much work is required to lift the pumpkin.
The required force is the weight of the pumpkin mg.
Solution: Apply the definition of work:
W = F d = (mg ) d
(
= (3.2 kg ) 9.81 m/s
0.80 m
2
)(0.80 m) =
25 J
Insight: A 3.2-kg pumpkin weighs 31 N, or about 7.1 pounds.
4.
Picture the Problem: This is a follow-up question to Guided Example 6.2. A 4900-kg lifeboat slides a distance of
5.0 m on a ramp that makes an angle of 60° with the vertical. The slide does not occur continuously. Instead, the
lifeboat slides halfway to the water, gets stuck for a moment, and then continues to the end of the ramp.
Strategy: Use the definition of work to answer the conceptual question.
Solution: The calculation of work does not depend upon time or whether the slide occurs continuously. It depends only
upon the force and the displacement vectors. Therefore, we conclude that the work done by gravity is exactly the same
as it was in Guided Example 6.2, namely, W = F d cos θ = (48, 000 N )(5.0 m ) cos 60 = 120, 000 J .
Insight: If the lifeboat were to stop and then be winched up the ramp for 1.0 m, the work done by gravity would be
Wup = − (48, 000 N )(1.0 m ) cos 60° = 24, 000 J during the time it is sliding upward. But if it were then released again,
gravity would do a total of 120,000 J of work on the lifeboat, from the time it is released to the time it enters the water.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–1
Chapter 6: Work and Energy
5.
Pearson Physics by James S. Walker
Picture the Problem: A wagon rolls horizontally, pulled by a force that
is directed upward at an angle.
F
θ
Strategy: Use the definition of work, keeping in mind the angle between
the force and the direction of motion.
Solution: Use the definition of work:
d
W = Fd cos θ = (16 N )(12 m ) cos 25o = 174 J = 0.17 kJ
Insight: Only the component of the force along the direction of the motion does any work. The vertical component of
the force reduces the normal force a little.
6.
Picture the Problem: This is a follow-up question to Guided Example 6.4 You push on a drum set to slide it into
position. As you do so, the frictional force acting on the drum set does negative work.
Strategy: Use the definition of work to answer the conceptual question.
Solution: The definition of work, W = F d cos θ , indicates that if the force vector points opposite to the displacement
vector, then θ = 180 and W = − F d . Because the friction force always opposes the direction of motion, we conclude
that yes, friction always does negative work.
Insight: If friction were to do positive work, it would be adding energy to the system. Wouldn’t it be great if friction
were a source of free energy?
7.
Picture the Problem: A pushing force and the force of friction act on a coffee mug
as it slides across a table.
Strategy: Use the definition of work to find the work done by each force. The total
work is the sum of the works done by each force. We need only consider the
horizontal forces because they are parallel to the displacement. The normal force
from the table and the weight of the coffee mug are each perpendicular to the
displacement and they do no work.
Solution: 1. Find the work done by friction:
Wfriction = − f d = − (μ m g ) d
(
d
F
f
)
= − (0.050)(0.12 kg ) 9.81 m/s 2 (0.15 m ) = − 0.00883 J
2. Find the work done by you:
Wapplied = F d = (0.10 N )(0.15 m ) = 0.015 J
3. Sum the works to find the total work:
Wtotal = Wfriction + Wapplied = − 0.00883 + 0.015 J = 0.0062 J
Insight: The fact that a positive work has been done on the coffee mug indicates that it has kinetic energy; that is, it is
still moving. If you stop pushing and the coffee mug comes to rest, the total work done on the mug will be zero.
8.
When the force and displacement vectors are in the same direction, work is calculated as force multiplied by the
displacement.
9.
When the force and displacement vectors are at an angle to each other, work is calculated as the component of the force
that is parallel to the displacement, multiplied by the displacement.
10. No, it is not possible to do work on an object that remains at rest because the displacement will be zero. Because work is
the parallel component of force multiplied by displacement, if the displacement is zero, the work will be zero.
11. (a) Because the friction force always opposes the direction of motion, the work done by friction will always be negative.
Therefore, there are no examples of positive work done by a frictional force. (b) Negative work is done by a frictional
force when a sliding book is brought to rest by friction. Another example is tidal friction, where friction between the
ocean water and the Earth’s crust does work on Earth as the tides go in and out. This removes a tiny bit of Earth’s
rotational kinetic energy and very gradually lengthens the day.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–2
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
12. Picture the Problem: A child lifts a dog in a basket vertically upward into a tree house.
Strategy: Multiply the force by the displacement because the two point in the same direction.
Solution: Calculate the work:
W = F d = (22 N )(4.7 m ) = 103 J = 0.10 kJ
mg
4.7 m
Insight: The applied force equals the weight as long as the basket does not accelerate. This is a
pretty small dog, as 22 N is equivalent to 4.9 lb.
13. Picture the Problem: A suitcase is pushed horizontally at an airport.
Strategy: Work is the force multiplied by the displacement because the two point in the same direction. Solve the work
expression for the force.
W = Fd
Solution: Solve the work expression for F:
⇒F=
W
32 J
=
= 34 N
d 0.95 m
Insight: A longer distance d would require a smaller force if the same amount of work was done.
14. Picture the Problem: A farmhand pushes the hay
horizontally while applying a downward force.
θ F
Strategy: Multiply the parallel component of the force
by the displacement.
Solution: Calculate the work:
3.9 m
W = F d cos θ = (88 N )(3.9 m ) cos 25 = 311 J = 0.31 kJ
Insight: The required force is significant, as 88 N is equivalent to 20 lb of force. Farming is hard work!
15. Picture the Problem: A pushing force and the force of friction act on a large box.
d
Strategy: Use the definition of work to find the work done by each force. The total
work is the sum of the works done by each force. We need only consider the
horizontal forces because they are parallel to the displacement. The normal force
from the floor and the weight of the large box are each perpendicular to the
displacement and they do no work.
Solution: 1. Find the work done by friction:
Wfriction = − f d = − (μ m g ) d
(
f
F
)
= − (0.21)(72 kg ) 9.81 m/s 2 (2.3 m ) = − 341 J
2. Find the work done by the person:
Wapplied = F d = (160 N )(2.3 m ) = 368 J
3. Sum the works to find the total work:
Wtotal = Wfriction + Wapplied = − 341 + 368 J = 27 J
Insight: The fact that a positive work has been done on the box indicates that it has kinetic energy; that is, it is still
moving. If you stop pushing and the box comes to rest, the total work done on the box will be zero.
16. Picture the Problem: A small goldfinch flies with a given speed.
Strategy: Use the definition of kinetic energy to answer the question.
KE = 12 m v 2 =
Solution: Apply the definition of kinetic energy:
1
2
(0.013 kg )(8.5 m/s)2 =
0.47 J
Insight: Note that the mass must be in kilograms for the units to work (no pun intended) out. A joule is a newton
multiplied by a meter, or 1 J = 1 N ⋅ m = 1 kg ⋅ m 2 /s 2 .
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–3
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
17. Picture the Problem: A boat travels along the surface of the water.
Strategy: Solve the definition of kinetic energy for the mass.
KE = 12 m v 2
Solution: Apply the definition of kinetic energy:
m=
2 KE 2 (15, 000 J )
=
= 1200 kg
v2
(5.0 m/s)2
Insight: The boat is small compared to an ocean going ship, but it is larger than your average fishing boat. A mass of
1200 kg is equivalent to 1.3 short tons, and is equivalent to a 22-foot boat with a cabin. An average 16-foot aluminum
fishing boat has a mass of about 400 kg and weighs about 900 lb.
18. Picture the Problem: A baseball has kinetic energy as it flies through the air.
Strategy: Solve the definition of kinetic energy for the speed.
KE = 12 m v 2
Solution: Apply the definition of kinetic energy:
2 KE
=
m
v=
2 (77 J )
0.15 kg
= 32 m/s
Insight: This is a quickly moving baseball, as 32 m/s is equivalent to 72 miles per hour.
19. Picture the Problem: This is a follow-up question to Guided Example 6.7. A student lifts a 4.10-kg box of books
vertically from rest with an upward force of 52.7 N. The distance of the lift is1.60 m.
Strategy: Use the work-energy theorem to answer the conceptual question..
Solution: 1. Guided Example 6.7 showed that the total work done on the box of books is 19.9 J. The work-energy
theorem says the total work done on an object equals the change in its kinetic energy. Therefore, we expect the change
in kinetic energy of the box to be equal to 19.9 J.
ΔKE = KEf − KEi = 12 mvf2 − 0
2. Calculate the change in kinetic energy.
In Guided Example 6.7 we found the final
speed of the box is 3.12 m/s:
=
1
2
(4.10 kg )(3.12 m/s)2 = 19.9 J
Insight: A 4.10-kg box of books weighs 9.0 lb. If you’ve ever helped to move a large number of books, you’ll
recognize that this is a small box of books. The average physics textbook weighs over 4.5 lb!
20. Picture the Problem: An object falls straight down under the influence of gravity.
Strategy: Use the dependence of kinetic energy upon mass and speed to answer parts (a) and (b). The work done by
gravity can be found from the change in the kinetic energy by using the work-energy theorem.
(0.40 kg )(6.0 m/s)2 =
Solution: 1. (a) Apply the definition of kinetic energy:
K = 12 mv 2 =
2. (b) Solve the definition of kinetic energy for speed:
v=
3. (c) Calculate W = Δ KE :
W = Δ KE = KEf − KEi = 25 J − 7.2 J = 18 J
2 KE
=
m
1
2
7.2 J
2 (25 J )
= 11 m/s
0.40 kg
Insight: As an object falls, the work done by gravity increases the kinetic energy of the object.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–4
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
21. Picture the Problem: An Acapulco cliff diver plunges straight downward due to the force of gravity.
Strategy: Solve the expression for gravitational potential energy for the weight of the diver. Let y = 0 correspond to the
surface of the water.
Solution: Solve PEgravity = m g h for mg:
PEgravity
PEgravity = m g h ⇒ mg =
h
=
25, 000 J
= 540 N = 0.54 kN
46 m
Insight: If you set PE = 0 at the top of the cliff, then PE = −25 kJ and y = − 46 m when the diver enters the water.
22. Picture the Problem: This is a follow-up question to Guided Example 6.11. A candy bar called the Mountain Bar has a
nutritional energy content of 8.87×105 J. An 81.0-kg mountain climber eats a Mountain Bar and converts all of its
energy to potential energy, gaining 1120 m in elevation. The mass of the mountain climber is then increased before
repeating the climb.
Strategy: Use the definition of gravitational potential energy to answer the conceptual question, then solve the
PEgravity expression for h.
Solution: 1. If the mass of the mountain climber is increased, more energy is needed to lift her mass. Because the
energy in the candy bar does not change, an increase in the amount of required energy per meter of elevation gain
means that the possible elevation gain will decrease if the mass of the mountain climber is increased.
2. Solve PEgravity = m g h for h:
PEgravity = m g h
PEgravity
h=
mg
=
8.87 × 105 J
(91.0 kg ) (9.81 m/s 2 )
= 994 m
Insight: This is 122 m less than the 1116 m the climber could gain with the same amount of energy and an 81.0 kg
mass. Increasing a mountain climber’s mass either increases the effort required or decreases the achievable altitude.
23. Picture the Problem: A ball is located a certain distance above the floor.
Strategy: Use the expression for gravitational potential energy, taking h = 0 at the floor.
Solution: Use PEgravity = m g h :
(
)
PEgravity = m g h = (0.25 kg ) 9.81 m/s 2 (1.3 m ) = 3.2 J
Insight: You may set PE = 0 wherever it is convenient, but it is often best to set h = 0 at the lowest point of the object’s
motion or, in this case, at the floor.
24. Picture the Problem: A diver is located a certain distance above the water.
Strategy: Solve the PEgravity expression for mass, taking h = 0 at the water level.
Solution: Solve PEgravity = m g h for m:
m=
PEgravity
gh
=
(
1800 J
)
9.81 m/s 2 (3.0 m )
= 61 kg
Insight: You may set PE = 0 wherever it is convenient, but it is often best to set h = 0 at the lowest point of the object’s
motion or, in this case, at the water level.
25. Picture the Problem: A spring with a known spring constant is stretched until it stores a given amount of energy.
Strategy: Solve the PEspring expression for the stretch distance x.
Solution: Solve PEspring = 12 k x 2 for x:
x=
2 PEspring
k
=
2 (0.22 J )
= 0.072 m = 7.2 cm
85 N/m
Insight: If the stretch distance were doubled to 14.4 cm, the energy stored would quadruple to 0.88 J.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–5
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
26. Picture the Problem: A hockey stick can be modeled as a spring that stores energy when it is flexed.
Strategy: Solve the PEspring expression for the spring constant k.
Solution: Solve PEspring = 12 k x 2 for k:
k=
2 PEspring
x2
=
2 (4.2 J )
(0.031 m )2
= 8700 N/m = 8.7 kN/m
Insight: If the stretch distance were doubled to 6.2 cm, the energy stored would quadruple to 16.8 J. That is, it would
store that much energy as long as the stick did not break first!.
27. The kinetic energy is proportional to the square of an object’s speed. Therefore, the kinetic energy of an object will
increase by a factor of 4 if its speed doubles, and increase by a factor of 9 if its speed triples.
28. The work-energy theorem states that the net work done on an object is equal to its change in kinetic energy. If positive
work is done on an object, its change in kinetic energy will be positive, which means that its kinetic energy will
increase.
29. Picture the Problem: A pitcher throws a ball at 40 m/s and the catcher stops it in her glove.
Strategy: Use the work-energy theorem to answer the conceptual question.
Solution: The ball loses kinetic energy from the action of the catcher, so we conclude that the work done on the ball by
the catcher is negative.
Insight: The answers are consistent with the previously described definition of work: The pitcher does positive work
on the ball by exerting a force in the same direction as the motion of the ball. The catcher does negative work on the ball
because the force exerted by the catcher is opposite in direction to the motion of the ball.
30. No, the work required to lift a box 1 meter off the ground is not the same on Earth as it is on the Moon. The box will
weigh about one-sixth as much on the Moon as it does on the Earth, and so six times more force, and six times more
work, is required to lift the box 1 meter on the Earth than on the Moon. That means the change in potential energy is
six times greater when the box is lifted on Earth than when it is lifted on the Moon.
31. The potential energy of a spring is proportional to the square of the stretch distance. Therefore, the potential energy
stored in a spring will increase by a factor of 4 if the stretch distance is doubled.
32. Picture the Problem: A bullet moves at high speed in a straight line.
Strategy: Calculate the kinetic energy according to its definition.
Solution: Use the definition of kinetic energy:
KE = 12 mv 2 =
1
2
(0.00950 kg )(1300 m/s )2 = 8030 J
= 8.03 kJ
Insight: The kinetic energy of the bullet is responsible for the damage it causes upon impact with another object.
33. Picture the Problem: A volleyball has kinetic energy as it flies through the air.
Strategy: Solve the definition of kinetic energy for the speed.
KE = 12 m v 2
Solution: Apply the definition of kinetic energy:
v=
2 KE
=
m
2 (7.8 J )
0.27 kg
= 7.6 m/s
Insight: This speed (about 17 mi/h) would correspond to a soft set during a warm-up drill.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–6
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
34. Picture the Problem: A runner accelerates horizontally and runs in a straight line.
Strategy: The work done equals the change in kinetic energy, according to the work-energy theorem.
Solution: Apply the work-energy theorem:
W = ΔK = 12 mvf2 − 12 mvi2 =
1
2
(73 kg ) ⎡⎣(7.5 m/s)2 − 0⎤⎦ = 2053 J
= 2.1 kJ
Insight: The runner’s kinetic energy comes from the forces his muscles exert on his center of mass over the distance
which his center of mass moves.
35. Picture the Problem: A pinecone is located a certain distance above the ground.
Strategy: Use the expression for gravitational potential energy, taking h = 0 at the ground.
Solution: Use PEgravity = m g h :
(
)
PEgravity = m g h = (0.14 kg ) 9.81 m/s 2 (16 m ) = 22 J
Insight: You may set PE = 0 wherever it is convenient, but it is often best to set h = 0 at the lowest point of the object’s
motion or, in this case, at the ground level.
36. Picture the Problem: A spring with an unknown spring constant is stretched a certain distance with a known amount of
force.
Strategy: First use Hooke’s law to find the spring constant, and then apply the expression for spring potential energy.
F
13 N
=
= 137 N/m
x 0.095 m
Solution: 1. Solve Hooke’s law for k:
F =kx ⇒k=
2. Now find PEspring = 12 k x 2 :
PEspring = 12 k x 2 =
1
2
(137 N/m )(0.095 m )2 =
0.62 J
Insight: The force was not 13 N the entire distance over which the spring was stretched. If it were, the work done on
the spring, and the potential energy stored in the spring, would be W = F d = (13 N )(0.095 m ) = 1.24 J, exactly twice
the actual amount of energy. It can be proved using calculus that this is always the case; that is, the energy stored in a
spring is exactly half the work done if the maximum force had been exerted over the entire distance.
37. Picture the Problem: This is a follow-up question to Guided Example 6.13. A player hits a 0.15-kg baseball over the
outfield fence. The ball leaves the bat with a speed of 36 m/s, and a fan in the bleachers catches it 7.2 m above the point
where it was hit. The question considers how the catch speed of 34 m/s would change if the mass of the ball were
increased while everything else remained the same.
Strategy: Use the principle of the conservation of mechanical energy to determine the dependence of the catch speed
upon the mass of the ball. Although the expression vf = 2 KEf m in step 4 of Guided Example 6.13 appears to show
that the catch speed is inversely proportional to the mass, we must recognize that the ball also would have had more
potential energy if it had greater mass but its trajectory was the same. Going back to the original expression for
conservation of energy will give a complete picture of the effect of changing the mass of the ball.
PEi + KEi = PEf + KEf
Solution: 1. Write an expression for the
conservation of energy and solve it for vf:
mghi + 12 mvi2 = mghf + 12 mvf2
(
mg (hi − hf ) = 12 m vf2 − vi2
2 g (hi − hf ) = v − v
2
f
)
2
i
vi2 + 2 g (hi − hf ) = vf
2. The resulting expression for the catch speed is independent of the mass of the ball. If the initial speed and initial
and final heights are the same, then the final speed of the ball will be the same as previously found, regardless of any
change in the mass of the ball.
Insight: While the kinetic energy of the ball is greater when its mass is greater, so is the required potential energy to
change its height. Both terms vary linearly with m, so changing the mass of the ball has no effect upon vf.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–7
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
38. Picture the Problem: As a tennis ball flies through the air and gains altitude, some of its initial kinetic energy is
converted into gravitational potential energy.
Strategy: Set the mechanical energy just after the bounce equal to the mechanical energy when it is caught. Let the
height be yi = 0 at the bounce, and find yf at the catch.
Solution: Set Ei = Ef and solve for yf :
Ei = Ef
KEi + PEi = KEf + PEf
1
2
mvi2 + 0 = 12 mvf2 + mghf
hf =
(16 m/s) − (12 m/s) = 5.7 m
1 2
vi − vf2 =
2g
2 9.81 m/s 2
(
2
)
2
(
)
Insight: A more massive ball would have more kinetic energy at the start, but would require more energy to change its
height by 5.7 m. The mass cancels out of the expression for the final height.
39. Picture the Problem: A crow drops a clam onto a rocky beach from a large height.
Strategy: The clam’s gravitational potential energy decreases as it falls downward, and its kinetic energy increases, but
its mechanical energy E remains constant. Set the mechanical energy just after the crow releases the clam equal to the
mechanical energy when it is 5.0 m above the ground in order to find its kinetic energy and speed. Assume that the crow
flies with negligible speed so that the clam essentially falls from rest.
Solution: 1. Set Ei = Ef and solve for KEf :
Ei = Ef
KEi + PEi = KEf + PEf
0 + mghi = KEf + mghf
(
)
mg (hi − hf ) = KEf = (0.11 kg ) 9.81 m/s 2 (9.8 − 5.0 m )
= 5.2 J
2. Solve KEf for vf:
1
2
mvf2 = KEf
vf =
2 KEf
=
m
2 (5.2 J )
= 9.7 m/s
0.11 kg
Insight: If the crow were flying with appreciable horizontal speed, the clam would have some initial kinetic energy due
to its horizontal motion, and a little additional kinetic energy at 5.0 m above the ground, because its horizontal motion
would not change. The clam’s change in kinetic energy, however, depends only upon the change in potential energy.
40. If there is no friction, there is no loss of mechanical energy. Any change of potential energy will result in the opposite
change in kinetic energy so that the sum of potential and kinetic energies is always constant. We conclude that if the
potential energy of an object decreases by 10 J, the kinetic energy of the object increases by 10 J.
41. In order for mechanical energy of a system to be conserved there must not be any loss of energy from the system. This
usually occurs by means of friction of some kind, which converts mechanical energy into thermal energy, resulting in a
net loss of mechanical energy.
42. A pole vaulter first converts the chemical energy from the food she ate into kinetic energy as she sprints down the track.
Her kinetic energy is largely converted to a compressional, spring-like potential energy as the pole bends. The pole
straightens out, converting its potential energy into gravitational potential energy. As the vaulter falls, her gravitational
potential energy is converted into kinetic energy, and finally, her kinetic energy is converted to compressional potential
energy as the cushioning pad on the ground is compressed. The pad converts all of the compressional potential energy
into thermal energy by means of friction, so the pole vaulter ends up sitting on the pad at rest, with zero mechanical
energy.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6–8
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
43. (a) The kinetic energy of the ball is a maximum at the instant it leaves your hand, when it has no gravitational potential
energy. It is maximum again at the instant it returns to your hand. (b) The kinetic energy of the ball is zero at the
maximum height, when it momentarily comes to rest. (c) The potential energy of the ball is a maximum at the
maximum height, when its kinetic energy is zero. (d) The potential energy of the ball is zero at the height it is released
from your hand.
44. Picture the Problem: A swimmer descends on a water slide through a vertical height of 2.31 m.
Strategy: As the swimmer descends along the slide her gravitational potential energy is converted into kinetic energy.
Set the loss in gravitational potential energy equal to the gain in kinetic energy by setting her change in mechanical
energy equal to zero, so that ΔE = Ef − Ei = 0 or Ef = Ei . Let h = 0 at the bottom of the slide, v = 0 at the top.
Ebottom = Etop
Solution: Set Ebottom = Etop
and solve for vbottom :
KEbottom + PEbottom = KEtop + PEtop
1
2
2
mvbottom
+ 0 = 0 + mghtop
(
)
vbottom = 2 ghtop = 2 9.81 m/s 2 (2.31 m ) = 6.73 m/s
Insight: If her mass were 40 kg, the swimmer would lose 906 J of potential energy and gain 906 J of kinetic energy.
45. Picture the Problem: An apple falls straight down from rest, accelerating at a rate of 9.81 m/s2.
Strategy: As the apple falls its gravitational potential energy is converted into kinetic energy. The sum of the
gravitational and kinetic energies equals the mechanical energy, which remains constant throughout the fall.
Use these principles to find the kinetic and gravitational potential energies, respectively.
(
)
Solution: 1. Find PEi when hi = 4.0 m:
PEi = mghi = (0.21 kg ) 9.81 m/s 2 (4.0 m ) = 8.24 J = 8.2 J
2. The apple falls from rest so vi = 0 m/s:
KEi = 0
3. The total energy is the sum of KE and PE:
E = KEi + PEi = E = 0 + 8.24 J = 8.2 J
4. Find PEf when hf = 3.0 m:
PEf = mghf = (0.21 kg ) 9.81 m/s 2 (3.0 m ) = 6.18 J = 6.2 J
5. The total energy remains 8.24 J always,
so subtract PE from E to find KE:
KEf = E − PEf = 8.24 − 6.18 J = 2.1 J
(
)
Insight: We bent the rules a bit on significant figures in step 5 to avoid rounding errors. This is especially a problem
when subtracting two large numbers in order to obtain a small number, as in step 5. As the apple falls to the ground,
its mechanical energy begins as entirely gravitational potential energy and ends as entirely kinetic energy.
46. Picture the Problem: A baseball glove rises vertically to height h, stops, and returns to the ground.
Strategy: Gravity does negative work on the glove, reducing its kinetic energy to zero when it reaches height h. Set
the work done by gravity equal to the change in kinetic energy between the ground and h/2. This is an application of
the work-energy theorem.
Solution: 1. Apply the work-energy theorem:
W = Fd = − (mg ) ( 12 h ) = K f − K i
K f = K i − 12 mgh
W = Fd = − (mg ) h = K f − K i = 0 − K i
2. Now set the work done by gravity when
the glove rises to height h equal to the initial
kinetic energy:
K i = mgh
3. Substitute the result into the first equation:
K f = K i − mgh 2 = mgh − mgh 2 = mgh 2 = K i 2 = K 2
Insight: The work done by gravity converts the glove’s kinetic energy into gravitational potential energy, but the total
mechanical energy of the glove remains constant throughout its flight.
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6–9
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
47. Power is the rate at which work is done, not the total amount of work done. Therefore, no, it is not correct to conclude
that engine 1 produces twice as much power as engine 2 unless we also know that engine 1 does twice the work of
engine 2 in the same period of time.
48. Picture the Problem: A runner does work against gravity as he runs up the stairs of the Empire State Building.
Strategy: The power required is the work required to change the runner’s elevation divided by the time elapsed.
The work required to change the elevation of an object is W = mgh.
Solution: Divide the work required by the time:
P=
(
)
2
W mgh (70.0 kg ) 9.81 m/s (1576 steps × 0.20 m/step )
=
=
t
t
(9 min × 60 s/min + 33 s)
P = 380 W × 1 hp 746 W = 0.51 hp
Insight: The energy and power required of the runner is much higher than this because muscles aren’t 100% efficient
at converting food energy into mechanical energy, and the body requires additional energy to stay warm, keep the heart
pumping, and so on.
49. Picture the Problem: This is a follow-up question to Guided Example 6.16. To pass a slow-moving truck, a driver
accelerates his 1300-kg fancy car from 17.9 m/s (40.0 mph) to 22.4 m/s (50.0 mph) by using a power of 3.05 ×104 W.
Strategy: The power required is the work required to change the car’s kinetic energy divided by the time elapsed.
Solve this power expression for the time elapsed.
Solution: Solve the power expression for t:
P=
W 12 mvf2 − 12 mvi2
=
t
t
mvf2 − 12 mvi2
=
t=
P
1
2
1
2
(1300 kg )
(22.4 m/s )2 − (17.9 m/s)2
3.05 × 104 W
t = 3.86 s
Insight: If the engine power remained constant (not a good assumption for an actual car) it would take a little more
time to go from 40.0 mph to 50.0 mph than it did to go from 30.0 mph to 40.0 mph because the change in kinetic
energy would be a little larger. This is because the kinetic energy is proportional to the square of the speed. We saw
in the example that the car’s kinetic energy changes by 9.16×104 J when it accelerates from 30.0 mph to 40.0 mph,
but it changes by 11.8×104 J when it accelerates from 40.0 mph to 50.0 mph.
50. Picture the Problem: A baseball is accelerated along a straight line as it is pitched.
Strategy: The work done on the ball by the pitcher changes the ball’s kinetic energy. The power required to accelerate
the ball is the work divided by the time.
Solution: 1. (a) Apply the
work-energy theorem:
W = 12 mvf2 − 12 mvi2 =
2. (b) Apply the power equation:
P=
1
2
(0.14 kg )(42.5 m/s)2 − 0 = 130 J
W
130 J
=
= 2200 W = 2.9 hp
t
0.060 s
Insight: The power generated by the pitcher is the rate at which the ball gains kinetic energy.
51. Picture the Problem: A load of bricks is lifted vertically upward by a small motor.
Strategy: The power required is the work required to change the elevation divided by the time. The work required to
change the elevation of an object is W = mgh .
Solution: Divide the work required by the time:
P=
W mgh (836 N )(10.7 m )
=
=
= 386 W = 0.517 hp
t
t
(23.2 s)
Insight: A ¾ hp motor would do the trick. Lifting faster than this would require more power.
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6 – 10
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
52. Power is the rate at which work is done. Therefore, if the rate at which work is done on an object is increased, the
power supplied to that object will increase.
53. The power delivered to an object is the applied force multiplied by the speed of the object. The power applied to
system 1 is 10 N of force multiplied by 5 m/s, or 50 watts. The power applied to system 2 is (20 N )(2 m/s ) = 40 W.
We conclude that a greater power is applied to system 1.
54. Power is the rate at which work is done, not the total amount of work done. Therefore, no, it is not correct to conclude
that engine 1 does twice as much work as engine 2 unless we also know that the two engines operate for the same
period of time.
55. Power is the product of force and velocity. Therefore, no, it is not correct to conclude that football player 1 produces
twice as much power as football player 2 unless we also know that the two football players exert their different forces
at the same velocity.
56. Picture the Problem: A bucket is lifted vertically upward from the bottom of a well.
Strategy: The power required is the work divided by the time.
Solution: Divide the work required by the time:
P=
(
)
2
W mgh (5.00 kg ) 9.81 m/s (12 m )
=
=
= 39 W
t
t
(15 s)
Insight: Lifting faster than this would require more power. If the rope’s mass were not ignored it would require
additional power since its center of mass is also lifted. If the force exceeded the weight, the bucket would accelerate.
57. Picture the Problem: A fly does work against gravity as it elevates its center of mass.
Strategy: The power required is the force times the velocity, where the force is just the weight of the fly.
Solution: Calculate the power:
(
)(
)
P = Fv = (mg ) v = 1.4 × 10-3 kg 9.81 m/s 2 (0.023 m/s )
P = 3.2 × 10 −4 W = 0.32 mW
Insight: The energy and power required of the fly is higher than this because it isn’t 100% efficient at converting food
energy into mechanical energy.
58. Picture the Problem: A kayaker paddles horizontally in a straight line at constant speed.
Strategy: The kayaker does positive work on the kayak as she paddles, and the resistive force of the water does
negative work at the same time. The two works are equal because the kinetic energy of the kayak does not change.
Therefore the force she exerts must be equivalent to the resistive force of the water.
P 50.0 W
=
= 33.3 N = 7.5 lb
v 1.50 m/s
Insight: Newton’s second law, F = ma, states that the net force on the kayak is zero because it is not accelerating.
That’s another way of saying that the magnitude of the paddling force equals the magnitude of the friction force.
Solution: Solve P = F v for F:
F=
59. Picture the Problem: A microwave oven delivers energy to an ice cube via electromagnetic waves.
Strategy: The power required is the energy delivered divided by the time elapsed.
W 32, 200 J
=
= 307 s = 5.11 min
P
105 W
Insight: Power can be regarded as the rate of energy transfer because work is essentially transferred energy.
In later chapters we will learn more about melting ice cubes and electromagnetic waves.
Solution: Solve P = W t for t:
t=
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6 – 11
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
60. The statement is false. In fact, only the component of force parallel to the direction of motion does work. If the force is
parallel to the direction of motion, the work is positive and the object gains kinetic energy. If the force is anti-parallel to
the direction of motion, the work is negative and the object loses kinetic energy.
61. In the equation W = F d cos θ the angle θ is the angle between the force vector and the direction of motion.
62. If the force is parallel to the direction of motion ( θ = 0 ), the work is positive Fd and the object gains kinetic energy. If
the force is anti-parallel to the direction of motion ( θ = 180 ), the work is −Fd and the object loses kinetic energy .
63. The statement is true. In order to do work on an object there must be a nonzero component of force parallel to the
direction of motion.
64. Picture the Problem: The International Space Station orbits the Earth in an approximately circular orbit at a certain
height above the Earth’s surface.
Strategy: According to the definition of work, the work done on an object is positive if the force and the direction of
motion are parallel, but zero if the force is perpendicular to the direction of motion.
Solution: The gravitational force exerted by the Earth is radial (toward the center of the circular orbit), but the
displacement is tangential, perpendicular to the force. We conclude that the work done by the Earth on the space
station is zero.
Insight: If the Earth were to do any work on the space station, its energy would change. Of course the rockets that lifted
the parts of the space station into space did do work on them, increasing their energy sufficiently to achieve orbit.
65. Picture the Problem: A pendulum bob swings from point A to point B
along the circular arc indicated in the figure at right.
Strategy: Apply the definition of work, which says that the work done on
an object is positive if the force and the displacement are along the same
direction, but zero if the force is perpendicular to the displacement.
Solution: 1. (a) As the pendulum bob swings from point A to point B, the
force of gravity points downward and a component of the displacement is
downward. Therefore, the work done on the bob by gravity is positive.
A
C
B
2. (b) As the pendulum bob swings, the force exerted by the string is radial (toward the pivot point) but the
displacement is tangential, perpendicular to the force. We conclude that the work done on the bob by the string is zero.
Insight: Because the work done by the string force is zero, the net work is positive and the bob gains kinetic energy as
it swings from A to B, consistent with the work-energy theorem.
66. Picture the Problem: A pendulum bob swings from point B to point C
along the circular arc indicated in the figure at right.
Strategy: Apply the definition of work, which says that the work done on
an object is positive if the force and the displacement are along the same
direction, but zero if the force is perpendicular to the displacement.
Solution: 1. (a) As the pendulum bob swings from point B to point C, the
force of gravity points downward and a component of the displacement is
upward. Therefore, the work done on the bob by gravity is negative.
C
A
B
2. (b) As the pendulum bob swings, the force exerted by the string is radial (toward the pivot point) but the
displacement is tangential, perpendicular to the force. We conclude that the work done on the bob by the string is zero.
Insight: Because the work done by the string force is zero, the net work is negative and the bob loses kinetic energy as
it swings from B to C, consistent with the work-energy theorem.
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6 – 12
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
67. Picture the Problem: A bone is lifted vertically by a dog.
Strategy: Work is the force multiplied by the displacement because the two point in the same direction. The force the
dog exerts is the weight mg of the bone.
(
)
W = F d = mg d = (0.75 kg ) 9.81 m/s 2 (0.11 m ) = 0.81 J
Solution: Apply the definition of work:
Insight: The heavier the bone, the more work the dog must do in order to lift it, and the more gravitational potential
energy would be possessed by the bone.
68. Picture the Problem: A weight is lifted vertically by a weightlifter.
Strategy: Work is the force multiplied by the displacement because the two point in the same direction. Solve the
definition of work for the force.
W = Fd
Solution: Solve the work expression for F:
⇒F=
W
9.8 J
=
= 82 N
d 0.12 m
Insight: The 82 N weight is equivalent to about 18 pounds.
69. Picture the Problem: You lift a pail of water vertically.
Strategy: Work is the force multiplied by the displacement because the two point in the same direction. The force you
exert is the weight mg.
Solution: Solve the work expression for d:
W = Fd
⇒d =
W 13 J
=
= 0.37 m
F 35 N
Insight: A larger force F would do the same amount of work in a shorter distance.
70. Picture the Problem: You push a book horizontally across a desk with a force that is directed at a small angle below
the horizontal.
Strategy: Work is the parallel component of force F cos θ multiplied by the displacement.
Solution: Apply the definition of work:
W = F d cos θ = (5.2 N )(0.45 m ) cos 21° = 2.2 J
Insight: If the 5.2-N force were entirely directed horizontally, you would have done work equal to
W = (5.2 N )(0.45 m ) cos 0° = 2.3J.
71. Picture the Problem: A paint can is lifted vertically.
Strategy: Multiply the force by the distance because the two vectors point in the same direction in
part (a). In part (b) the distance traveled is zero, and in part (c) the force and distance are
antiparallel.
Solution: 1. (a) Apply the definition of work:
W = Fd = mgd
(
mg
1.8 m
)
W = (3.4 kg ) 9.81 m/s 2 (1.8 m ) = 60 J
2. (b) Now the distance traveled is zero:
W= 0
3. (c) Now the force and distance are antiparallel:
W = − mgd = − 60 J = − 0.060 kJ
Insight: The applied force equals the weight as long as the paint can does not accelerate. The can gains gravitational
potential energy as it is lifted and loses gravitational potential energy as it is lowered.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6 – 13
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
72. Picture the Problem: A water skier is pulled horizontally.
Strategy: Multiply the force by the distance because the two point along the same direction.
Solution: 1. (a) The work is positive because the force is along the direction of motion (θ= 0°).
W = Fd = (120 N )(65 m ) = 7800 J = 7.8 kJ
2. (b) Apply the definition of work:
Insight: While the work done by the rope is positive, the work done by friction is negative, so as long as the skier
moves at constant speed, she does not gain or lose kinetic energy.
73. Picture the Problem: A water skier is pulled horizontally.
Strategy: Take the negative of the force times the distance because in this case the two point in opposite directions;
the force exerted by the rope on the boat is backward while the boat travels forward.
Solution: 1. (a) The work is negative because the force on the boat is opposite the direction of motion (θ= 180°).
W = Fd cos θ = (120 N )(65 m ) cos180o = −7800 J = −7.8 kJ
2. (b) Apply the definition of work:
Insight: While the works done on the boat by the rope and by friction are negative, the work done by the motor is
positive. As long as the boat moves at constant speed it doesn’t gain or lose kinetic energy.
74. Picture the Problem: A wagon rolls horizontally but is pulled by a force
that is directed upward at an angle.
F
θ
Strategy: Use the definition of work, keeping in mind the angle between
the force and the direction of motion.
d
W = Fd cos θ = (16 N )(12 m ) cos 25o = 174 J = 0.17 kJ
Solution: Apply the definition of work:
Insight: Only the component of the force along the direction of the motion does any work. The vertical component of
the force reduces the normal force a little.
75. Picture the Problem: A packing crate slides horizontally but is pulled
by a force that is directed upward at an angle.
F
Strategy: Use the definition of work, keeping in mind the angle
between the force and the direction of motion.
d
W = Fd cos θ = (120 N )(18 m ) cos 43° = 1580 J = 1.6 kJ
Solution: Apply the definition of work:
Insight: Only the component of the rope force along the direction of the motion does any work. The vertical component
of the force reduces the normal force a little. If the crate is pulled at constant speed, it does not gain or lose kinetic
energy and we can conclude the friction does −1.6 kJ of work on the crate.
76. Picture the Problem: A boat and skier are both moving toward
the right, but the rope pulls the skier at an angle.
Strategy: Solve the definition of work equation for the angle
between the force and the direction of motion.
Solution: 1. Solve the definition of work for the angle:
W = Fd cos θ ⇒
2. Calculate the angle:
θ = cos −1 ⎜
cos θ =
W
Fd
⎡
⎤
2500 J
⎛W ⎞
= cos −1 ⎢
⎥ = 37
⎟
⎝ Fd ⎠
⎣ (75 N )(42 m ) ⎦
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6 – 14
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
Insight: Only the component of the force along the direction of the motion does any work. The work the boat does on
the skier is balanced by the negative work friction does on the skier, so that the kinetic energy of the skier is constant.
77. The kinetic energy of the package is not increasing because the elevator rises with constant speed. The package is
gaining elevation, however. The work done on the package by the elevator floor increases its gravitational potential
energy, not its kinetic energy. The work-energy theorem still holds because the two forces acting on the package, the
elevator floor force and gravity, do equal and opposite amounts of work, producing no change in kinetic energy.
78. No. Because it is not accelerating we can conclude that the net force acting on the object is zero, but there could be
several forces acting on the object in such a way that the net work done on the object by all the forces is zero. In this
way Newton’s second law (the net force must be zero because it is not accelerating) agrees with the work-energy
theorem (the net work must be zero because it does not gain or lose kinetic energy).
79. Picture the Problem: Four joggers have a variety of masses and speeds.
Strategy: Use the definition of kinetic energy to determine the relative magnitudes of the kinetic energies.
Solution: 1. Calculate the kinetic energies of each jogger.
2. K A =
1
2
(m)(v )2 = 12 mv 2
4. K C =
1
2
(3m) ( 12 v )
2
=
3
4
(
1
2
mv 2
)
3. K B =
1
2
( m) (3v)
5. K D =
1
2
(4m) ( 12 v )
1
2
2
2
=
9
2
(
1
2
mv 2
)
= 12 mv 2
6. By comparing the magnitudes of the kinetic energies we arrive at the ranking C < A = D < B.
Insight: Even with three times the mass of jogger A, jogger C has only three-fourths the kinetic energy because the
kinetic energy is proportional to the square of the speed.
80. No. Mass is always positive, and even if a negative velocity were included in the calculation, it is squared because
KE = 12 mv 2 . We conclude that the kinetic energy is always a positive quantity.
81. Picture the Problem: The work W accelerates a car from 0 to 50 km/h.
Strategy: Use the work-energy theorem to answer the conceptual question.
Solution: 1. (a) The work done on the car is equal to its change in kinetic energy, which in turn is proportional to the
square of its velocity. Therefore, if the kinetic energy of the car is W when its speed is 50 km/h, its kinetic energy will
be 9W when its speed is 150 km/h. We conclude that the work required to accelerate the car from 50 km/h to 150 km/h
is 9W − W = 8W.
2. (b) The best explanation is B. The final speed is three times the speed that was produced by the work W.
Statement A is false because the kinetic energy depends on the speed squared, and the work is proportional to the
change in kinetic energy. Statement C is true but irrelevant; KE is calculated from speed, not change in speed.
Insight: By using a similar analysis you can show that a car that is traveling 35 mph has twice the kinetic energy as a
car that is traveling 25 mph. In other words, it takes just as much work to increase the car’s speed from 25 to 35 mph
as it did to accelerate from 0 to 25 mph. The extra 10 mph makes a big difference!
82. Picture the Problem: Two identical balls are released from the same height, but ball 1 is dropped from rest while ball 2
is thrown downward with an initial velocity.
Strategy: Apply the concept of gravitational potential energy to answer the conceptual question.
Solution: 1. (a) Changes in gravitational potential energy mgh depend only upon any changes in an object’s weight mg
or changes in its altitude h. Gravitational potential energy does not depend upon speed. We conclude that in this case
the change in gravitational potential energy of ball 1 is equal to the change in gravitational potential energy of ball 2.
2. (b) The best explanation is B. The gravitational potential energy depends only on the mass of the ball and its initial
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6 – 15
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
height above the ground. Statement A is partly true (ball 1 does have greater total energy than ball 2) but the amount
of gravitational potential energy that is converted into kinetic energy is determined by m and h. Statement C is true but
irrelevant.
Insight: If ball 1 were thrown upward it would convert some of its initial kinetic energy into gravitational potential
energy. However, the net change in gravitational potential energy from the release point to the landing point would
still remain the same as for ball 2.
83. Picture the Problem: A pendulum bob swings from point A to point B to
point C along the circular arc indicated in the figure at right.
Strategy: Apply the definition of gravitational potential energy, which
says that the energy depends only upon the weight mg of an object and its
altitude h above some reference level where h = 0.
Solution: 1. (a) The height h of the pendulum bob at point C is the same as
its height at point A. We conclude that the gravitational potential energy at
point C is equal to the gravitational potential energy at point A.
A
C
B
2. (b) The height h of the pendulum bob at point C is greater than its height at point B. We conclude that the
gravitational potential energy at point C is greater than the gravitational potential energy at point B.
Insight: The work done by the Earth’s gravity is positive if the bob swings from point A to point B because a component
of the displacement is downward and the force is downward. Because the work done by the string force is zero, the net
work is positive and the bob gains kinetic energy as it swings from A to B, consistent with the work-energy theorem.
We can also say that it converts its gravitational potential energy into kinetic energy as it swings from A to B.
84. The potential energy of a compressed spring is also positive because the force required to compress it is along the same
direction as the compression distance. That means the work done on the spring is positive and the spring stores potential
energy. This is true whether the spring is stretched or compressed.
85. Picture the Problem: A fragment of Skylab moves at high speed in a straight line.
Strategy: Calculate the kinetic energy of the fragment using its definition.
Solution: Use the definition of KE:
KE = 12 mv 2 =
1
2
(1770 kg )(120 m/s)2 = 1.27 × 107
J = 12.7 MJ
Insight: The energy came from the work the rocket motor did in order to place Skylab into orbit. This work gave
Skylab a combination of kinetic energy and gravitational potential energy.
86. Picture the Problem: A bowling ball is lifted to a shelf above the floor.
Strategy: Calculate the gravitational potential energy of the bowling ball using its definition.
Solution: Use the definition of PE:
(
)
PEgravity = mgh = (7.3 kg ) 9.81 m/s 2 (1.7 m ) = 122 J = 0.12 kJ
Insight: The net work done on the bowling ball is zero, because the work done by gravity is −122 J. Thus the bowling
ball gains gravitational potential energy but not kinetic energy, consistent with the work-energy theorem.
87. Picture the Problem: A baseball has kinetic energy as it moves through the air.
Strategy: Use the definition of kinetic energy to calculate the ball’s speed.
Solution: Solve the definition of KE for v:
KE = 12 mv 2
v=
⇒ 2 KE m = v 2
2 KE
=
m
2 (18 J )
0.15 kg
= 15 m/s
Insight: The ball is moving pretty quickly, as this speed is equivalent to 35 mi/h.
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6 – 16
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
88. Picture the Problem: A bird sits in a tree an appreciable distance above the ground.
Strategy: Use the definition of gravitational potential energy to find the bird’s height.
PEgravity = mgh
Solution: Use the definition of PE:
h=
PEgravity
mg
=
6.6 J
(0.12 kg ) (9.81 m/s 2 )
= 5.6 m
Insight: If it fell from the tree the bird would gain 6.6 J of kinetic energy and hit the ground at 10.5 m/s (23.5 mi/h!)
unless it spread its wings and flew.
89. Picture the Problem: A spring is compressed by an applied force.
Strategy: Calculate the spring potential energy by using its definition.
Solution: Use the definition of spring PE:
PEspring = 12 k x 2 =
1
2
(92 N/m )(0.028 m )2 = 0.036 J
= 36 mJ
Insight: If this potential energy were converted to the kinetic energy of a 0.080-kg pinball, it would travel at 0.95 m/s.
You’ll need a stiffer spring than this one in order to build a pinball launcher that can produce some excitement!
90. Picture the Problem: A spring is stretched by an applied force, and then it is compressed.
Strategy: Use the stretch force and distance to find the spring constant of the spring, and then calculate the spring
potential energy by using its definition.
F
27 N
=
= 614 N/m
x 0.044 m
Solution: 1. Find the spring constant:
F =kx ⇒k=
2. Use the definition of spring PE:
PEspring = 12 k x 2 =
1
2
(614 N/m)(0.035 m )2 =
0.38 J
Insight: It does not matter whether the spring is stretched or compressed; in both cases it stores positive potential
energy.
91. Picture the Problem: A spring is stretched by an applied force.
Strategy: Use the definition of spring potential energy to find the spring constant.
Solution: Solve the definition of spring PE for k:
PEspring = 12 k x 2
k=
2 PEspring
x
2
=
2 (0.053 J )
(0.026 m )2
= 157 N/m = 0.16 kN/m
Insight: A spring with a larger spring constant could store the same amount of energy with a smaller stretch distance.
92. Picture the Problem: A car slows down as it rolls horizontally a distance of 32 m through a sandy stretch of road.
Strategy: The kinetic energy of the car is reduced by the amount of work done by friction. The work done is the force
times the distance, so once we know the work done and the distance, we can find the force.
Solution: 1. (a) The net work done on the car must have been negative because the kinetic energy decreased.
2. (b) The work done by friction equals the
average force of friction times the distance the
car traveled. Use the work-energy theorem to
find the magnitude of the force:
W = − Fd = ΔKE = 12 mvf2 − 12 mvi2 so that F = −
F=−
1
2
(
(1100 kg ) 122 − 192 m 2 /s 2
32 m
1
2
(
m vf2 − vi2
)
d
) = 3700 N = 3.7 kN
Insight: Kinetic friction always does negative work because the force is always opposite to the direction of motion.
The vector force exerted on the car would be −3.7 kN if the vector displacement is taken to be +32.0 m.
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6 – 17
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
93. Picture the Problem: A bicycle rolls horizontally on level ground, slows down, and comes to a stop.
Strategy: According to the work-energy theorem, the work done by the brakes equals the change in the kinetic energy
of the bicycle. In part (b) use the definition of work to find the magnitude of the braking force.
Solution: 1. (a) Calculate W = ΔKE :
W = ΔKE = 12 mvf2 − 12 mvi2 = 0 − 12 (65 + 8.8 kg )(14 m/s )
2
= −7200 J = −7.2 kJ
W
W
7200 J
= 2100 N = 2.1 kN ≅ 460 lb
d
Δx 3.5 m
Insight: Kinetic friction always does negative work because the force is always opposite to the direction of motion.
The large force is required to bring the bicycle to a rather sudden stop from 14 m/s (31 mph) in only 3.5 m (11 ft).
2. (b) The force can be found
from the definition of work:
F=
=
=
94. Picture the Problem: A runner slides horizontally on level ground over a distance of 3.4 m and comes to rest.
Strategy: The work done by friction equals the negative of the kinetic energy the runner had just before the slide.
It also equals the force exerted by friction multiplied by the distance of the slide.
W = ΔKE = KEf − KEi = 0 − 12 mvi2
Solution: 1. (a) Calculate W = ΔKE :
W = − 12 (62 kg )(4.5 m/s ) = −628 J = − 0.63 kJ
2
W = − Fd = − (μ k mg ) d so that μ k = −
2. (b) The work done by friction equals
the average force of friction multiplied
by the distance the player slid:
μk = −
(−587 J )
(62.0 kg ) (9.81 m/s2 ) (3.40 m)
W
mgd
= 0.284
Insight: Kinetic friction always does negative work because the force is always opposite to the direction of motion.
95. Picture the Problem: An object moves along a straight line with known speeds and energies..
Strategy: Use the definition of kinetic energy, together with the initial speed and kinetic energy, to determine the mass
of the object. Use that mass to find the kinetic energy of the object at t = 5.0 s, and then use the work-energy theorem to
find the work done on the object between t = 0 s and t = 5.0 s.
Solution: 1. (a) Solve the definition of KE for m:
2 (14 J )
2 KE
KE = 12 mv 2 ⇒ m = 2 =
= 2.3 kg
v
(3.5 m/s )2
2. (b) Find the KE at t = 5.0 s:
KE = 12 mv 2 =
1
2
(2.3 kg )(4.7 m/s )2 =
25 J
3. (c) Calculate W = ΔKE :
W = ΔKE = KEf − KEi = 25 J − 14 J = 11 J
Insight: The work done by a force caused the object to gain kinetic energy. The work-energy theorem is one of the
most important and fundamental results in physics. It is also a very handy tool for problem solving.
96. Picture the Problem: A pendulum bob swings from point A to point B, losing altitude
and thus losing gravitational potential energy. See the figure at right.
Strategy: Use the geometry of the problem to find the change in altitude h of the
pendulum bob, and then find its change in gravitational potential energy.
Solution: 1. Find the height h of
the pendulum bob at point A:
h = L − L cos θ = L (1 − cos θ )
= (1.2 m )(1 − cos 35 ) = 0.217 m
h
2. Use h to find the change in
gravitational potential energy:
ΔPE = PEB − PEA = 0 − mgh
(
)
= 0 − (0.33 kg ) 9.81 m/s 2 (0.217 m )
ΔPE = − 0.70 J
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6 – 18
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
Insight: Note that Δ h is negative because the pendulum swings from A to B. Likewise, Δ h is positive and the
pendulum gains gravitational potential energy if it swings from B to A.
97. Picture the Problem: An object moves with no friction or air resistance.
Strategy: The sum of the gravitational and kinetic energies equals the mechanical energy, which remains constant
whenever an object moves without friction or air resistance. Use these principles to find the kinetic and gravitational
potential energies, respectively.
Solution: 1. Find the total mechanical energy:
E = KEi + PEi = 10 + 20 J = 30 J
2. Subtract the PE to find KE:
KE = E − PE = 30 − 15 J = 15 J
3. Subtract KE to find PE:
PE = E − KE = 30 − 5 J = 25 J
Insight: The maximum possible value of either the kinetic energy or the potential energy is 30 J.
98. Picture the Problem: An object moves with no friction or air resistance.
Strategy: The sum of the gravitational and kinetic energies equals the mechanical energy, which remains constant
whenever an object moves without friction or air resistance. The maximum possible value for either the kinetic or
potential energy, then, is the value of the total mechanical energy.
Solution: 1. Find the total mechanical energy:
E = KEi + PEi = 10 + 30 J = 40 J
2. The KE is maximum when PE = 0:
KE = E − PE = 40 − 0 J = 40 J
3. The PE is maximum when KE = 0:
PE = E − KE = 40 − 0 J = 40 J
Insight: If the initial potential energy is reduced to 15 J, the kinetic energy would increase to 25 J, ensuring that the
total mechanical energy remained 40 J.
99. Picture the Problem: Two identical balls are released from the same height. You throw a ball upward and let it fall to
the ground, but your friend drops her ball straight down to the ground.
Strategy: Apply the concept of conservation of energy to answer the conceptual question.
Solution: 1. (a) For a ball in freefall the gravitational potential energy is converted into kinetic energy. If the ball is
moving upward its kinetic energy is converted into gravitational potential energy. Therefore, any change in the kinetic
energy of the ball must come at the expense of its gravitational potential energy (as long as there are no other forces on
the ball). The gravitational potential energies of the two balls change by the same amount. We conclude that the change
in kinetic energy of your ball is equal to the change in kinetic energy of your friend’s ball.
2. (b) The best explanation is C. The change in gravitational potential energy is the same for each ball, which means
that the change in kinetic energy must also be the same. Statement A is true but fails to compare the two balls, and
statement B is true but the changes in kinetic and gravitational potential energies are independent of the time elapsed.
Insight: If you were to throw your ball downward it would begin and finish with more kinetic energy than your friend’s
ball, but the net change in gravitational potential energy (and kinetic energy) from the release point to the landing point
would still remain the same for each ball.
100. Picture the Problem: Three balls are thrown upward with the same
initial speed, but at different angles relative to the horizontal.
Strategy: Use the concept of the conservation of mechanical energy
to evaluate the statements. Note that when each of the three balls
increases its altitude by the same amount, they each convert the same
amount of kinetic energy into gravitational potential energy.
Solution: 1. (a) “Ball 3 has the lowest speed” is incorrect because all three balls will have the same kinetic energy
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6 – 19
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
(if they have the same mass) and the same speed at the dashed level. Ball 1’s velocity is vertical and ball 3’s velocity is
horizontal, but they have the same speed.
2. (b) “Ball 1 has the lowest speed” is incorrect because all three balls will have the same speed at the dashed level.
3. (c) “All three balls have the same speed” is correct because all three balls will have the same kinetic energy
(if they have the same mass), having had identical kinetic energies at the launch and having each converted
the same amount of kinetic energy into gravitational potential energy.
4. (d) “The speed of the balls depends on their mass” is incorrect because both gravitational potential energy and kinetic
energy depend linearly upon mass. The above statements were evaluated while supposing the masses were the same, but
even if they were different, the speeds would remain the same. A more massive ball would begin with a larger kinetic
energy and convert a larger amount into gravitational potential energy.
Insight: You can use a mathematical expression to support the claim in part (d). Set the mechanical energies at the
launch and at the dashed level (at altitude h) equal to each other: 12 mvi2 + 0 = 12 mv 2f + mgh . Now solve this expression
for v f to discover that v f = vi2 + 2 gh , independent of the mass of the ball.
101. Picture the Problem: As the ball flies through the air and gains altitude
some of its initial kinetic energy is converted into gravitational potential
energy.
Strategy: Set the mechanical energy at the start of the throw equal to the
mechanical energy at its highest point. Let the height be hi = 0 at the start
of the throw, and find hf at the highest point.
Solution: Set Ei = Ef and solve for yf :
Ei = Ef
KEi + PEi = KEf + PEf
1
2
mvi2 + 0 = 12 mvf2 + mghf
(8.30 m/s ) − (7.10 m/s) = 0.942 m
1 2
hf =
vi − vf2 =
2g
2 9.81 m/s 2
(
2
)
2
(
)
Insight: A more massive ball would have more kinetic energy at the start, but would require more energy to change
its height by 0.942 m, so the mass cancels out.
102. Picture the Problem: The rock falls straight down from rest, accelerating at a rate of 9.81 m/s2.
Strategy: The mechanical energy of the rock is initially stored as gravitational potential energy, but it is converted
into kinetic energy as the rock falls. Set the mechanical energy equal to zero initially, so that hi = 0 and vi = 0 .
The gravitational potential energy will decrease and have a negative value, but the kinetic energy will increase and
have a positive value, so that the sum of the two will remain zero. The mechanical energy remains zero throughout.
Solution: 1. (a) The initial kinetic energy
is zero because the rock falls from rest:
KEi a = 12 mvi2a = 0 J
2. Set Ef = Ei after the rock falls to
yfa = −2.00 m:
Ef a = Ei a = 0 = KEf a + PEf a = KEf a + mghf a
3. Calculate Δ K :
ΔKE = KEf a − KEi a = 113 − 0 J = 113 J
4. (b) The initial kinetic energy is 113 J:
KEi b = 113 J
(
)
KEf a = − mghf a = − (5.76 kg ) 9.81 m/s 2 (−2.00 m ) = 113 J
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6 – 20
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
KEi b + PEi b = KEf b + PEf b
5. Set Ef = Ei after the rock falls to
hf b = − 4.00 m:
KEi b + mghi b = KEf b + mghfb
(
KEf b = K i b + mg hi b − hf b
(
)
)
= 113 J + (5.76 kg ) 9.81 m/s 2 ⎡−
⎣ 2.00 m − (− 4.00 m )⎤⎦
= 226 J
6. Calculate Δ KE :
ΔKE = KEf b − KEi b = 226 − 113 J = 113 J
Insight: You can set the potential energy to any value you wish. For instance, we could claim that hi a = 12.00 m,
hf a = 10.00 m, and hf b = 8.00 m and the results would be exactly the same. It is the change in the potential energy,
not its particular value, that matters in physics!
103. Picture the Problem: The trajectory of the rock is depicted at right.
Strategy: The rock starts at height hi, rises to hmax , comes briefly to rest, then falls down to the base of the cliff at
h = 0. Set the mechanical energy at the point of release equal to the mechanical energy at the base of the cliff and at the
maximum height hmax in order to find vi and hmax .
Solution: 1. (a) Set Ei = Ef
and solve for vi :
KEi + PEi = KEf + PEf
1
2
mvi2 + mghi = 12 mvf2 + 0
vi = vf 2 − 2 ghi
=
vi
(29 m/s )2 − 2 (9.81 m/s 2 ) (32 m )
ymax
vi = 15 m/s
2. (b) Now set Eymax = Ef
KEh-max + PEh-max = KEf + PEf
and solve for hmax :
2
f
0 + mghmax = mv + 0 ⇒ hmax
1
2
=
h
v2
= f
2g
(29 m/s)2
(
2 9.81 m/s 2
)
= 43 m
Insight: In part (a) the initial energy is a combination of potential and kinetic, but becomes all kinetic just before impact
with the ground. In part (b) the rock at the peak of its flight has zero kinetic energy; all of its energy is potential energy.
104. Picture the Problem: A block slides without friction and then
compresses a spring, as shown at the right.
Strategy: Because there is no friction, we can set the initial mechanical
energy (when the block slides freely) equal to the final mechanical energy
(when the spring is fully compressed). Use the resulting equation to find
the compression distance d of the spring.
Solution: Set Ei = Ef
and solve for d:
KEi + PEi = KEf + PEf
1
2
mvi2 + 0 = 0 + 12 k d 2
mvi2
=d=
k
(3.7 kg )(2.2 m/s)2
3200 N/m
= 0.075 m = 7.5 cm
Insight: In this situation the kinetic energy of the sliding block is converted into spring potential energy.
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6 – 21
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
105. Picture the Problem: A compressed spring pushes a block, which then
slides without friction, as shown at the right.
Strategy: Because there is no friction, we can set the initial mechanical
energy (when the spring is fully compressed) equal to the final
mechanical energy (when the block slides without friction). Use the
resulting equation to find the compression distance d of the spring.
Solution: Set Ei = Ef
and solve for vf:
KEi + PEi = KEf + PEf
0 + 12 k d 2 = 12 mvf2 + 0
k d2
= vf =
m
(1400 N/m)(0.042 m )2
1.3 kg
= 1.4 m/s
Insight: In this situation the spring potential energy is converted into the kinetic energy of the sliding block.
106. Picture the Problem: A pendulum bob swings from point B to point A (see the figure
at right) and gains altitude and thus gravitational potential energy.
Strategy: Use the geometry of the problem to find the change in altitude h of the
pendulum bob, and then find its change in gravitational potential energy mgh. Apply
conservation of energy between points B and A to find the speed at A.
Solution: 1. (a) Find the height
change h of the pendulum bob:
2. Use h to find Δ PE :
h = L − L cos θ = L (1 − cos θ )
h
Δ PE = mgh = mgL (1 − cos θ )
(
)
= (0.33 kg ) 9.81 m/s 2 (1.2 m )(1 − cos 35° )
Δ PE = 0.70 J
3. (b) Set EB = EA and solve for vA :
KEB + PEB = KEA + PEA
1
2
mvB2 = 12 mvA2 + Δ PE
vA = vB2 −
2 Δ PE
=
m
(2.4 m/s)2 −
2 (0.70 J )
= 1.2 m/s
0.33 kg
Insight: Note that Δ PE m = mgh m = gh independent of mass. That means the speed vA is independent of mass.
107. Picture the Problem: A pendulum bob swings from point B to point A (see the figure
at right) and gains altitude and thus gravitational potential energy.
Strategy: Use its definition to find the kinetic energy of the bob at point B. Use the
geometry of the problem to find the maximum change in altitude hmax of the pendulum
bob, and then use PE = mgh to find its maximum change in gravitational potential
energy. Apply conservation of energy between points B and the endpoint of its travel
to find the maximum angle θ max the string makes with the vertical.
Solution: 1. (a) Use the definition
of kinetic energy to find KEB :
KEB = 12 mvB2 =
1
2
(0.33 kg )(2.4 m/s)2 =
0.95 J
h
2. (b)Mechanical energy is conserved because there is no friction. If we take the potential
energy at point B to be zero, we can say that all of the bob’s kinetic energy will become
potential energy when the bob reaches its maximum height and comes momentarily to
rest. Therefore the change in potential energy between point B and the point where the
bob comes to rest is 0.95 J.
3. (c) Find the height
change hmax of the pendulum bob:
hmax = L − L cos θ max = L (1 − cos θ max )
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6 – 22
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
ΔPE = mghmax = mgL (1 − cos θ max )
4. Use PE = mgh and the result
of part (b) to solve for θ max :
⎛
⎝
θ max = cos −1 ⎜1 −
⎡
⎤
0.95 J
ΔPE ⎞
−1
⎢
⎥
cos
1
=
−
= 41°
2
mgL ⎟⎠
⎢⎣ (0.33 kg ) 9.81 m/s (1.2 m ) ⎥⎦
(
)
Insight: The pendulum bob cannot swing any farther than 41° because there is not enough energy available to raise the
mass to a higher elevation.
108. Picture the Problem: The motions of the masses in the device are
depicted in the figure at right:
Strategy: Mechanical energy is conserved because there is no
friction. Set Ei = Ef and solve for vf . The speeds of each mass
must always be the same because they are connected by a rope.
Solution:
1. (a) Set
KEi + PEi = KEf + PEf
Ei = Ef and
0 + 0 = 12 m1vf2 + 12 m2 vf2 + m1 gy1 + m2 gy2
solve for vf :
0 = 12 (m1 + m2 ) vf2 + m1 gh + m2 g (− h )
(m2 − m1 ) gh = 12 (m1 + m2 ) vf 2
vf =
2. (b) Use the expression
from part (a) to find vf :
⎛ m − m1 ⎞
2 gh ⎜ 2
⎝ m1 + m2 ⎟⎠
⎛ 4.1 kg − 3.7 kg ⎞
vf = 2 9.81 m/s 2 (1.2 m ) ⎜
= 1.1 m/s
⎝ 3.7 kg + 4.1 kg ⎠⎟
(
)
Insight: The mass m2 loses more gravitational potential energy than m1 gains, so there is extra energy available to give
the system kinetic energy. We bent the rules for significant figures a bit in step 2 because by the rules of subtraction,
4.1 − 3.7 kg = 0.4 kg, only one significant figure.
109. The power produced by an engine equals the work done divided by the amount of time during which the work is
performed. The time required to do work W is therefore T = W / P. If engine 2 can produce half the power of engine 1,
it will take twice as much time to do the same work: T2 = W ( 12 P ) = 2W P = 2T .
110. The power produced by an engine equals the work done divided by the amount of time during which the work is
performed. The time required to do work W is therefore T = W / P. If engine 2 can produce half the power of engine 1,
it will take twice as much time to do the same work, and six times longer to do three times the work:
T2 = (3W ) ( 12 P ) = 6W P = 6T .
111. Picture the Problem: Forces of different magnitudes do different amounts of work in different intervals of time.
Strategy: Use the definition of power to determine the ranking of the powers.
Solution: 1. Calculate the powers produced by each force:
2. PA =
WA
5J
=
= 0.50 W
Δ tA 10 s
3. PB =
WB 3 J
=
= 0.60 W
Δ tB 5 s
4. PC =
WC
6J
=
= 0.33 W
Δ tC 18 s
5. PD =
WD
25 J
=
= 0.20 W
Δ tD 125 s
6. By comparing the magnitudes of the powers we arrive at the ranking PD < PC < PA < PB .
Insight: If the amount of work and the time interval are each doubled, the power that is produced remains the same.
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6 – 23
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
112. Picture the Problem: Four forces do different amounts of work while producing different amounts of power.
Strategy: Use the definition of power to determine the ranking of the elapsed times.
Solution: 1. Calculate the elapsed times in each case:
2. TA =
WA
40 J
=
= 0.50 s
PA 80 W
3. TB =
WB 35 J
=
= 7.0 s
PB 5 W
4. TC =
WC
75 J
=
= 3.0 s
PC 25 W
5. TD =
WD
60 J
=
= 2.0 s
PD 30 W
6. By comparing the magnitudes of the times we arrive at the ranking TA < TD < TC < TB .
Insight: If the power is doubled, the time required to do a certain amount of work is cut in half.
113. Picture the Problem: This is a units conversion problem.
Strategy: Use the conversion factors given in the inside front cover of the book.
Solution: Convert the units:
W = Pt = (1 kW × 1000 W/kW )(1 h × 3600 s/h ) = 3.6 × 106 J = 3.6 MJ
Insight: Electrical energy is sold in units of kWh, which are really units of energy in multiples of 3.6 million joules.
114. Picture the Problem: You walk leisurely up a flight of stairs, gaining elevation.
Strategy: The power required is the work necessary to elevate your center of mass divided by the time. The work
required to change the elevation of your center of mass m by a distance h is W = mgh.
Solution: Suppose you climb one flight
of stairs (3 m) in 10 seconds (you run up
the stairs quickly). Estimate your mass
to be 70 kg (about 150 lb at sea level):
P=
W mgh
=
t
t
(70 kg ) 9.81 m/s2 (3.0 m )
(
)
1 hp
= 0.27 hp
10 s
746 W
Insight: The energy and power required of you is much higher than this because muscles aren’t 100% efficient
at converting food energy into mechanical energy, and your body requires additional energy to stay warm, keep
the heart pumping, and so on.
P=
= 200 W ×
115. Picture the Problem: You produce power as you lift a box straight upward.
Strategy: The power required is the force required to lift the box multiplied by the speed with which it is lifted.
Solve the power expression for the speed of the box as you lift it.
P 72 W
=
= 0.82 m/s
F 88 N
Insight: Weight lifters can generate a large amount of power as they lift very heavy weights from the waist to above the
head in less than a second.
Solution: Solve P = Fv for the speed:
P= Fv ⇒v=
116. Picture the Problem: Water is pumped vertically upward to bail out a sinking ship.
Strategy: The power required is the work required to change the elevation divided by the time. The work required to
change the elevation of an object is W = mgh.
Solution: Divide the work required by the time:
P=
(
)
2
W mgh (12 kg ) 9.81 m/s (2.1 m )
=
=
t
t
(1.00 s )
P = 247 W × 1 hp 746 W = 0.33 hp
Insight: A 1/3-hp motor would do the trick. Pumping faster than this would require more power.
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6 – 24
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
117. Picture the Problem: A human-powered aircraft flies horizontally at constant speed.
Strategy: The power required is the work required to fly the plane (against friction) divided by the time.
Solution: 1. (a) Solve the
power equation for W:
W = Pt = (0.30 hp × 746 W/hp )(2 h × 3600 s/h + 49 min × 60 s/min )
2. (b) Convert W into
units of candy bars:
2.3 × 106 J ×
W = 2.3 × 106 J = 2.3 MJ
1 Cal 1 candy bar
×
= 2.0 candy bars
4186 J
280 Cal
Insight: The energy required of the pilot is much higher than this because muscles aren’t 100% efficient at converting
food energy into mechanical energy, and the body requires additional energy to stay warm, keep the heart pumping, and
so on.
118. Picture the Problem: A grandfather clock is powered by a weight that slowly descends straight down.
Strategy: The power delivered is the force (the weight) multiplied by the speed.
Solution: 1. (a) Apply the equation P = F v:
0.760 m
⎛
⎞
P = Fv = mgv = (4.35 kg ) 9.81 m/s 2 ⎜
⎝ 3.25 d × 86400 s/d ⎟⎠
(
)
P = 1.15 × 10− 4 W = 0.115 mW
2. (b) To increase the power delivered you must either increase the force or the velocity. In this case, the time it takes
for the mass to descend should be decreased so the velocity will increase and so will the delivered power.
Insight: The weight delivers energy to the clock by doing work. The downward force gravity exerts on the clock is
parallel to its displacement, so it is doing positive work on the clock.
119. It depends. If your bed is below waist level, yes, you must do work to get out of bed, because your leg muscles must lift
your body upward against gravity. If you have a very high bed, perhaps the top bunk of a bunk bed, then no, you do not
need to do any work to get out of bed. Instead, gravity does work on your body as your mass travels downward. In that
case you actually gain kinetic energy by getting out of bed!
120. Picture the Problem: A leaf falls to the ground with constant speed.
Strategy: Consider the value of the mechanical energy of the leaf in order to answer the question.
Solution: Because the leaf’s speed does not change, its kinetic energy remains constant. Its gravitational potential
energy, however, decreases as the leaf descends to a lower elevation. We conclude that the value of KEi + PEi is greater
than the value of KEf + PEf.
Insight: As the leaf falls, air friction is changing its mechanical energy into thermal energy (it heats up the air and
the leaf slightly), so that its mechanical energy is constantly decreasing.
121. Picture the Problem: As a ball is dropped from rest, its potential energy
decreases and its kinetic energy increases.
Strategy: Mechanical energy is conserved because there is no friction.
Therefore, any decrease in potential energy is offset by an increase in
kinetic energy, and the mechanical energy of the ball remains constant.
Solution: 1. (a) The potential energy decreases, so curve B is the potential
energy.
2. (b) The kinetic energy increases, so curve C is the kinetic energy.
3. (c) The total mechanical energy remains constant, so curve A is the total mechanical energy.
Insight: The speed of the ball increases linearly with time, but the kinetic energy is proportional to the square of the
speed, so it increases with the square of the time. That is why curve C has a parabolic shape.
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6 – 25
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
122. Picture the Problem: The potential energy in a spring increases
as it is stretched.
Strategy: The potential energy of a spring equals 12 k x 2 . Use this
formula to calculate the potential energy for each stretch amount x.
Solution: The potential energy varies from 0 J when it is stretched by
0 cm to PE = 12 k x 2 =
1
2
(310 N/m )(0.040 m )2 = 0.25 J
when it is
stretched by 4.0 cm. A plot of PE versus x is shown at the right.
Insight: The farther a spring is stretched, the greater the force required. Because the work done on the spring is the
force multiplied by the distance, the work done increases with the square of the stretch distance.
123. Picture the Problem: A load of bricks is lifted vertically upward at constant speed by a motor.
Strategy: The power required is the work required to change the elevation divided by the time. The work required to
change the elevation of an object is W = mgh .
Solution: Divide the work required by the time:
P=
W mgh (836 N )(10.7 m )
=
=
= 386 W = 0.517 hp
t
t
(23.2 s)
Insight: A ¾ hp motor would do the trick. Lifting faster than this would require more power.
124. Picture the Problem: The brain converts chemical energy into electrical energy and finally into heat energy.
Strategy: The power consumed by the brain equals the rate at which it converts chemical energy into other forms.
In this case the chemical energy comes from a 280-Cal candy bar.
Solution: Solve P = W t for t:
t=
W 280 Cal × 4186 J/Cal
1h
=
= 53,300 s ×
= 15 h
P
22 W
3600 s
Insight: The power required to lift a gallon (3.6 kg) of milk a distance of 1.0 m in 0.80 s is mgh t = 44 W, about twice
the power consumed by the normally functioning brain.
125. Picture the Problem: The energy required to run the Atmos clock is compared to the energy consumed by a light bulb.
Strategy: Find the energy consumed by the light bulb and divide by the given ratio.
Solution: 1. Solve P = Wbulb t for Wbulb :
Wbulb = Pt = (60 W )(24 h × 3600 s/h ) = 5.18 × 106 J
Wbulb
5.18 × 106 J
=
= 0.0216 J = 21.6 mJ
240 × 106
240 × 106
Insight: Incandescent light bulbs actually aren’t very efficient, producing more heat than visible light. The Atmos clock
had to be designed and built very carefully in order to run on such a small amount of energy.
2. Use a ratio to find WAtmos :
WAtmos =
126. Picture the Problem: A box is pushed horizontally in a straight line at constant speed.
Strategy: You do positive work on the box as it slides, and friction does negative work at the same time. The two works
are equal because the kinetic energy of the box is constant. Therefore the force you must generate equals the friction
force. The power you produce is equal to the force multiplied by the speed.
Solution: Substitute for F = μ k mg
in the equation P = F v :
(
)
P = Fv = (μ k mg ) v = (0.55)(67 kg ) 9.81 m/s 2 (0.50 m/s )
P = 180 W = 0.18 kW
Insight: Newton’s second law, F = ma, states that the net force on the box must be zero because it is not accelerating.
That’s another way of saying that the magnitude of the pushing force equals the magnitude of the friction force.
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6 – 26
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
127. Picture the Problem: A mop head is pushed downward into the floor as
it slides horizontally across the floor.
F
θ
Strategy: Use the definition of work, keeping in mind the angle between
the force and the direction of motion.
d
W = Fd cos θ = (43 N )(0.50 m ) cos 55o = 12 J
Solution: 1. (a) Apply the definition of work:
2. (b) If the angle is increased to 65°, a smaller component of the force will be along the direction of motion and
therefore the work done by the janitor will decrease.
Insight: If the angle is increased to 65° the work decreases to 9.1 J. Only the component of the force along the direction
of the motion does any work.
128. Picture the Problem: A plane and the glider it is towing are at different
altitudes but both are moving horizontally.
Strategy: Solve the definition of work for the angle between the force and
the direction of motion.
Solution: 1. Solve the definition
of work for the angle:
W
W = Fd cos θ or cos θ =
Fd
2. Calculate the angle:
θ = cos −1 ⎜

d
θ

F
⎡ 2.00 × 105 J ⎤
⎛W ⎞
o
= cos −1 ⎢
⎥ = 57.4
⎟
⎝ Fd ⎠
⎣ (2560 N )(145 m ) ⎦
Insight: Only the component of the force along the direction of the motion does any work.
129. Picture the Problem: Some chocolate chip cookie dough is stirred with a spoon.
Strategy: The power required to stir the dough is the force times the velocity.
Solution: 1. (a) Apply P = F v:
P = F v = (21 N )(0.23 m/s ) = 4.8 W
2. (b) Solve P = W t for W:
W = Pt = (4.8 W )(1.5 min × 60 s/min ) = 430 J = 0.43 kJ
Insight: The work you do with all that stirring is converted into heat by frictional forces within the dough.
130. Picture the Problem: A particle moves under the influence of a conservative force. The particle’s potential and kinetic
energies are to be considered at points A, B, and C.
Strategy: Use the conservation of mechanical energy to answer the questions. At each point the kinetic and
gravitational potential energies will sum to the same total energy.
Solution: 1. (a) At point B the particle is at rest and has no kinetic energy, but it has 25 J of potential energy. We
conclude that the total energy of the particle is 25 J. That means the potential energy at point A is 25 J − 12 J = 13 J.
2. (b) At point C the total energy is 25 J and the potential energy is 5 J, so the kinetic energy is 25 J − 5 J = 20 J.
Insight: If there were a point D at which the particle had 40 J of total energy, we would conclude that 15 J of positive
work was done on the particle between points C and D.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6 – 27
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
131. Picture the Problem: When a ball of mass m is dropped from rest from a height h, its kinetic energy just before landing
is KE. A second ball of mass 4m is dropped from rest from a height h/4.
Strategy: Apply the concept of conservation of energy to answer the conceptual question.
Solution: 1. (a) The change in gravitational potential energy for each ball is the same because mgh = (4m ) g ( 14 h ).
We conclude that just before ball 2 lands its kinetic energy is also KE .
2. (b) The best explanation is A. The two balls have the same initial energy. Statements B and C are each false because
neither scenario would conserve mechanical energy.
Insight: If the second ball were dropped from the same height h as the first ball, it would have 4 times the initial
potential energy and would land with a kinetic energy of 4KE, but its speed would be the same as the first ball.
132. Picture the Problem: A meteorite is slowed to a stop while traveling in a straight line into the trunk of a car.
Strategy: The work done on the meteorite by the car equals the negative of the initial kinetic energy of the meteorite.
The work can then be used to find the force exerted on the meteorite by the car.
W = Fd = 12 mvf2 − 12 mvi2 = 0 − 12 mvi2
Solution: Solve the work equation for the
force. The average force is negative if we
take the initial velocity of the meteorite to
be in the positive direction:
mvi2
(12 kg )(550 m/s)
=−
2d
2 (0.22 m )
2
F=−
F = −8.3 × 106 N = −927 tons!
Insight: The kinetic energy of the meteorite is used to deform the metal trunk of the car as well as heat up the meteorite
and the trunk by means of frictional forces.
133. Picture the Problem: A gallon of milk is lifted at a rate that requires 22 W of power.
Strategy: The power required to lift the milk is the force (the weight of the milk) multiplied by the speed.
Use this relationship to find the speed of the lift and the time required to lift the milk 1.0 m.
Solution: 1. (a) Solve P = Fv for v:
P= Fv ⇒v=
2. (b) Solve d = vt for t:
d = vt ⇒ t =
P
P
22 W
=
=
= 0.62 m/s
F mg (3.6 kg ) 9.81 m/s 2
(
)
d
1.0 m
=
= 1.6 s
v 0.62 m/s
Insight: If the milk were lifted 1.0 m in 0.50 s, it would require 70 W of power.
134. Picture the Problem: A jet is accelerated horizontally by a catapult.
Strategy: The work done by the catapult on the jet equals the change in kinetic energy of the jet. Since the kinetic
energy depends upon the mass, we can find the mass of the jet if we know the amount of work that was done. The
power is the work divided by the time.
7
2W 2 7.6 × 10 J
Solution: 1. (a) Solve the workW = 12 mvf2 − 0 ⇒ m = 2 =
= 2.9 × 104 kg = 32 tons!
2
energy theorem W = Δ KE for m:
vf
72
m/s
(
)
(
)
W 7.6 × 107 J
=
= 3.8 × 107 W = 38 MW = 51, 000 hp!
t
2.0 s
Insight: The work done by the catapult goes into the kinetic energy of the jet. We’re ignoring any energy removed by
frictional forces.
2. (b) Apply P = W t to find P:
P=
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6 – 28
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
135. Picture the Problem: A boat and skier are both moving toward the
right but the rope is pulling at an angle.
Strategy: Use the work equation to calculate the work done by the
rope on the skier. The rope does positive work on the skier as it skis,
and friction does negative work at the same time. The two works are
equal because the kinetic energy of the skier is constant.
Solution: 1. (a) Apply
the work equation:
W = Fd cos θ = F (vt ) cos θ = (90.0 N )(14 m/s )(10.0 s ) cos 35o = 10,300 J = 10 kJ
2. (b) The work done by friction must be opposite the work done by the rope: Wfriction = −10 kJ
Insight: Newton’s second law, F = ma, states that the net force on the skier must be zero because the skier is not
accelerating. That’s another way of saying that the magnitude of the towing force equals the magnitude of the friction
force, which implies that the two works are equal and opposite.
136. Picture the Problem: A spider crawls up a window at an angle, as pictured at right.
Strategy: Use the work equation, keeping in mind the angle between the force exerted
by the spider (vertically upward) and the displacement (upward at 25° from the vertical).
Solution: 1. Apply
the work equation:
W = Fd cos θ = mg y cos θ = mg (vt ) cos θ
2. Now use the power equation:
P=
W mg (vt ) cos θ
=
= mgv cos θ
t
t
P = 1.8 × 10−3 kg 9.81 m/s 2 (0.023 m/s ) cos 25o
(
)(
P = 3.7 × 10
−4
)
W = 0.37 mW
Insight: Only the component of the force along the direction of the motion does any work.
The spider is gaining gravitational potential energy as it elevates its center of mass. Yes,
that spider does look curiously like a beetle…
137. Picture the Problem: A baseball is accelerated along a straight line by a pitcher.
Strategy: The work done on the ball by the pitcher changes the ball’s kinetic energy. The power required to accelerate
the ball is the work divided by the time.
Solution: 1. (a) Apply the
work-energy theorem:
W = 12 mvf2 − 12 mvi2 =
2. (b) Apply the power equation:
P=
1
2
(0.14 kg )(25.5 m/s)2 − 0 =
46 J
W
46 J
=
= 610 W = 0.82 hp
t
0.075 s
3. (c) The power required to accelerate the ball to the same speed in less time would be more than the previous case
because the same amount of work would have to be done in a smaller amount of time.
Insight: The power generated by the pitcher is the rate at which the ball gains kinetic energy.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6 – 29
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
138. Picture the Problem: The human heart produces power in order to push the blood through your veins and arteries.
Strategy: Power is work (or energy) per time. The work necessary to elevate your center of mass is W = mgh .
Solution: 1. (a) Solve P = W t for W:
W = Pt = (1.33 W )(24 h × 3600 s/h ) = 1.15 × 105 J = 115 kJ
2. (b) Use the energy W to increase your
body’s potential energy PE = mgh :
h=
W
1.15 × 105 J
=
= 170 m = 550 ft!
mg (70 kg ) 9.81 m/s 2
(
)
Insight: You wouldn’t be able to climb quite this high because muscles aren’t 100% efficient at converting food energy
into mechanical energy. Besides, you still need some energy to keep your heart pumping!
139. Picture the Problem: A sled slides down a hill without friction.
Strategy: As the sled descends the hill its gravitational potential energy is converted into kinetic energy. Find the loss
in gravitational potential energy by setting the change in mechanical energy equal to zero, so that ΔE = Ef − Ei = 0 or
Ef = Ei . Let h = 0 at the bottom of the hill and either v = 0 or v = 1.50 m/s at the top.
Solution: 1. (a) When the sled reaches the bottom of the hill after starting with a speed of 1.50 m/s, it will arrive at
the bottom of the hill with a speed less than 9.00 m/s. That’s because the kinetic energy at the top of the hill is small
compared to the gain in kinetic energy as the sled descends the hill. Another way to say the same thing is to recognize
that the energies add but the speeds do not.
2. (b) Set Ebottom = Etop and solve for PEtop ,
letting PEbottom = KEtop = 0 :
Ebottom,1 = Etop,1
KEbottom,1 + PEbottom = KEtop,1 + PEtop
1
2
3. Now find vbottom,2 after the second run using
the same approach:
2
mvbottom,1
+ 0 = 0 + PEtop
KEbottom, 2 + PEbottom = KEtop, 2 + PEtop
1
2
2
2
2
1
1
mvbottom,
2 + 0 = 2 mvtop, 2 + 2 mvbottom,1
2
2
vbottom, 2 = vtop,
2 + vbottom,1
=
(1.50 m/s )2 + (7.50 m/s)2
= 7.65 m/s
Insight: Note that the sled is not going 1.50 m/s faster than the 7.50 m/s it would be traveling if it started from rest.
That’s because the 45 J of kinetic energy it has at the start (if the sled and passengers have a mass of 40 kg) is tiny
compared with the 1125 J of kinetic energy it gains on the way down. The energies add but the speeds do not.
140. Picture the Problem: An airplane takes off, changing both its gravitational potential and kinetic energies.
Strategy: The airplane engines do nonconservative work on the airplane to change its mechanical energy, including its
kinetic and gravitational potential energies. Let PE = 0 on the runway and let the airplane start from rest. Use the given
altitude and speed values to find the change in its mechanical energy. Then set its kinetic energy change equal to its
gravitational potential energy change and solve for vf .
Solution: Use the given hf and vf
values to find ΔE :
ΔE = ΔPE + ΔKE
(
= mg (hf − 0) + 12 m vf2 − 02
(
)
)
= (1865 kg ) 9.81 m/s 2 (2420 m ) + 12 (1865 kg )(96.5 m/s )
2
ΔE = 4.43 × 107 J + 8.68 × 106 J = 5.30 × 107 J = 53.0 MJ
Insight: Note that the change in potential energy (44.3 MJ) is 5.1 times larger than the change in kinetic energy
(8.68 MJ). The primary struggle for airplanes is gaining altitude, not speed.
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6 – 30
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
141. Picture the Problem: A child slides from rest at point A and
lands at point B as indicated in the figure at the right.
Strategy: Use the conservation of mechanical energy to find
the horizontal speed of the child at the bottom of the slide in
terms of h. Then use t = 2 ybottom g , the time it takes a
projectile to fall a distance ybottom from rest, to find the speed
the child should have in order to land 2.50 m down range.
Set the speeds equal to each other and solve for h.
KEA + PEA = KEbottom + PEbottom
Solution: 1. Use EA = Ebottom to find vbottom :
1
2
2
mvA2 + mgyA = 12 mvbottom
+ mgybottom
2
0 + mg (h + 1.50 m ) = 12 mvbottom
+ mg (1.50 m )
2 gh = vbottom
2. Use x = vbottom t = vbottom 2 ybottom g to find the appropriate
x = vbottom 2 ybottom g ⇒ vbottom = x g 2 ybottom
vbottom for the child to land 2.50 m down range:
3. Set the two velocities equal to each other and solve for h:
2 gh = x g 2 ybottom
2 gh = x 2 g 2 ybottom
(2.50 m) = 1.04 m
x2
h=
=
4 ybottom 4 (1.50 m )
2
Insight: These are the sorts of calculations an engineer might make to determine how high to build a slide so that a
child will land in a certain place. However, an engineer should account for the nonzero friction when making his design.
142. Picture the Problem: The physical situation is depicted in the figure.
Strategy: Use the conservation of mechanical energy together with the
geometry of the problem to find the speed of the skateboarder at point A.
Let hA = 0 and vB = 0 .
Solution: Set EA = EB
to find vA :
KEA + PEA = KEB + PEB
1
2
mvA2 + 0 = 0 + mghB
(
)
vA = 2 ghB = 2 9.81 m/s 2 (2.64 m )
= 7.20 m/s
Insight: The solution assumes there is no friction as the skateboarder travels around the half pipe. In real life the
skater’s speed must exceed 7.20 m/s at point A because friction will convert some of his kinetic energy into heat.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6 – 31
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
143. Picture the Problem: A block slides from rest at point A and is launched
horizontally at point B as indicated in the figure at right.
Strategy: Use the conservation of mechanical energy to find the
horizontal speed of the block at the bottom of the ramp. Then use the
equations for projectile motion to find the landing site of the block, noting
that it is launched horizontally.
Solution: 1. Set EA = EB to find vB :
KEA + PEA = KEB + PEB
0 + mghA = 12 mvB2 + mghB
2 g (hA − hB ) = vB
2. Use d = vB t = vB 2hB g to find d:
d = vB 2hB g = ⎡⎣ 2 g (hA − hB )⎤⎦ (2hB g ) = 4hB (hA − hB )
= 4 (0.25 m )(1.5 − 0.25 m )
d = 1.1 m
Insight: If there were friction between the block and the ramp it would convert some of the kinetic energy of the block
into heat. The block would then be launched with less speed and would land a shorter distance d from the ramp.
144. Answers will vary. Students should research the power that humans produce in everyday life and make a table that
compares the powers generated from various activities. The power can be calculated as either the work divided
by the time elapsed, or as the force multiplied by the velocity. Verify that the values seem reasobable.
145. Although the mechanical energy of a projectile is conserved (as long as there is no friction), there is no limitation on
the values of the potential and kinetic energies. They are constantly changing, such that any gain in potential energy is
balanced by a loss of kinetic energy. When a ball is falling, its potential energy is decreasing and its kinetic energy is
increasing, but the sum of the two energies, the mechanical energy, remains constant. Similarly, as a ball rises upward
its potential energy is increasing and its kinetic energy is decreasing, but the mechanical energy remains constant.
146. Picture the Problem: The power required for Microraptor gui to
fly, as a function of speed, is depicted in the graph at right.
Strategy: Examine the lower horizontal line that shows the
estimated 9.8-W power output of Microraptor. The two locations
where it intersects the graph of the power required for flight
indicate the minimum and maximum flight speeds the creature
could achieve.
Solution: The lower horizontal line at 9.8 W power output
intersects the graph of the power required for flight at 7.7 m/s
and 15 m/s, indicating the creature could fly at speeds between
7.7−15 m/s, which corresponds to choice B.
Insight: If Microraptor could only generate 7.5 W of power, it would not have been able to fly at all! A minimum of
8.1 W is required, as can be seen from the minimum of the graph of the power required for flight.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6 – 32
Chapter 6: Work and Energy
Pearson Physics by James S. Walker
147. Picture the Problem The power required for Microraptor gui to
fly, as a function of speed, is depicted in the graph at right.
Strategy: Examine the upper horizontal line that shows the
hypothetical 20-W power output of Microraptor. The two
locations where it intersects the graph of the power required
for flight indicate the minimum and maximum flight speeds the
creature could achieve if it could generate that much power.
Solution: The upper horizontal line at 20-W power output
intersects the graph of the power required for flight at 2.5 m/s
and 25 m/s, indicating the creature could fly at speeds between
2.5 − 25 m/s, which corresponds to choice C.
Insight: The Microraptor would be able to hover (fly at 0 m/s) if it could generate at least 25 W of power.
148. Picture the Problem: The power required for Microraptor gui to
fly, as a function of speed, is depicted in the graph at right.
Strategy: A minimum of 8.1 W of power would be required
for the Microraptor to fly at 10 m/s. Use this power value to
determine the energy expended if the creature flew at 10 m/s
for 1.0 minute.
Solution: Multiply the power
by the time to find the work. The
calculated answer is choice C.
W = Pt
= (8.1 W )(60 s )
W = 490 J
Insight: In a later chapter we will learn about thermal energy and food calories. If Microraptor could somehow use
the energy of 100 food calories (4.2×105 J) solely for flying at 10 m/s, it could fly for over 14 hours!
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6 – 33