Download ENERGY AND WORK

ENERGY AND WORK
10
Q10.1. Reason: The brakes in a car slow down the car by converting its kinetic energy to thermal energy in the
brake shoes through friction. Cars have large kinetic energies, and all of that energy is converted to thermal energy in
the brake shoes, which causes their temperature to increase greatly. Therefore they must be made of material that can
tolerate very high temperatures without being damaged.
Assess: This is an example of an energy conversion. All of the car’s kinetic energy is converted to thermal energy
through friction. To get an appreciation of how much kinetic energy is absorbed by the brake shoes, consider instead
the energy explicit in stopping the car by hitting a stationary object instead!
Q10.2. Reason: When you hit a nail with a hammer to pound it into some object, many processes are at work. For
example, some small amount of energy goes into temporarily increasing the nail’s kinetic energy as it moves into the
object. Part of the energy goes into permanently deforming the object to accept the nail. An appreciable portion of the
initial kinetic energy of the hammer is converted to thermal energy through, for example, friction between the nail
and the object as the nail moves into the object. Some gets directly converted to kinetic energy of the molecules that
make up the nail (see section 11.3 for an atomic view of thermal energy and temperature) from the collision between
the nail and the hammer.
Assess: If you ever try hammering nails, the thermal energy generated is appreciable. Note that energy can be
transformed directly into kinetic energy of atoms or molecules that make up an object. As a simpler example,
banging a hammer on a solid object directly will increase the temperature of the both the hammer and the object.
Q10.3. Reason: We must think of a process that increases an object’s kinetic energy without increasing any
potential energy. Consider pulling an object across a level floor with a constant force. The force does work on the
object, which will increase the object’s kinetic energy. Since the floor is level the gravitational potential energy does
not change. The other form of potential energy possible is that stored in a spring, which is also zero here.
Assess: For there to be no potential energy change, the object in question must remain at the same height.
Q10.4. Reason: Here we must increase potential energy without increasing kinetic energy. Consider lifting an
object at constant speed. Consider the object plus the earth as the system. The force does work that increases the
gravitational potential energy of the object, while the kinetic energy does not increase because the velocity of the
object remains the same. Another possibility is the compression of a spring by an applied force at constant velocity.
Note that constant velocity is not necessary for the change in kinetic energy between the beginning and end of a
process to be zero. Lifting an object or compressing a spring in any way, as long as the initial velocity is equal to the
final velocity at the end of the process leads to no change in net kinetic energy. Any kinetic energy gained during the
process is lost when the object is brought the rest.
Assess: Kinetic energy does not change if an object has the same velocity at the beginning and end of a process.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10-1
10-2
Chapter 10
Q10.5. Reason: The system must convert kinetic energy directly to potential energy with no external force doing
work. For gravitational potential energy we must change the height of the object. One simple example would be
rolling a ball up a hill. The initial kinetic energy is converted to gravitational potential energy as the ball increases its
height. The ball loses kinetic energy while it gains potential energy. Another example is rolling a ball into an
uncompressed spring on level ground. As the ball compresses the spring, the system gains potential energy, while
losing kinetic energy. Since there are no forces external to the systems in these examples, no work is done on the
systems by the environment.
Assess: As long as no forces external to the system are applied, work done on a system is zero.
Q10.6. Reason: We need a process that converts kinetic energy to work without any change in potential energy.
Consider a block sliding on level ground, to which is attached a cord you are holding on to. As the block slides, it
exerts a force on your hand by virtue of its kinetic energy. As the block pulls your hand, it is doing work on you. The
kinetic energy of the block will decrease as it continues to exert the force on your hand.
Assess: To have a change in gravitational potential energy you must have a change in height.
Q10.7. Reason: Here we need to convert potential energy to kinetic energy without any work done on the system.
Consider dropping a ball from a height. The ball’s gravitational energy is converted to the kinetic energy of the ball
as it falls. Another example would be releasing a ball at the end of a compressed spring. The potential energy stored
in the compressed spring is converted to the kinetic energy of the ball as the spring stretches to its equilibrium length.
Since no external forces act on a system, the work on the system is zero.
Assess: Many examples in the problem section will involve just this type of conversion of potential energy to kinetic
energy. If no forces from the environment act on a system, the work done on the system is zero.
Q10.8. Reason: We need a process that converts work totally into thermal energy without any change in the
kinetic or potential energy. Consider moving a block of wood across a horizontal rough surface at constant speed.
Because the surface is horizontal there is no change in potential energy, and because the speed is constant there is no
change in kinetic energy. All the work done in moving the block across the rough horizontal surface is transferred
into thermal energy.
Assess: To have a change in gravitational energy you must have a change in height and to have a change in kinetic
energy you must have a change in speed.
Q10.9. Reason: We need a process that converts potential energy totally into thermal energy without changing the
kinetic energy. Consider a wood block sliding down a rough inclined surface at a constant speed. The gravitational
potential energy is decreasing and the kinetic energy is constant. All the decrease in gravitational potential energy
becomes an increase in thermal energy.
Assess: Gravitational potential energy decreases because there is a change in the height of the block. The kinetic
energy does not change because the speed of the block is constant.
Q10.10. Reason: We need a process that converts kinetic energy totally into thermal energy without changing the
gravitational potential energy. Consider a wood block sliding across a rough horizontal surface and slowing to a stop.
The gravitational potential energy is not changing and all the kinetic energy is being transferred into thermal energy.
All the decrease in kinetic energy becomes an increase in thermal energy.
Assess: Gravitational potential energy does not change because there is a change in height of the block. The kinetic
energy decreases as the block slows to a stop.
Q10.11. Reason: The energies involved here are kinetic energy, gravitational potential energy, elastic potential
energy, and thermal energy. For the system to be isolated, we must not have any work being done on the system and
no heat being transferred into or out of the system. The ball’s kinetic and elastic energy is changing, so we should
consider it part of the system. Since its gravitational potential energy is changing, we need to also consider the earth
as part of the system. Thermal energy will be generated in the ball and floor when the ball hits the floor, so we must
consider both to be part of the system.
Assess: In order to have an isolated system no work can be done on the system and all forces must be internal.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Energy and Work
10-3
Q10.12. Reason: The tension force is perpendicular to the direction of motion of the mass. The work done by
tension is W = Td cosθ = Td cos90° = 0. No work is done by tension.
Assess: The work done by a force depends not only on the force and the displacement, but on the angle between
them. When the force is perpendicular to the displacement, the force does no work at all.
Q10.13. Reason: (a) The work done is W = Fd. Both particles experience the same force, so the greater work is
done on the particle that undergoes the greater displacement. Particle A, which is less massive than B, will have the
greater acceleration and thus travel further during the 1 s interval. Thus more work is done on particle A. (b) Impulse
is FΔt. Both particles experience the same force F for the same time interval Δt = 1 s. Thus the same impulse is
delivered to both particles. (c) Both particles receive the same impulse, so the change in their momenta is the same,
that is, mA (vf )A = mB (vf )B . But because mA < mB , it must be that (vf ) A > (vf )B . This result can also be found from
kinematics, as in part (a).
Assess: Work is the product of the force and the displacement, while impulse is the product of the force and the time
during which the force acts.
Q10.14. Reason: (a) The work done is W = Fd. Both particles experience the same force and move the same
distance, so the work done on them is the same. From the work-kinetic energy theorem we know that the kinetic
energy of each puck has changed by the same amount. They both started from rest, so they end up with the same
kinetic energy.
(b) The kinetic energies are equal, but the speeds are not.
v
mB
2mA
1
mA vA2 = 12 mBvB2 ⇒ A =
=
= 2
2
vB
mA
mA
So the speed of puck A is 1.41 times the speed of puck B.
Assess: This makes intuitive sense.
Q10.15. Reason: Neglecting frictional losses, the work you do on the jack is converted into gravitational potential
energy of the car as it is raised. The work you do is Fd, where F is the force you apply to the jack handle and d is
the 20 cm distance you move the handle. This work goes into increasing the potential energy by an amount
mgh = wh, where w is the car’s weight and h = 0.2 cm is the change in the car’s height. So Fd = wh so that
F/w = h/d .
Assess: Because the force F you can apply is so much less than the weight w of the car, h must be much less than d.
Q10.16. Reason: The ball of mass m has an initial potential energy of mgh and the ball of mass 2m has an initial
potential energy if 2mgh. The ball of mass 2m has twice the initial potential energy of the ball of mass m. There is no
initial kinetic energy. Since energy is conserved, as the balls hit the floor the ball of mass 2m still has twice as much
energy as the ball of mass m. As the balls hit the floor the energy is totally kinetic. So the ball of mass 2m hits the
floor with twice the kinetic energy of the ball of mass m.
Assess: No energy was lost, rather it was changed from potential to kinetic energy.
Q10.17. Reason: (a) If the car is to go twice as fast at the bottom, its kinetic energy, proportional to v 2 , will be
four times as great. You thus need to give it four times as much gravitational potential energy at the top. Since
gravitational potential energy is linearly proportional to the height h, you’ll need to increase the height of the track by
a factor of four. (b) Using considerations of conservation of energy, as in part (a), we see that the speed of the car at
the bottom depends only on the height of the track, not its shape.
Assess: Kinetic energy is proportional to the square of the velocity.
Q10.18. Reason: (a) Because the elastic potential energy of a spring is proportional to the square of the distance
it’s compressed, its potential energy will increase by a factor of four when this distance is doubled. (b) The ball now
has four times the energy it had on the first shot. When the ball is released, conservation of energy tells us that the
ball will then have four times as much kinetic energy. Kinetic energy depends on the square of the speed, so the ball
need travel only twice as fast to have four times the energy.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10-4
Chapter 10
Assess: Elastic potential energy and kinetic energy both depend on the square of the displacement and speed,
respectively. So doubling the displacement leads to a doubling of the speed.
Q10.19. Reason: Because both rocks are thrown from the same height, they have the same potential energy. And
since they are thrown with the same speed, they have the same kinetic energy. Thus both rocks have the same total
energy. When they reach the ground, they will have this same total energy. Because they’re both at the same height at
ground level, their potential energy there is the same. Thus they must have the same kinetic energy, and hence the
same speed.
Assess: Although Chris’s rock was thrown angled upward so that it slows as it first rises, it then speeds up as it
begins to fall, attaining the same speed as Sandy’s as it passes the initial height. Sandy’s rock will hit the ground first,
but its speed will be no greater than Chris’s.
Q10.20. Reason: By the time the blocks reach the ground, they have transformed identical amounts of
gravitational potential energy into translational kinetic energy of the blocks and rotational kinetic energy of the
cylinders. But the moment of inertia of a hollow cylinder is higher than that of a solid cylinder of the same mass, so
more of the energy of the system is in the form of rotational kinetic energy for the hollow cylinder than for the solid
one. This leaves less energy in the form of translational kinetic energy for the hollow cylinder. But it is the
translational kinetic energy that determines the speed of the block. So the block moves more slowly for the system
with the hollow cylinder, and so its block reaches the ground last.
Assess: The energy is shared between the rotating cylinder and the falling block. The more energy the cylinder has,
the less is available for the block.
Q10.21. Reason: As you land, the force of the ground or pad does negative work on your body, transferring out
the kinetic energy you have just before impact. This work is − Fd, where d is the distance over which your body
stops. With the short stopping distance involved upon hitting the ground, the force F will be much greater than it is
with the long stopping distance upon hitting the pad.
Assess: For a given amount of work, the force is large when the displacement is small.
Q10.22. Reason: If the crate slid down the ramp at a constant speed without Jason pushing back on it (as it could if
the angle were adjusted just right) then, by conservation of energy, the increase in thermal energy would exactly
equal the loss in gravitational potential energy. But in fact Jason is pushing on the crate in a direction opposite its
motion. Thus he is doing negative work on the crate, removing energy from the system. So some of the loss of
potential energy that would otherwise have gone into increasing thermal energy is instead removed from the system,
so the increase in thermal energy is actually less than the loss on potential energy.
Assess: A numerical example with made-up numbers might help us visualize this. Suppose that as the crate
slides down the ramp it loses 100 J of gravitational potential energy, so that ΔU g = −100 J. And suppose that the
(negative) work done by Jason is –20 J. Then the work-energy equation, Δ Eth + ΔU g = W is Δ Eth + ( −100 J) =
− 20 J, so that Δ Eth = 80 J, which is less than the 200 J of potential energy loss.
Q10.23. Reason: When the coaster is at the top U = mgy relative to the ground. That amount of energy equals the
kinetic energy at the bottom. Halfway down (or up) the potential energy is half of what it was at the top, so the
kinetic energy must also be half of what it is at the bottom. If v′ is the speed at the halfway height, then
1
2
1
2
2
m(v ')2 1 ⎛ v ' ⎞
1
1
2
v'
(30 m/s) = 21 m/s
= ⇒
= ⇒ =
⇒ v' =
2
2
2
m(v) 2 2 ⎜⎝ v ⎟⎠
v
So the correct choice is C.
Assess: Even though the height is half the total, the speed is not half of 30 m/s.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Energy and Work
10-5
Q10.24. Reason: Work is defined by W = Fd when the force is parallel to the displacement, as it is in this case.
Since you and your friend each carry suitcases of the same mass up the same flights of stairs, you both exert the same
force on the suitcases over the same vertical distance and therefore do the same amount of work. Your friend takes
longer than you to get up the flight of stairs, so you expend a greater amount of power since P = W /Δt. The correct
choice is C.
Assess: For a given amount of work, differing amounts of power are expended depending on how quickly the work
is done.
Q10.25. Reason: Assuming the woman raises the weight at constant velocity, the force she exerts must equal the
weight of the object. Since the mass of the object is 20 kg, its weight is about w = mg = (20 kg)(9.80 m/s 2 ) = 200 N.
Since she lifts it 2 m, the work done is W = Fd = (200 N)(2 m) = 400 J. She does this work in 4 s, so the power she
exerts is P = W /Δt = (400 J)/(4 s) = 100 W. The correct choice is A.
Assess: Power is defined as the rate of doing work. This seems like a reasonable amount of power for the woman to
expend.
Q10.26. Reason: Since kinetic energy is proportional to the square of the velocity of an object, an object with
twice the velocity will have four times the amount of kinetic energy. In this question, all the kinetic energy is
converted to elastic potential energy in the spring. The potential energy stored in a spring is proportional to the
square of the compression from its equilibrium position. Since we start with four times the kinetic energy, four times
as much energy is stored in the spring. But since the energy stored in the spring is proportional to the square of the
compression, the compression is only twice the compression previously, or 2(2.0 cm) = 4.0 cm. The correct choice
is C.
Assess: Kinetic energy is proportional to the square of the velocity of an object and the potential energy of a spring
is proportional to the square of the displacement from the equilibrium position.
Q10.27 Reason: The potential energy at the top of the first ramp is equal to the kinetic energy at the bottom, so
that mgh = (1/2)mv02 . By conservation of energy, the energy when the block is moving up the second ramp at speed
1
v0 /2 is equal to the initial energy mgh, so we have mgh′ + 2 m
mgh′ + 14 ×
(
1
2
)
()
v0
2
2
= mgh, which we can rewrite as
mv02 = mgh. From the conservation of energy applied to the first ramp, we can rewrite this as
mgh′ + mgh = mgh, or h′ = 3h/4. The correct choice is C.
1
4
Assess: Because the block had energy mgh at the top of the ramp, it has energy mgh at all points along both ramps.
Similarly, its energy is (1/2)mv02 at all points along both ramps.
Q10.28. Reason: As the ball falls, energy is conserved since the only force doing work is the force of gravity.
Since gravity is conserved we may write
or
ΔE = 0
ΔK + ΔU g = 0
As the ball falls we have
Δ K = mv 2 /2 and ΔU g = − mgh = − mg ( L − L cos30 °) = − mgL(1 − cos30 °)
Combining these obtain
mv 2 /2 = mgL (1 − cos30 °)
or
v = 2 gL(1 − cos30°) = 3.6 m/s
The correct response is C.
Assess: The key to the problem is to realize that energy is conserved and then find the change in kinetic and
potential energy. A speed of 3.6 m/s is reasonable.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10-6
Chapter 10
Problems
P10.1. Prepare: Equation 10.6 gives the work done by a force F on a particle. The work is defined as
W = Fd cos(θ ), where d is the particle’s displacement. Since there is a component of the lifting force in the
direction of the displacement, we expect the work done in both parts of this problem to be nonzero. Assume you lift
the book steadily, so that the force exerted on the book is constant.
Solve: (a) Refer to the diagram. We are assuming the book does not accelerate, so the force you exert on the book is
exactly equal to the force of gravity on the book. Fhand on book = − Fgravity on book
The total displacement of the book is 2.3 m − 0.75 m = 1.55 m (keeping one extra significant figure for this
intermediate result).
The work done by gravity is then
Wgravity on book = wd cos(θ ) = (2.0 kg)(9.80 m/s2 )(1.55 m)cos(180°) = −30 J
(b) The work done by hand is Whand on book = Fhand on book d cos(θ ).
⇒ Whand on book = (2.0 kg)(9.80 m/s2 )(1.55 m)cos(0°) = +30 J
Assess: Note that the only difference is in the sign of the answer. This is because the two forces are equal, but act in
opposite directions. The work done by gravity is negative because gravity acts opposite to the displacement of the
book. Your hand exerts a force in the same direction as the displacement, so it does positive work. We should expect
the total work to be zero from Equation 10.4 since energy is conserved in this process. Referring to the results, we
see that the work by your hand cancels the work done by gravity and the total work is zero as expected.
P10.2. Prepare: Note that not all the forces in this problem are parallel to the displacement. Equation 10.6 gives
the work done by a constant force which is not parallel to the displacement: W = Fd cos(θ ), where W is the work
done by the force F at an angle θ to the displacement d. Here the displacement is exactly downwards in the same
direction as w. We will take all forces as having four significant figures (as implied by T2 = 1295 N).
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Energy and Work
10-7
Solve: The angle between the force w and the displacement is 0°, so
Ww = wd cosθ = (2500 N)(5 m)cos(0°) = 12.5 kJ
The angle between the force T1 and the displacement is 90° + 60° = 150°.
WT = T1d cosθ = (1830 N)(5 m)cos(150°) = −7.92 kJ
1
The angle between the tension T2 and the displacement is 90° + 45° = 135°.
WT = T2 d cosθ = (1295 N)(5 m)cos(135°) = −4.58 kJ
2
Assess: Note that the displacement d in all these cases is directed downwards and that it is always the angle
between the force and displacement used in the work equation. For example, the angle between T1 and d is 150°,
not 60°.
P10.3. Prepare: Note that not all forces act in the same direction as the displacement. We must use Equation 10.6:
W = Fd cos(θ ) for each force. W is the work done by a force of magnitude F on a particle and d is the particle’s
displacement. The crate is moving directly to the right. We assume all forces are given to three significant figures,
including the 500 N force since the other two forces are given to three significant figures.
Solve: For the force f k , the displacement is exactly opposite the force, so
W fk = f k d cos(180°) = (500 N)(3 m)(−1) = −1.50 kJ
For the tension T1:
WT1 = T1d cos(20°) = (326 N)(3 m)(0.9397) = 0.919 kJ
For the tension T2 :
WT2 = T2d cos(30°) = (223 N)(3 m)(0.8660) = 0.579 kJ
Assess: Negative work done by the force of kinetic friction f k means that 1.50 kJ of energy has been transferred out
of the crate and converted to heat. The other two forces have components along the displacement, and therefore do
positive work to move the crate.
P10.4. Prepare: We will use the definition of work, Equation 10.6 to calculate the work done. The sidewalk and
escalators exert a normal force on you, and may exert a force to propel you forward. We will assume that the
escalator propels you at constant velocity, as the sidewalk does.
Solve: (a) The escalator moves you across some distance like the sidewalk, but it also moves you upwards. See the
following diagram.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10-8
Chapter 10
The force exerted on you by the escalator is the normal force, which is equal to your weight.
n = w = mg = (60 kg)(9.80 m/s2 ) = 588 N
which should be reported as 590 N to two significant figures.
Unlike the sidewalk case, there is a component of the displacement parallel to the normal force. The angle between
the force and displacement is cos −1 (4.5/7) = 50°, so d cos(θ ) = 4.5 m. Then
W = Fd cosθ = (588 N)(4.5 m) = 2.6 kJ
(b) Refer to the figure.
Here, the displacement is in the opposite direction compared to part (b), so the angle between the force and the
displacement is now 180° − 50° = 130°. So
W = Fd cosθ = (588 N)(7.0 m)cos(130°) = −2.6 kJ
Assess: In part (a), since the force has no component in the direction of your displacement, the force does no work.
In part (b), there is a component of the force along and in the same direction as the displacement, so the force does
positive work. In part (c), the component of the force along the displacement is in the opposite direction to the
displacement, so the force does negative work.
P10.5. Prepare: Equation 10.6 is the definition of work when the force and displacement are not parallel, as is the
case in this problem.
Solve: (a) The boy is standing still in this case, so the displacement is zero. W = Fd cos(θ ) = (F )(0 m)cos(θ ) = 0 J.
The work done on the boy by the string is exactly zero Joules.
(b) The displacement is non-zero in this case, so we expect the work done to be non-zero. Refer to the following
figure. The angle between the force and the displacement is 180° − 30° = 150°.
The work is
W = Fd cos(θ ) = (4.5 N)(11 m)cos(150°) = −43 J
(c) The angle between the force and displacement in this case is 30° (Look at the figure and imagine the direction of
the displacement vector is reversed). The work is
W = Fd cos(θ ) = (4.5 N)(11 m)cos(30°) = 43 J
Assess: For there to be work done, the displacement must not be zero. If there is no displacement there is no work
done. Note that the answers to parts (b) and (c) have opposite signs. This is because the displacement is exactly
opposite in those cases for the same direction of the force.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Energy and Work
10-9
P10.6. Prepare: Equation 10.6 is the definition of work when the force and displacement are not parallel, as is the
case in this problem.
Solve: Because Paige pushes back the angle between the force and the displacement is actually 160°.
W = Fd cosθ = (68 N)(3.5 m)cos160 = −220 N
The amount of work Paige did is therefore 220 N.
Assess: The cosine of the angle was negative showing that the angle was greater than 90 degrees.
P10.7. Prepare: The kinetic energy for any object moving of mass m with velocity v is given in Equation 10.8:
1
K = mv 2 .
2
Solve: For the bullet,
KB =
1
1
m v 2 = (0.010 kg)(500 m /s)2 = 1.3 kJ
2 B B 2
For the bowling ball,
K BB =
1
1
m v 2 = (10 kg)(10 m/s) 2 = 0.50 kJ
2 BB BB 2
Thus, the bullet has the larger kinetic energy.
Assess: Kinetic energy depends not only on mass but also on the square of the velocity. The previous calculation
shows this dependence. Although the mass of the bullet is 1000 times smaller than the mass of the bowling ball, its
speed is 50 times larger, which leads to the bullet having over twice the kinetic energy of the bowling ball.
P10.8. Prepare: Use the definition of kinetic energy, Equation 10.8, to set up an equation such that the kinetic
energy of the car is equal to that of the truck.
Solve: For the kinetic energy of the compact car and the kinetic energy of the truck to be equal,
1
1
mT
20 000 kg
K C = K T ⇒ mCvC2 = mT vT2 ⇒ vC =
vT =
(25 km/h) = 110 km/h
2
2
mC
1000 kg
To match the kinetic energy of the truck, the car needs a velocity of 110 km/h (to two significant figures).
Assess: Note that the smaller mass needs a greater velocity for its kinetic energy to be the same as that of the larger
mass. Though the truck has 20 times the mass, the car only needs about four times the velocity of the truck to have
the same kinetic energy. This is because kinetic energy is proportional to the mass, but proportional to the square of
the velocity.
P10.9. Prepare: In order to work this problem, we need to know that the kinetic energy of an object is given by
K = mv 2 /2.
Solve: The problem may be solved in a qualitative manner or in a quantitative manner. Since some students think one
way and some the other, we will use both methods.
(a) First, in a qualitative manner. Since the kinetic energy depends on the square of the speed, the kinetic energy will
be doubled if the speed is increases by a factor of
2. This is true because ( 2 ) 2 = 2. Then the new speed is
2 (10 m / s) = 14 m / s.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10-10
Chapter 10
Second, in a more quantitive manner. Use a subscript 1 for the present case where the speed is 10 m/s and a subscript
2 for the new case where the speed is such that the kinetic energy is doubled.
K1 = mv12 /2 and K 2 = mv22 /2
We want
K 2 = 2K1
Inserting expressions for K1 and K 2 obtain
mv22
mv 2
= 2 1 or v2 = 2v1 = 14 m/s
2
2
(b) First, in a qualitative manner, if the speed is doubled and the kinetic energy depends on the square of the speed,
the kinetic energy will increase by a factor of four.
Second, in a more quantitative manner, use a subscript 1 for the present case and a subscript 2 for the new case where
the speed is doubled.
K1 = mv12 /2 and K 2 = mv22 /2
We want v2 = 2v1 . Inserting v2 into K 2 , we obtain
⎛ mv 2 ⎞
mv22 m(2v1 ) 2
=
= 4 ⎜ 1 ⎟ = 4 K1
2
2
⎝ 2 ⎠
This expression clearly shows that the kinetic energy is increased by a factor of four when the speed is doubled.
Assess: The key to the problem is to know that the kinetic energy depends on the speed squared. After that, we can
approach the problem in a qualitative or a quantitative manner. You may prefer one method over the other, but you
should be able to work in either mode.
K2 =
P10.10. Prepare: Use the definition of kinetic energy.
Solve:
K = 12 mv 2 = 12 (60 kg)(33 m/s) 2 = 33 kJ
P10.11. Prepare: Use the definition of kinetic energy.
Solve: The man and the bullet have the same kinetic energy.
1
mm vm2 = 12 mb vb2 ⇒
2
vm =
mb
8.0 g
vb =
(400 m/s) = 4.0 m/s
mm
80 kg
Assess: We expected the man to need much less speed than the bullet to have the same kinetic energy.
P10.12. Prepare: We will assume that all the work Sam does goes into stopping the boat. We can use conservation
of energy as expressed in Equation 10.7 to calculate the work done from the change in kinetic energy.
Solve: Refer to the before and after representation of Sam stopping a boat. Equation 10.7 becomes
1
1
mvf 2 − mvi 2 = W
2
2
Since the boat is at rest at the end of the process, vf = 0 m/s. Therefore, the final kinetic energy is zero. The work
done on the boat is then
1
1
W = − mvi 2 = − (1200 kg)(1.2 m/s) 2 = − 0.86 kJ
2
2
The amount of work is 0.86 kJ.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Energy and Work
10-11
Assess: Note that the work done by Sam on the boat is negative. This is because the force Sam exerts on the boat
must be opposite to the direction of motion of the boat to slow it down.
P10.13. Prepare: Use the law of conservation of energy, Equation 10.4, to find the work done on the particle. We
will assume there is no change in thermal energy of the ball.
Solve: Consider the system to be the plastic ball. Since there is no change in potential, thermal or chemical energy of
the ball, and there is no heat leaving or entering the system, the conservation of energy equation becomes
1
1
1
1
W = ΔK = mvf2 − mvi2 = m(vf2 − vi2 ) = (0.020 kg)[(30 m/s)2 − (−30 m/s)2 ] = 0 J
2
2
2
2
Assess: Note that no work is done on the ball in reversing its velocity. This is because negative work is done in
slowing the ball down to rest, and an equal amount of positive work is done in bringing the ball to the original speed
but in the opposite direction.
P10.14. Prepare: The definition of kinetic energy for objects rotating around a stationary axis is given by
Equation 10.9. In Equation 10.9, the rotational velocity should be in units of radians/second.
Solve: The turntable turns once every 4.0 s. So its rotational velocity is
⎛ 1 rev ⎞ ⎛ 2π rad ⎞
= 1.57 rad/s
ω =⎜
⎝ 4.0 s ⎟⎠ ⎜⎝ rev ⎟⎠
This should be reported as 1.6 rad/s to two significant figures. We keep an extra significant figure for substitution in
the next step:
1
1
K rot = Iω 2 = (0.040 kg · m2 )(1.57 rad/s)2 = 0.049 J
2
2
Assess: This is a reasonable result for such a low rotational velocity and moment of inertia.
P10.15. Prepare: Energy is stored in the flywheel by virtue of the motion of the particles and is given by Equation
10.9. In this equation, units for rotational velocity must be rad/s.
Solve: Using Equation 10.9,
1
2K
K rot = I ω 2 , so I = 2rot
ω
2
We need to convert ω to proper units, radians/s. Since ω = 20 000 rev/min and there are 2π rad/rev and
60 s/min,
⎛
ω = ⎜ 20 000
⎝
rev ⎞⎛ 1 min ⎞⎛ 2π rad ⎞
⎟⎜
⎟
min ⎟⎜
⎠⎝ 60 s ⎠⎝ rev ⎠
So.
(2)(4.0 × 106 J)
= 1.8 kg ⋅ m 2
2
⎡⎛
⎤
rev ⎞⎛ 1 min ⎞⎛ 2π rad ⎞
⎢⎜ 20 000 min ⎟⎜ 60 s ⎟⎜ rev ⎟ ⎥
⎠⎝
⎠⎝
⎠⎦
⎣⎝
Assess: The flywheel can store this large amount of energy even though it has a low moment of inertia because of its
high rate of rotation.
I=
P10.16. Prepare: As the height of the hiker changes so does his potential energy. Gravitational potential energy is
given by Equation 10.13. We only need to calculate the difference in potential energy. The reference point for
measuring the hiker’s height is arbitrary.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10-12
Chapter 10
We choose the origin of the coordinate system chosen for this problem at sea level so that the hiker’s position in
Death Valley is y0 = − 85.0 m.
Solve: The hiker’s change in potential energy from the bottom of Death Valley to the top of Mt. Whitney is
ΔU = (U g )f − (U g )i = mgyf − mgyi = mg ( yf − yi )
= (65.0 kg)(9.80 m/s 2 )[4420 m − (−85.0 m)] = 2.87 × 106 J
Assess: This is a large amount of energy, as expected. Note that ΔU is always independent of the origin of the
coordinate system.
P10.17. Prepare: In part (a) we can simply use the definition of kinetic energy in Equation 10.8. We then use this
result in part (b) to find the height the car must be dropped from to obtain the same kinetic energy. The car is falling
under the influence of gravity. We can use conservation of energy to calculate its kinetic energy as a result of the fall.
The sum of kinetic and potential energy does not change as the car falls.
Solve: (a) The kinetic energy of the car is
1
1
KC = mCvC2 = (1500 kg)(30 m/s)2 = 6.75 × 105 J ≈ 6.8 × 105 J
2
2
We keep one additional significant figure here for use in part (b).
(b) Refer to the diagram.
Here we set K f equal to K C in part (a) and place our coordinate system on the ground at yf = 0 m. At this point,
the car’s potential energy (U g )f is zero, its velocity is vf , and its kinetic energy is K f . At position yi , vi = 0 m/s,
so K i = 0 J, and the only energy the car has is (Ug )i = mgyi . Since the sum K + U g is unchanged by
motion, Kf + (Ug )f = Ki + (Ug )i . This means
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Energy and Work
10-13
K f + mgyf = K i + mgyi ⇒ K f + 0 = Ki + mgyi
( K f − Ki )
(6.75 × 105 J − 0 J)
=
= 46 m
mg
(1500 kg)(9.80 m / s 2 )
(c) To check if this result depends on the car’s mass, rewrite the result of part (b) leaving m as a variable, and check
if it cancels out.
( K − Ki ) 12 mvf2 − 12 mvi2 (vf2 − vi2 )
yi = f
=
=
mg
mg
2g
Since m cancels out, the distance does not depend upon the mass.
Assess: A car traveling at 30 m/s is traveling at 108 km/h or 67 mi/h. At that speed, it has the same amount of energy
as from being dropped 46 m, which is 151 ft, or from the top of an approximately 19 story building!
⇒ yi =
P10.18. Prepare: We need to know how to find kinetic energy ( K = mv 2 /2) and gravitational potential energy
(U g = mgh) and be aware that we want the potential energy to change by an amount equal to the kinetic energy.
Equate the potential energy to the kinetic energy
mv 2
= mgh
2
And solve for h to obtain
v2
(11 m/s)2
h=
=
= 6.2 m
2g 2(9.8 m/s 2 )
Assess: This is as high as a two-story building.
Solve:
P10.19. Prepare: In this case the bike racer will increase his gravitational potential energy by an amount
ΔU g = mgh, where h is the change in his vertical position.
Solve: The change in the bike racer’s gravitation potential energy as he executes the climb is
ΔU g = mgh = mgd sin 4.3° = 63 kJ
Assess: The bike racer would increase his gravitational potential energy this same about by climbing up a flight of
stairs that increased his vertical position by 90 m. This allows one to conclude that stair climbing is excellent
exercise.
P10.20. Prepare: The wrecking ball will increase its gravitational potential energy by an amount ΔU g = mgh,
where h is the change in the vertical position.
Solve: The increase in gravitational potential energy of the wrecking ball is determined by
ΔU g = mgh = mg ( L − L cos 25°) = mgL(1 − cos 25 °) = 14 kJ
Assess: This is a reasonable change in gravitational potential energy for such a massive object.
P10.21. Prepare: Assume an ideal spring that obeys Hooke’s law. Equation 10.15 gives the energy stored in a
spring. The elastic potential energy of a spring is defined as U s = 12 kx 2 , where x is the magnitude of the stretching or
compression relative to the unstretched or uncompressed length. ΔU s = 0 when the spring is at its equilibrium length
and x = 0.
Solve: We have U s = 200 J and k = 1000 N/m. Solving for x:
x=
2U s / k = 2(200 J) / (1000 N/m) = 0.63 m
Assess: In the equation for the elastic potential energy stored in a spring, it is always the distance of the stretching of
compression relative to the unstretched or equilibrium length.
P10.22. Prepare: Assume an ideal spring that obeys Hooke’s law. Equation 10.15 gives the potential energy stored
in an ideal spring. The elastic potential energy of a spring is defined as U s = 12 kx 2 , where x is the magnitude of the
stretching or compression relative to the unstretched or uncompressed length.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10-14
Chapter 10
Solve: We have x = 20 cm = 0.20 m
and
k = 500 N / m. This means
1 2 1
kx = (500 N/m)(0.20 m) 2 = 10 J
2
2
Assess: Since x is squared, U s is positive for a spring that is either compressed or stretched. U s is zero when the
spring is in its equilibrium position.
Us =
P10.23. Prepare: We will assume the knee extensor tendon behaves according to Hooke’s Law and stretches in a
straight line. The elastic energy stored in a spring is given by Equation 10.15, U s = 12 kx 2 .
Solve: For athletes,
U s,athlete =
1 2 1
kx = (33 000 N/m)(0.041 m) 2 = 27.7 J
2
2
For non-athletes,
1 2 1
kx = (33 000 N/m)(0.033 m) 2 = 18.0 J
2
2
The difference in energy stored between athletes and non-athletes is therefore 9.7 J.
Assess: Notice the energy stored by athletes is over 1.5 times the energy stored by non-athletes.
U s,non-athlete =
P10.24. Prepare: Since the floor is horizontal, there is no change in gravitational potential energy. If Marissa drags
the bag across the floor at a constant speed, there is no change in kinetic energy and hence no net work done on the
bag. If this is the case, then WMarissa + Wfriction = 0 and ΔEthermal = WMarissa = −Wfriction
Solve: The thermal energy created by Marissa may be determined by:
ΔEthermal = WMarissa = −Wfriction = − Ffriction d cosθ = − μkinetic Nd cos(180 °) = μ kinetic mgd = 4.7 × 102 J
Assess: If Marissa accomplished this in 30 s (which is a reasonable time) the power delivered would light a 150 watt
light bulb. This is a reasonable amount.
P10.25. Prepare: Since the gravitational potential energy and the kinetic energy of the car do not change, all the
work Mark does on the car goes into thermal energy.
Solve: The thermal energy created in the tires and the road may be determined by:
Δ Et = WMark = FMark d cos 0 ° = (110 N)(150 m)cos0° = 16.5 kJ
Assess: All the work Mark does in pushing the car, becomes thermal energy of the tires and road. Since Mark is
pushing in the direction the car is moving, the angle between the direction of F and d is 50°
P10.26. Prepare: In this case all the work done on the crate by gravity goes into thermal energy. The work done by
gravity depends on the angle between the force of gravity and the displacement of the crate, which is 55 degrees.
Solve: The thermal energy created may be determined by:
ΔEt = Wgravity = Fgravity L cos55 ° = (900 N)(12 m)cos 55 ° = 6.2 kJ
Assess: This seems like a lot of thermal energy, however the crate is heavy (about 200 lbs) and the ramp is long so
we should expect a large number. We could also solve this problem as follows:
Δ ETh = Wgravity = −ΔU gravity = − (− mgh) = Fgravity L sin 35° = Fgravity L cos55 °
P10.27. Reason: The force of gravity and the force of friction are doing work on the child. Since the child slides at
a constant speed, the net work (which is the change in kinetic energy) is zero. This allows us to write:
Wg + W f = Δ K = 0 or W f = −Wg
Knowing how the work done by gravity is related to the change in gravitational potential energy and how the work
done by friction is related to the force of friction, we can determine the force of friction.
Solve: Writing expressions for the work done by friction obtain
W f = −Wg = −(−ΔU g ) = ΔU g = Mgh and W f = F f Lcos180° = − F f L
Combining these and solving for the force of friction obtain
Ff = − Mgh / L = −(25 kg)(9.8 m / s 2 )(3.0 m) /(7.0 m) = −1.0 × 102 N
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Energy and Work
10-15
The minus reminds us that the force of friction opposes the motion of the object, so the magnitude is 100 N.
Assess: A 100 N force of friction for a child sliding down a playground slid is a reasonable number.
P10.28. Prepare: This is a case of free fall, so the sum of the kinetic and gravitational potential energy does not
change as the ball rises and falls.
The figure shows a ball’s before-and-after pictorial representation for the three situations in parts (a), (b) and (c).
Solve: The quantity K + U g is the same during free fall: K f + U gf = Ki + U gi . At the top of its trajectory where the
ball turns around the ball’s velocity is 0 m/s. We have
1
1
(a) mv12 + mgy1 = mv02 + mgy0
2
2
⇒ y1 = (v02 − v12 ) / 2 g = [(10 m/s) 2 − (0 m/s) 2 ]/(2 × 9.80 m/s 2 ) = 5.1 m
5.1 m is therefore the maximum height of the ball above the window. This is 25 m above the ground.
1 2
1 2
(b) mv2 + mgy2 = mv0 + mgy0
2
2
Since y2 = y0 = 0, we get for the magnitudes v2 = v0 = 10 m/s.
(c)
1 2
1
mv3 + mgy3 = mv02 + mgy0 ⇒ v32 + 2 gy3 = v02 + 2 gy0 ⇒ v32 = v02 + 2 g ( y0 − y3 )
2
2
⇒ v32 = (10 m/s) 2 + 2(9.80 m/s 2 )[0 m − (− 20 m)]
Taking the square root, the magnitude of v3 is equal to 22 m/s.
Assess: Note that the ball’s speed as it passes the window on its way down is the same as the speed with which it
was tossed up, but in the opposite direction.
P10.29. Prepare: The only force acting on the ball during its trip is gravity. The sum of the kinetic and
gravitational potential energy for the ball, considered as a particle, does not change during its motion. Use Equation
10.4. Note that at the top of its trajectory when the ball turns around, the velocity of the ball is zero. Assume there is
no friction.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10-16
Chapter 10
The figure shows the ball’s before-and-after pictorial representation for the two situations described in parts (a) and
(b).
Solve: Since energy is conserved, the quantity K + U g is the same during the entire trip. Thus, K f + U gf = Ki + U gi .
(a)
(b)
1 2
1
mv1 + mgy1 = mv02 + mgy0 ⇒ v02 = v12 + 2 g ( y1 − y0 )
2
2
⇒ v02 = (0 m/s)2 + 2(9.80 m/s 2 )(10 m − 1.5 m) = 167 m 2 /s 2 ⇒ v0 = 13 m/s
1 2
1
mv2 + mgy2 = mv02 + mgy0 ⇒ v22 = v02 + 2 g ( y0 − y2 )
2
2
⇒ v22 = 167 m2 /s 2 + 2(9.80 m/s2 )(1.5 m − 0 m) ⇒ v2 = 14 m/s
Assess: An increase in speed from 13 m/s to 14 m/s as the ball falls through a distance of 1.5 m is reasonable. Also,
note that mass does not appear in the calculations that involve free fall since both gravitational potential energy and
kinetic energy are proportional to mass. The mass cancels out in the equations.
P10.30. Prepare: Since the ramp is frictionless, the sum of the puck’s kinetic and gravitational potential energy
does not change during its sliding motion. Use Equation 10.4 for the conservation of energy.
Solve: The quantity K + U g is the same at the top of the ramp as it was at the bottom. The energy conservation
equation K f + U gf = K i + U gi is
1 2
1
mvf + mgyf = mvi2 + mgyi ⇒ vi2 = vf2 + 2 g ( yf − yi )
2
2
⇒ vi2 = (0 m/s) 2 + 2(9.80 m/s 2 )(1.03 m − 0 m) = 20.1 m 2 /s 2 ⇒ vi = 4.5 m/s
Assess: An initial push with a speed of 4.5 m/s ≈ 10 mph to cover a distance of 3.0 m up a 20° ramp seems
reasonable. Note that a ramp of any angle to the same final height would lead to the same final velocity for the puck.
Note that the mass cancels out in the equation since both kinetic energy and gravitational potential energy are
proportional to mass.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Energy and Work
10-17
P10.31. Prepare: The following figure shows a before-and-after pictorial representation of the rolling car. The car
starts at rest from the top of the hill since it slips out of gear. Since we are ignoring friction, energy is conserved. The
total energy of the car at the top of the hill is equal to total energy of the car at any other point.
Solve: The energy conservation equation then becomes
Ki + (U g )i = K f + (U g )f
or
1
1
mvi 2 + mgyi = mvf 2 + mgyf
2
2
The car starts from rest, so vi = 0 m/s, which gives K i = 0 J. Taking the bottom of the hill as the reference point for
gravitational potential, yf = 0 m and so U f = 0 J.
The energy conservation equation becomes
1
mvf 2 = mgyi
2
Canceling m and solving for vf,
vf = 2 gyi = (2)(9.80 m/s 2 )(50 m) = 31 m/s
Assess: Note that the problem does not give the shape of the hill, so the acceleration of the car is not necessarily
constant. Constant acceleration kinematics can’t be used to find the car’s final speed. However, energy is conserved
no matter what the shape of the hill. Note that the mass of the car is not needed. Since kinetic energy and
gravitational potential energy are both proportional to mass, the mass cancels out in the equation. The final speed of
the car, after traveling to the bottom of the 50 m hill is 31 m/s which is nearly 70 mi/h!
P10.32. Prepare: Assume there is zero rolling friction since friction is not mentioned in the problem. The sum of
the kinetic and gravitational potential energy, therefore, does not change during the car’s motion. Consider the other
side of the hill to be the zero for gravitational potential energy.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10-18
Chapter 10
Solve: (a) The initial energy of the car is
1
1
K 0 + U g0 = mv02 + mgy0 = (1500 kg)(10 m/s) 2 = 7.5 × 104 J
2
2
The energy of the car at the top of the hill is
K1 + U g1 = K1 + mgy1 = K1 + (1500 kg)(9.80 m/s 2 )(5.0 m) = K1 + 7.4 × 10 4 J
If the car just wants to make it to the top, then K1 = 0. That is, the car has no velocity at the top of the hill. In other
words, a minimum energy of 7.4 × 104 J is needed to get to the top. Since this energy is less than the available
energy of 7.5 × 104 J, the car can make it to the top.
(b) The conservation of energy equation K 0 + U g0 = K 2 + U g 2 is
7.5 × 104 J =
1 2
1
mv2 + mgy2 ⇒ 7.5 × 104 J = (1500 kg)v22 + (1500 kg)(9.80 m/s 2 )(−5.0 m)
2
2
⇒ v2 = 14 m/s
Assess: A higher speed on the other side of the hill is reasonable because the car has increased its kinetic energy and
lowered its potential energy compared to its starting values. Note that the shape of the hill is irrelevant because
gravitational potential energy depends only on height.
P10.33. Prepare: Consider the spring as an ideal spring that obeys Hooke’s law. We will also assume zero rolling
friction during the compression of the spring, so that mechanical energy is conserved. At the maximum compression
of the spring, 60 cm, the velocity of the cart will be zero.
The figure shows a before-and-after pictorial representation. The “before” situation is when the cart hits the spring in
its equilibrium position. We put the origin of our coordinate system at this equilibrium position of the free end of the
spring. This give x1 = xe = 0 and x2 = 60 cm.
Solve: The conservation of energy equation K 2 + U s 2 = K1 + U s1 is
1 2 1 2 1 2 1 2
mv2 + kx2 = mv1 + kx1
2
2
2
2
Using v2 = 0 m/s, x2 = 0.60 m, and x1 = 0 m gives:
⎛ 250 N/m ⎞
⎛ k ⎞
1 2 1
kx2 = m v12 ⇒ v1 = ⎜
⎟ (0.60 m) = 3.0 m/s
⎟⎟ x2 = ⎜⎜
⎜
2
2
10 kg ⎟⎠
⎝ m⎠
⎝
Assess: Elastic potential energy is always measured from the unstretched or uncompressed length of the spring.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Energy and Work
10-19
P10.34. Prepare: Model the jet plane as a particle, and the spring as an ideal that obeys Hooke’s law. We will also
assume zero rolling friction during the stretching of the spring, so that mechanical energy is conserved.
The figure shows a before-and-after pictorial representation. The “before” situation occurs just as the jet plane lands
on the aircraft carrier and the spring is in its equilibrium position. We put the origin of our coordinate system at the
right free end of the spring. This gives x1 = xe = 0 m. Since the spring stretches 30 m to stop the plane, x2 = 30 m.
Solve: The conservation of energy equation K 2 + U s 2 = K1 + U s1 for the spring-jet plane system is
1 2 1 2 1 2 1 2
mv2 + kx2 = mv1 + kx1
2
2
2
2
Using v2 = 0 m/s, x1 = xe = 0 m, and x2 = 30 m yields
⎛ 60 000 N/m
⎛ k ⎞
1 2 1 2
kx2 = mv1 ⇒ v1 = ⎜
⎟⎟ ( x2 ) = ⎜⎜
⎜
2
2
⎝ m⎠
⎝ 15 000 kg
Assess: A landing speed of 60 m/s or ≈ 130 mph is reasonable.
⎞
⎟⎟ (30 m) = 60 m/s
⎠
P10.35. Prepare: This is a case of free fall, so the sum of the kinetic and gravitational potential energy does not
change as the rock is thrown. Assume there is no friction. The direction the rock is thrown is not known.
The coordinate system is put on the ground for this system, so that yf = 16 m. The rock’s final velocity vf must be at
least 5.0 m/s to dislodge the Frisbee.
Solve: The energy conservation equation for the rock K f + U gf = K i + U gi is
1 2
1
mvf + mgyf = mvi2 + mgyi
2
2
This equation involves only the velocity magnitudes and not the angle at which the rock is to be thrown to dislodge
the Frisbee. This equation is true for all angles that will take the rock to the Frisbee 16 m above the ground and
moving with a speed of 5.0 m/s.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10-20
Chapter 10
Using the previous equation we get
vf2 + 2 gyf = vi2 + 2 gyi
vi = vf2 + 2 g ( yf − yi ) = (5.0 m/s)2 + 2(9.80 m/s 2 )(16 m − 2.0 m) = 17 m/s
Assess: Kinetic energy is defined as K = 12 mv 2 and is a scalar quantity. Scalar quantities do not have directional
properties. Note also that the mass of the rock is not needed.
P10.36. Prepare: We will use conservation of energy, Equation 10.4, to calculate the increase in thermal energy.
Assume the initial velocity of the fireman to be zero.
Solve: Consider the figure. Using ground level as the reference for gravitational potential energy, the conservation of
energy equation becomes
ΔK + ΔU g + ΔEth = 0 ⇒ ΔEth = −(ΔK + ΔU g )
The change in his gravitational potential energy is
Δ U g = mgyf − mgyi = 0 − (80 kg)(9.80 m/s 2 )(4.2 m) = − 3.3 kJ
The change in his kinetic energy is
ΔK = K f − Ki = 0 J −
1
(80 kg)(2.2 m/s) 2 = 0.19 kJ
2
So
Δ Eth = − (190 J − 3300 J) = +3.1 kJ.
Assess: Note that most of the gravitational potential energy is converted to thermal energy.
P10.37. Prepare: The thermal energy of the slide and the child’s pants changes during the slide. If we consider the
system to be the child and slide, total energy is conserved during the slide. The energy transformations during the
slide are governed by the conservation of energy equation, Equation 10.4.
Solve: (a) The child’s kinetic and gravitational potential energy will be changing during the slide. There is no heat
entering or leaving the system, and no external work done on the child. There is a possible change in the thermal
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Energy and Work
10-21
energy of the slide and seat of the child’s pants. Use the ground as reference for calculating gravitational potential
energy.
1
K i = K 0 = mv02 = 0 J U i = U g 0 = mgy0 = (20 kg)(9.80 m/s 2 )(3.0 m) = 590 J
2
1
1
W = 0 J K f = K1 = mv12 = (20 kg)(2.0 m/s)2 = 40 J U f = U g1 = mgy1 = 0 J
2
2
At the top of the slide, the child has gravitational potential energy of 590 J. This energy is transformed partly into the
kinetic energy of the child at the bottom of the slide. Note that the final kinetic energy of the child is only 40 J, much
less than the initial gravitational potential energy of 590 J. The remainder is the change in thermal energy of the
child’s pants and the slide.
(b) The energy conservation equation becomes Δ K + Δ U g + ΔEth = 0. With Δ U g = − 590 J and ΔK = 40 J, the
change in the thermal energy of the slide and of the child’s pants is then 590 J − 40 J = 550 J.
Assess: Note that most of the gravitational potential energy is converted to thermal energy, and only a small amount
is available to be converted to kinetic energy.
P10.38. Prepare: The only force doing work on the puck is friction. Knowing that the change in kinetic energy of the
puck is equal to the work done on the puck, we can determine how far the puck will slide before coming to rest.
Solve: The change in kinetic energy of the puck is
ΔK = K f − K i = − mvi2 /2
The work done on the puck by the force of friction is
Wf = Ff d cos180° = − μ Nd = − μ mgd
Knowing that
Wf = ΔK
obtain
− μmgd = − mvi2 /2
or
d = vi2 /2μ g = (5.0 m/s)2 /2(0.05)(9.8 m/s2 ) = 26 m
Assess: This is a reasonable distance for a puck traveling 5 m/s to slide across the ice before coming to rest.
P10.39. Prepare: Call the system the tube, rider, and slope (but not the rope or rope puller). The snow is not
frictionless. Assume the tow rope is parallel to the slope. Use the work-energy equation and W = Fd .
Solve: Since there are no springs or chemical reactions involved the work-energy equation is
W = ΔK + ΔU g + ΔEth
We are told the towing takes place at a constant speed so ΔK = 0. Solve for the change in thermal energy.
ΔEth = W − ΔU g = Fd − mg Δy = (340 N)(120 m) − (80 kg)(9.8 m/s 2 )(30 m) = 17 kJ
Assess: W is the work done on the system by the rope.
P10.40. Prepare: Call the system the bike, rider. Use the work-energy equation and W = Fd . Assume the cyclist
and air do not heat up: ΔEth = 0 .
Solve: Since there are no springs or chemical reactions involved the work-energy equation is
K f + (U g )f + ΔEth = Ki + (U g )i + W
1
2
mvf2 + 0 + 0 = 12 mvi2 + mgh − FD L
vf2 = vi2 + 2 gh −
2 FD L
2(12 N)(450 m)
= (12 m/s) 2 + 2(9.8 m/s 2 )(30 m) −
m
70 kg
vf = 24 m/s
Assess: 24 m/s is quite fast (54 mph) for a cyclist, but the drag was small.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10-22
Chapter 10
P10.41. Prepare: This is a one-dimensional collision that obeys the conservation laws of momentum. Since the
collision is perfectly elastic, mechanical energy is also conserved. Equation 10.20 applies to perfectly elastic
collisions.
Solve: Using Equation 10.20,
(v1x )f =
m1 − m2
50 g − 20 g
(v1x )i =
(2.0 m/s) = 0.86 m/s
m1 + m2
50 g + 20 g
(v1x )f =
2m1
2(50 g)
(v1x )i =
(2.0 m/s) = 2.9 m/s
m1 + m2
50 g + 20 g
Assess: These velocities are of a reasonable magnitude. Since both these velocities are positive, both balls move
along the positive x direction. This makes sense since ball 1 is more massive than ball 2 and ball 2 is initially at rest.
P10.42. Prepare: For elastic collisions, Equation 10.20 applies. In this case of a one-dimensional collision, the
momentum conservation law is obeyed whether the collision is perfectly elastic or perfectly inelastic. However, in an
inelastic collision energy is not conserved.
Solve: (a) Using Equation 10.20,
(v1x )f =
m1 − m2
(100 g) − (300 g)
(v1x )i =
(10 m/s) = − 5.0 m/s
m1 + m2
(100 g) + (300 g)
( v2 x ) f =
2m1
2(100 g)
(v1x )i =
(10 m/s) = + 5.0 m/s
m1 + m2
(100 g) + (300 g)
(b) For the perfectly inelastic collision, both balls travel with the same final speed vf . The momentum conservation
equation becomes
(m1 + m2 )vf = m1 (v1 )i + m2 (v2 )i
100 g
)(10 m/s) = 2.5 m/s
100 g + 300 g
Assess: In the case of the perfectly elastic collision, the two balls bounce off each other with a speed of 5.0 m/s and
in opposite directions. This is reasonable, since the smaller mass ball, which is initially moving, hits the larger mass
ball, which is stationary. In the case of the perfectly inelastic collision, the balls stick together and move together at
2.5 m/s which is less than the intial velocity of the smaller mass ball as expected.
⇒ vf = (
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Energy and Work
10-23
P10.43. Reason: For a perfectly inelastic collision, the two collision objects stick together after the collision and
energy is not conserved. Since we are given no information about outside forces acting during the collision, we will
assume there are none and that momentum is conserved. Knowing these two pieces of information we can solve the
problem.
Solve: Conserving momentum we obtain the velocity of the compound object: mv = 2mV or V = v/2
The initial kinetic energy (the kinetic energy of the incident glider) is
K i = mv 2 /2
The final kinetic energy (the kinetic energy of the combined two gliders) is
K f = (2m)V 2 /2 = (2m)(v/2)2 /2 = mv 2 /4
The final kinetic energy is some fraction f of the initial kinetic energy or K f = fK i
Solving for the fraction f , obtain
K f (mv 2 /4) 1
=
=
Ki (mv 2 /2) 2
Knowing that the final kinetic energy is one-half the initial kinetic energy, we may conclude that one-half the first
glider’s kinetic energy is transformed into thermal energy during the collision.
Assess: After a quick first glance at this problem, one might conclude that nothing is given and that the problem can
not be solved. After thinking about the concepts involved, the problem can be solved and a numerical value obtained
even though no values are given.
f =
P10.44. Prepare: Equation 10.20 governs elastic collisions with one object initially at rest.
Solve: (a) Consider the incoming ball to be much more massive than the stationary ball. Then m1 >> m2 .
Consider this in Equation 10.20,
m − m2
(v1x )f = 1
(v1x )i
m1 + m2
(v2 x )f =
2m1
(v1x )i
m1 + m2
If m1 >> m2 , m1 − m2 ≈ m1 and m1 + m2 ≈ m1 so these equations become
(v1x )f ≈ (v1x )i
(v2 x )f ≈ 2(v1x )i
Then
(v1x )f ≈ 200 m/s
(v2 x ) f ≈ 400 m/s
(b) If m1 << m2 , m1 − m2 ≈ − m2 and m1 + m2 ≈ m2 so the equations become
(v1x )f ≈ − (v1x )i
(v2 x )f ≈
m1
m
(v1x )i ≈ 0 m/s since 1 is very small
m2
m2
And then
(v1x )f ≈ − 200 m/s
(v2 x ) f ≈ 0.0 m/s
Assess: These results make sense. In the first case when the incoming ball has the much greater mass, its velocity is
practically unaffected, while the much less massive ball shoots off with a very high velocity. In the second case,
when the initially stationary ball has much more mass, it remains approximately stationary, while the much less
massive incoming ball simply bounces back.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10-24
Chapter 10
P10.45. Prepare: We can use the definition of work, Equation 10.5, to calculate the work you do in pushing the
block. The displacement is parallel to the force, so we can use W = Fd. Since the block is moving at a steady speed,
the force you exert must be exactly equal and opposite to the force of friction.
Solve: (a) The work done on the block is W = Fd where d is the displacement. We will find the displacement
using kinematic equations. The displacement in the x-direction is
d = (1.0 m/s)(3.0 s) = 3.0 m
We will find the force using Newton’s second law of motion. Consider the preceding diagram.
The equations for Newton’s second law along the x and y components are
( F ) y = n − w = 0 N ⇒ n = w = mg = (10 kg)(9.80 m/s2 ) = 98.0 N
( F ) x = F − f k = 0 N ⇒ F = f k = μk n = (0.60)(98 N) = 58.8 N
⇒ W = Fd = (58.8 N)(3.0 m) = 176 J, which should be reported as 1.8 × 102 J to two significant figures.
An extra significant figure has been kept in intermediate calculations.
(b) The power required to do this much work in 3.0 s is
W 176 J
=
= 59 W
P=
3.0 s
t
Assess: This seems like a reasonable amount of power to push a 10 kg block at 1.0 m/s. Note that this power is
almost what a standard 60 W lightbulb requires!
P10.46. Prepare: We can apply conservation of energy to calculate the work done and then use this to calculate
the power supplied by the motor using Equation 10.22.
Solve: The tension in the cable does work on the elevator to lift it. Because the cable is pulled by the motor, we say
that the motor does the work of lifting the elevator.
(a) The energy conservation equation is K i + U i + W = K f + U f + Δ Eth . Using K i = 0 J, K f = 0 J, and Δ Eth = 0 J gives
W = (U f − U i ) = mg ( yf − yi ) = (1000 kg)(9.80 m/s 2 )(100 m) = 9.8 × 105 J
(b) The power required to give the elevator this much energy in a time of 50 s is
W 9.8 × 105 J
=
= 2.0 × 104 W
P=
50 s
Δt
Assess: Since 1 horsepower (hp) is 746 W, the power of the motor is 27 hp. This is a reasonable amount of power to
lift a mass of 1000 kg to a height of 100 m in 50 s.
P10.47. Prepare: The work done on the car while it is accelerating from rest to the final speed is the change in kinetic
energy. Knowing the work done and the time to do this work we can determine the power associated with this work.
Solve: The change in kinetic energy of the car is
1
1
W = Δ K = K f − K i = mvf 2 = (1000 kg)(30 m/s) 2 = 4.5 × 105 J
2
2
since the initial kinetic energy is zero.
The power associated with this work is
W 4.5 × 105 J
P=
=
= 45 kW
Δt
10 s
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Energy and Work
10-25
Assess: This is reasonable. In most cars only a small fraction of the work done by the engine goes into propelling the
car.
P10.48. Prepare: By conservation of energy, we are aware that the work done on the spring is equal to its change
in elastic potential energy. Knowing the work done on the spring and the time to do this work we can determine the
power associated with this work.
Solve: The work done by you on the spring is
1
1
W = Δ U s = k ( x 2f − xi2 ) = (5000 N/m)[(0.040) 2 − (0) 2 ] = 4.0 J
2
2
The power associated with this work is
W
4.0 J
=
= 13 W
P=
Δ t 0.30 s
Assess: This is a reasonable amount of power to compress a spring this relatively short distance.
P10.49. Prepare: Use the definition of power and solve for the time interval.
Solve: The change in energy is the change in gravitational potential energy.
ΔE
Δ E mg Δ y (85 kg)(9.8 m/s 2 )(1100 m)
P=
⇒ Δt =
=
=
= 2036 s ≈ 34 min
Δt
P
P
450 W
Assess: This is a reasonable length of time for a cylcist in a climbing stage in the Tour de France.
P10.50. Prepare: Since the car is traveling at a constant speed, the force that the car’s engine provides to move the
car forward must equal the total force opposing the car’s motion. We will use Equation 10.23 to calculate the power,
since we are given the velocity of the car and will calculate the force.
Solve: (a) Refer to the diagram. Since the force provided by the engine must equal the drag force,
Fengine = Fdrag = 500 N
(b) The power is
P = Fv = (500 N)(23 m/s) = 1.2 × 104 W
Refer to the diagram. Now in addition to the drag force, there is a component of the weight of the car that is opposing the
motion of the car. From the diagram, the component of the weight which is opposite the direction of motion of the car is
Fgravity, along the slope = mg sin(θ ) = (710 kg)(9.80 m/s 2 )sin(2°) = 243 N
The total force the engine must overcome is now 500 N + 243 N = 743 N.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10-26
Chapter 10
The power required from the engine for this is
P = Fv = (743 N)(23 m/s) = 1.7 × 104 W
Assess: Note that this form of the definition of power is more convenient to use since the velocity is given.
P10.51. Prepare: The two forces acting on the elevator are its weight and the force F due to the motor. Since the
elevator is moving with constant velocity, the net force on the elevator is zero.
Solve: Since the net force on the elevator is zero, F + w = 0 N. So
F = − w = 2500 N
The power due to this force acting on the elevator moving with constant velocity can be calculated using Equation 10.23.
P = Fv = (2500 N)(8.0 m/s) = 2.0 × 104 W
Assess: One horsepower (hp) is 746 W, so the power of the motor is 26.8 hp. This is a reasonable amount of power
to lift an elevator.
P10.52. Prepare: Since the sofa is moving with constant velocity, the net force on it is zero.
Solve: Since the net force on the sofa is zero, in the vertical direction n − w = 0 N ⇒ n = mg. Now in the horizontal
direction
FScott − f k = 0 ⇒ FScott = fk = μk n = μk mg
The work Scott does on the sofa is his pushing force multiplied by the distance moved (assumed in the same
direction as the push).
WScott = FScott d = μ k mgd = (0.23)(80 kg)(9.8 m/s 2 )(2.0 m) = 360 J
Assess: Scott could push with a force of 360 N.
P10.53. Prepare: The elevator is not moving with constant velocity, so there is a non-zero net force. The tension in
the cable is greater than the weight of the elevator.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Energy and Work
10-27
Solve: We need to find the tension in the cable and use that in the work equation
W = Fd = Td = m(a + g )d = (550 kg)(1.2 m/s2 + 9.8 m/s2 )(15 m) = 91 kJ
Assess: This seems to be a reasonable amount of work on a large elevator.
P10.54. Prepare: Use the definitions of kinetic energy (Equation 10.8), gravitational potential energy (Equation
10.13), work (Equation 10.5), and conservation of energy (Equation 10.3) in this problem.
Solve: (a) The work done by the force can be calculated with Equation 10.5 since the force is parallel to the
displacement of the box.
W = Fd = (10 N)(2.0 m) = 20 J
(b) We know the box starts from rest, so K i = 0 J. We are given that the speed of the box at the top of the ramp is
0.80 m/s, so we can calculate the change in kinetic energy.
1
1
ΔK = Kf − Ki = mvf 2 = (2.3 kg)(0.80 m/s)2 = 0.74 J
2
2
(c) Taking the reference for gravitational potential energy to be the bottom of the ramp, U i = 0 J. We need to find
the final height of the box to calculate the final gravitational potential energy. Refer to the preceding diagram. Since
the ramp is 2.0m long and at an angle of 17° the final height of the box is
yf = (2.0 m)sin(17 °) = 0.58 m
So the change in gravitational potential energy is
ΔUg = (Ug )f − (Ug )i = mgyf = (2.3 kg)(9.80 m/s2 )(0.58 m) = 13 J
(d) From the conservation of energy equation we have
Δ K + Δ U g + ΔEth = W ⇒ ΔEth = W − Δ K − Δ U g = 20 J − 0.74 J − 13 J = 6.3 J
Assess: Note that much of the work goes into overcoming gravity and the friction in the ramp, giving a relatively
small increase in kinetic energy.
P10.55. Prepare: Assuming that the track offers no rolling friction, the sum of the skateboarder’s kinetic and
gravitational potential energy does not change during his rolling motion.
The vertical displacement of the skateboarder is equal to the radius of the track.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10-28
Chapter 10
Solve: (a) The quantity K ⋅ U g is the same at the upper edge of the quarter-pipe track as it was at the bottom. The
energy conservation equation K f + U gf = Ki + U gi is
1 2
1
mvf + mgyf = mvi2 + mgyi ⇒ vi2 = vf2 + 2 g ( yf − yi )
2
2
vi2 = (0 m/s) 2 + 2(9.80 m/s 2 )(3.0 m − 0 m) = 58.8 m/s ⇒ vi = 7.7 m/s
(b) If the skateboarder is in a low crouch, his height above ground at the beginning of the trip changes to 0.75 m. His
height above ground at the top of the pipe remains the same since he is horizontal at that point. Following the same
procedure as for part (a),
1 2
1
mvf + mgyf = mvi2 + mgyi ⇒ vi2 = vf2 + 2 g ( yf − yi )
2
2
vi2 = (0 m/s) 2 + 2(9.80 m/s 2 )(3.0 m − 0.75 m) = 44.1 m/s ⇒ vi = 6.6 m/s
Assess: Note that we did not need to know the skateboarder’s mass, as is the case with free-fall motion. Note that the
shape of the track is irrelevant.
P10.56. Prepare: We will take the system to be the flea plus the earth. Right after the flea jumps, it has kinetic
energy. This is transformed to potential energy and thermal energy as it moves upward.
Solve: (a) If there is no air resistance, none of the initial kinetic energy is transformed to thermal energy, so we can
write K i + U i = K f + U f . We want the initial point to be when the flea takes off; at this point, the kinetic energy
K i = 12 mv 2 is a maximum. We take the zero of potential energy to be at the ground, so U i = 0. We will take the final
point to be the highest point of the motion. Here, K f = 0, and the potential energy U f = mgh is a maximum.
Solving, we find
K = 1 mvi2 = mgh = (5.0 × 10−4 kg)(9.80 m / s 2 )(0.40 m) = 2.0 mJ
2
(b) If some of the initial kinetic energy is transformed to thermal energy, the final potential energy, and the final
height, will be less. When air resistance is a factor, there is a loss to thermal energy and the final height is half the
height with no air resistance. As potential energy is proportional to height above the ground, half the energy, 50%, is
lost to thermal energy. So only half the initial kinetic energy has been converted to potential energy.
Assess: As shown here, air resistance and wind speed are a real concern for a flea.
P10.57. Prepare: We will take the system to be the car and the road. We can use the equation for an isolated
system with friction. We’ll call the stopping point yf = 0.
Solve:
K f + (U g )f + ΔEth = Ki + (U g )i
mvf2 + mgyf + f k d = 12 mvi2 + mgyi
The car comes to a stop, so the final kinetic energy is zero. The final potential energy is also zero. From the triangle
yi = d sin θ .
1
2
0 + 0 + f k d = 12 mvi2 + mgd sin θ
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Energy and Work
10-29
Solve for d.
f k d − mgd sin θ = 12 mvi2 ⇒
2
1
mvi2
2 (1500 kg)(20 m/s)
=
= 28 m
4
f k − mg sin θ (1.2 × 10 N) − (1500 kg)(9.8 m/s 2 )sin 5
Assess: 28 m seems like a reasonable stopping distance from that speed.
d=
1
2
P10.58. Prepare: Assume the chain to be massless. In the absence of frictional and air-drag effects, the sum of the
kinetic and gravitational potential energy does not change during the swing’s motion.
Solve: The quantity K + U g is the same at the highest point of the swing as it is at the lowest point. That is,
K 0 + U g0 = K1 + U g1. The preceding equation is
1 2
1
mv0 + mgy0 = mv12 + mgy1 ⇒ v12 = v02 + 2 g ( y0 − y1 )
2
2
⇒ v12 = (0 m/s) 2 + 2 g ( y0 − 0 m) ⇒ v1 = 2 gy0
We see from the pictorial representation that
y0 = L − L cos 45 ° = (3.0 m) − (3.0 m)cos 45 ° = 0.88 m
⇒ v1 = 2 gy0 = 2(9.80 m/s 2 )(0.88 m) = 4.2 m/s
Assess: We did not need to know the swing’s or the child’s mass. Also, a maximum speed of 4.2 m/s is reasonable.
P10.59. Prepare: We will need to use Newton’s laws here along with the definition of work (Equation 10.5).
Assume you lift the box with constant speed.
Solve: (a) You lift the box with constant speed so the force you exert must equal the weight of the box. So
F = mg = (20 kg)(9.80 m/s 2 ) = 196 N. The work done by this force is then W = Fd = (196 N)(1.0 m)=196 J
which should be reported as 0.20 kJ to two significant figures
(b)
Refer to the preceding diagram. Since the box moves at constant speed, the force that is required to push the box up
the ramp must exactly equal the component of the gravitational force along the slope.
F = mg sin(θ ) = (20 kg)(9.80 m/s 2 )sin(30 °) = 98 N
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10-30
Chapter 10
(c) Since the height of the ramp is 1.0 m and the angle of the ramp is 30 , the length of the ramp is the length of the
hypotenuse in the diagram, which is
y
1.0 m
=
= 2.0 m
y = L sin θ ⇒ L =
sin θ sin30
(d) We will use the result of parts (b) and (c) here. The force is parallel to the displacement of the block, so we can
use Equation 10.5 again. The work done by the force to push the block up the ramp is
W = Fd = (98 N)(2.0 m) = 196 J which should be reported as 0.20 kJ to two significant figures.
This is exactly the same result as part (a), where the block is lifted straight up.
Assess: We could have expected that the answers to parts (d) and (a) would be the same. In both cases the force we exert
opposes gravity. We know that gravitational potential energy depends only on the change in height of an object, and not the
exact path the object follows to change its height. Note that the answer doesn’t even depend on the shape of the ramp.
P10.60. Prepare: This is case of free-fall projectile motion, so the sum of the kinetic and gravitational potential
energy does not change as the cannon ball falls.
The figure shows a before-and-after pictorial representation. To express the gravitational potential energy, we put the
origin of our coordinate system on the ground below the fortress.
Solve: Using yf = 0 and the equation K i + U gi = K f + U gf we get
1 2
1
mvi + mgyi = mvf2 + mgyf ⇒ vi2 + 2 gyi = vf2
2
2
vf = vi2 + 2 gyi = (80 m/s) 2 + 2(9.80 m/s 2 )(10 m) = 81 m/s
Assess: Note that we did not need to use the tilt angle of the cannon, because kinetic energy is a scalar. Also note
that using the energy conservation equation, we can find only the magnitude of the final velocity, not the direction of
the velocity vector. Note that this method is much easier than using kinematics to calculate the final speed of the ball.
P10.61. Prepare: Since the hill is frictionless, mechanical energy will be conserved during the sledder’s trip. To
make it over the next hill, the sledder’s velocity must be greater than or equal to zero at the top of the hill. The
minimum velocity the sledder can have at the top of the second hill is 0 m/s to just make it over. The corresponding
velocity at the top of the initial hill will be the minimum the sledder needs to just make it over the next hill.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Energy and Work
10-31
Solve: Consider the before and after pictorial representation. We will use the sledder’s initial height as the reference
for gravitational potential energy. Since there is no friction, the conservation of energy equation, Equation 10.4 reads
1 2
1
mvi + mgyi = mvf2 + mgyf ⇒ vi2 = 2 gyf
2
2
vi = 2 gyf = 2(9.80 m/s 2 )(1.7 m) = 5.8 m/s
Where we have used yi = 0 m, and vf = 0m/s for the sledder to just make it over the second hill. Note that since we
are using the top of the first hill as the reference of gravitational potential energy, we must use the height of the top of
the second hill above the first for yf , yf = 5.9 m − 4.2 m = 1.7 m.
Assess: Note the shape of the hill doesn’t matter, only the difference in height between the first and second hill is
needed, as expected for gravitational potential energy. Since the second hill is higher than the first, we expect that the
sledder needs the additional kinetic energy at the initial hill to make up for the additional potential energy needed at
the top of the second hill.
P10.62. Prepare: Assume an ideal spring that obeys Hooke’s law. The mechanical energy K + U s + U g is
conserved during the launch of the ball and during the ball’s drop to the floor. However, since we are not given the
ball’s final velocity before it hits the floor, we will use kinematics to calculate the ball’s velocity when it leaves the
spring.
This is a two-part problem. In the first part, we use projectile equations to find the ball’s velocity v2 as it leaves the
spring. This will yield the ball’s kinetic energy as it leaves the spring. Using conservation of energy for the second
part of the trip, we can find the ball’s final velocity when it hits the floor.
Solve: Using the equations of kinematics,
v2t2 = x2 = 5.0 m ⇒ t2 = (5.0 m/ v2 )
This is an equation that gives the time of flight of the ball in terms of the unknown velocity of the ball as it leaves the
spring. Use the height of the table as the origin of the coordinate system for this part of the problem. Since the
velocity of the ball has no vertical component when it leaves the spring,
1
y2 = − a y t 2 2
2
1
− 1.5 m = − (9.80 m / s 2 )(t2 ) 2
2
2
⎛ 5.0 m ⎞
Substituting the value for t3 , − 1.5 m = −(4.90 m/s ) ⎜
⎟
⎝ v2 ⎠
⇒ v2 = 9.0 m /s
2
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10-32
Chapter 10
This is the velocity of the ball as it leaves the spring.
The conservation of energy equation for the part of the motion when the spring is launching the ball is
K 2 + U s 2 = K1 + U s1 is
1 2 1
1
1
mv2 + k (0 m)2 = mv12 + k (Δx)2
2
2
2
2
Using v1 = 0 m/s, Δ x = 0.2 m, we get
1 2
1
mv2 + 0 J = 0 J + k ( Δ x ) 2 ⇒ ( Δ x) 2 k = mv22
2
2
(0.20 m) 2 k = (0.020 kg)(9.0 m/s) 2 ⇒ k = 41 N/m
Assess: Note that we can ignore gravitational potential energy because the spring is horizontal.
P10.63. Prepare: Assume an ideal spring that obeys Hooke’s law. There is no friction, and therefore the
mechanical energy K + U s + U g is conserved. At the top of the slope, as the ice cube is reversing direction, the
velocity of the ice cube is 0 m/s.
The figure shows a before-and-after pictorial representation. We have chosen to place the origin of the coordinate
system at the position where the ice cube has compressed the spring 10.0 cm. That is, y0 = 0.
Solve: The energy conservation equation K 2 + U s2 + U g 2 = K 0 + U s0 + U g0 is
1 2 1
1
1
mv2 + k ( xe − xe )2 + mgy2 = mv02 + k ( x − xe )2 + mgy0
2
2
2
2
Using v2 = 0 m/s, y0 = 0 m, and v0 = 0 m/s,
mgy2 =
1
k ( x − xe ) 2
(25 N/m)(0.10 m) 2
k ( x − xe ) 2 ⇒ y2 =
=
= 25.5 cm
2
2 mg
2(0.050 kg)(9.80 m/s 2 )
The distance traveled along the incline is y2 / sin 30° = 51 cm.
Assess: The net effect of the launch is to transform the potential energy stored in the spring into gravitational
potential energy. The block has kinetic energy as it comes off the spring, but we did not need to know this energy to
solve the problem since energy is conserved during the whole process.
P10.64. Prepare: We will take the system to be the person plus the earth. When a person drops from a certain
height, the initial potential energy is transformed to kinetic energy. When the person hits the ground, if they land
rigidly upright, we assume that all of this energy is transformed into elastic potential energy of the compressed leg
bones. The maximum energy that can be absorbed by the leg bones is 200 J; this limits the maximum height.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Energy and Work
10-33
Solve: (a) The initial potential energy can be at most 200 J, so the height h of the jump is limited by mgh = 200 J
For m = 60 kg, this limits the height to
h = 200 J/mg = 200 J/(60 kg)(9.80 m/s2 ) = 0.34 m
(b) If some of the energy is transformed to other forms than elastic energy in the bones, the initial height can be
greater. If a person flexes her legs on landing, some energy is transformed to thermal energy. This allows for a
greater initial height.
Assess: There are other tissues in the body with elastic properties that will absorb energy as well, so this limit is
quite conservative.
P10.65. Prepare: This is a two-part problem. The slide is frictionless, so mechanical energy is conserved. We will
calculate the final velocity of the people as they exit the slide and then use that result to calculate how far they travel
from the exit before they hit the water.
Solve: Refer to the diagram. Setting the reference for gravitational potential energy to be zero at the bottom of the
slide, the energy conservation equation becomes
1
mgy1 = mv12 ⇒ v1 = 2 gy1 = 2(9.80 m/s2 )(3.0 m) = 7.67 m/s
2
Note that this result does not depend on the person’s mass. We keep an additional significant figure here for the
second part of the calculation.
After they leave the slide, they are falling under the influence of gravity. Their initial velocity in the y direction is
zero. The time it takes for them to fall to the water can be calculated with ordinary kinematics.
1
Δ y = v0 y Δ t − aΔt 2 , with v0y = 0 m/s gives
2
1
Δ y = − 1.2 m= − (9.80 m/s 2 )Δ t 2 or
2
(2)(1.2 m)
Δt =
= 0.50 s
9.80 m/s 2
Using v1 = 7.67 s from the first part of the problem, we find
Δ x = v1Δ t = (7.67 m/s)(0.50 s) = 3.8 m
The mass of the person was not necessary for this part of the calculation either.
Assess: Though this is a two-part problem mechanical energy is conserved throughout the whole process. However
we could not use conservation of energy to solve the problem since we are not given the final velocity of the person
before they hit the water, which is necessary for the conservation of energy equation. Note that it does not matter
what the mass of the person is, they will always travel 3.8 m from the exit of the tube before hitting the water.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10-34
Chapter 10
P10.66. Prepare: Since there is no friction, total mechanical energy is conserved.
Solve: (a) As she swings, her height above the cliff increases since the rope doesn’t stretch. Her initial kinetic
energy is being converted to gravitational potential energy during the swing.
(b) Refer to the diagram. When she is directly above the opposite side of the ravine, she has moved 3.0 m
horizontally while the rope is also swinging her upwards. Using the Pythagorean theorem, we can find the distance
between the branch and her new height. l = (5.0 m) 2 − (3.0 m) 2 = 4.0 m. Therefore, she is 5.0 m − 4.0 m = 1.0 m
above the cliff.
(c) To make it to the other side of the ravine, she must have enough kinetic energy to be converted to the equivalent
gravitational potential energy of her additional 1.0 m of height. The minimum initial velocity she will need will be
when she just makes it to the other side of the ravine with no kinetic energy (all her initial kinetic energy being
converted to potential energy of her height above the cliff). Using the cliff as the reference for gravitational potential
energy, the conservation of energy equation reads
K1 + (Ug )1 = K2 + Ug (Ug )2
1
mv12 = mgy2
2
v1 = 2gy2 = 2(9.80 m/s2 )(1.0 m) = 4.4 m/s
Assess: We calculated for the case where all her initial kinetic energy is converted to gravitational potential energy
at the other side of the ravine. Note she could start with a greater initial kinetic energy, which would also get her to
the other side of the ravine. In this case, when she’s above the other side of the ravine, she will have some additional
kinetic energy instead of just making it to the other side.
P10.67. Prepare: Assume an ideal spring that obeys Hooke’s law. This is a two-part problem. The first part, when
the bullet embeds itself in the block, is a perfectly inelastic collision. In a perfectly inelastic collision, the momentum
is conserved while energy is not conserved. In the second part of the problem, when the bullet and block hit the
spring, there is no friction. Since there is no friction after the bullet enters the block, the mechanical energy of the
system (bullet + block + spring) is conserved during that part of the motion.
We place the origin of our coordinate system at the end of the spring that is not anchored to the wall.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Energy and Work
10-35
Solve: (a) Momentum conservation for perfectly inelastic collision states pf = pi . This means
⎛ m ⎞
v
(m + M )vf = m(vi ) m + M (vi ) M ⇒ (m + M )vf = mvB + 0 kg m/s ⇒ vf = ⎜
⎝ m + M ⎟⎠ B
where we have used vB for the initial speed of the bullet. This is velocity of the bullet and block after the bullet
embeds itself in the block. Now, when the bullet and block hit the spring and compress it, mechanical energy is
conserved. The mechanical energy conservation equation K1 + U s1 = K e + U se as the bullet-embedded block
compresses the spring is:
1
1
1
1
m(v ′f ) 2 + k ( x1 − xe ) 2 = ( m + M )(vf ) 2 + k ( xe − xe ) 2
2
2
2
2
2
1 2 1
( m + M ) kd 2
⎛ m ⎞ 2
kd = ( m + M ) ⎜
vB + 0 J ⇒ vB =
⎟
⎝m+M ⎠
m2
2
2
(b) Using the preceding formula with m = 5.0 g, M = 2.0 kg, k = 50 N/m, and d = 10 cm,
0 J+
vB =
(0.0050 kg + 2.0 kg)(50 N/m)(0.10 m) 2
= 200 m/s
(0.0050 kg) 2
which should be reported as 2.0 × 102 m/s to two significant figures.
(c) The fraction of energy lost is (initial energy − final energy)/(initial energy), which is
1 2 1
2
2
mvB − (m + M )vf2
m+ M ⎛ vf ⎞
m+M ⎛ m ⎞
2
2
=1−
=1−
⎜
⎟
1 2
m ⎝ vB ⎠
m ⎜⎝ m + M ⎟⎠
mvB
2
m
0.0050 kg
=1−
=1−
= 99.8%
m+M
(0.0050 kg + 2.0 kg)
where we have kept an additional significant figure.
Assess: During the perfectly inelastic collision 99.8% of the bullet’s energy is lost. The energy is transformed into
the energy needed to deform the block and bullet and to the thermal energy of the bullet and block combination.
P10.68.
Prepare: During the collision, as Lisa leaps onto the bobsled, the momentum of the
Lisa + bobsled system is conserved. The collision is perfectly inelastic so energy is not conserved during this part of
the problem. We will assume the ramp to be frictionless, so that after Lisa is on the sled, the mechanical energy of the
system (Lisa + bobsled + spring) is conserved. We will also assume the spring to be an ideal one that obeys
Hooke’s law.
We place the origin of our coordinate system directly below the bobsled’s initial position as in the diagram.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10-36
Chapter 10
Solve: (a) For the first part of the problem, momentum is conserved. Momentum conservation in Lisa’s collision
with the bobsled states p1 = p0 , or
(mL + mB )v1 = mL (v0 ) L + mB (v0 ) B ⇒ (mL + mB )v1 = mL (v0 ) L + 0
⎛ mL ⎞
⎛
⎞
40 kg
⇒ v1 = ⎜
⎟ (v0 ) L = ⎜
⎟ (12 m/s) = 8.0 m/s
⎝ 40 kg + 20 kg ⎠
⎝ mL + mB ⎠
After Lisa is on the bobsled, energy is conserved.
The energy conservation equation: K 2 + U s2 + U g 2 = K1 + U s1 + U g1 is
1
1
1
1
(mL + mB )v22 + k ( x2 − xe )2 + (mL + mB ) gy2 = (mL + mB )v12 + k ( xe − xe )2 + (mL + mB ) gy1
2
2
2
2
Using v2 = 0 m/s, k = 2000 N/m, y2 = 0 m, y1 = (50 m) sin 20° = 17.1 m (keeping an additional significant figure for
this intermediate result), v1 = 8.0 m/s, and (mL ⋅ mB) = 60 kg, we get
1
1
0 J + (2000 N/m)( x2 − xe )2 + 0 J = (60 kg)(8.0 m/s)2 + 0 J + (60 kg)(9.80 m/s2 )(17.1 m)
2
2
Solving this equation yields ( x2 − xe ) = 3.5 m.
(b) As long as the ice is slippery enough to be considered frictionless, we know from conservation of mechanical
energy that the speed at the bottom depends only on the vertical descent Δy. Only the ramp’s height h is important,
not its shape or angle.
Assess: It’s important to note that during the Lisa’s leap onto the sled mechanical energy is not conserved though
momentum is conserved. However, energy is conserved once Lisa is on the sled. Due to this, this problem must be
split into two separate calculations using the two separate principles of momentum and energy conservation.
P10.69. Reason: If we consider the system of interest to be the two masses and the pulley, the only outside force
doing work on the system is gravity. Since gravity is a conservative force, energy is conserved and we may write
ΔK + ΔU g = ΔE = 0 or ΔK = −ΔU g .
Solve: Knowing ΔK = −ΔU g , we may write
( M A + M B )v 2 / 2 = − (− M B gh)
or
v = [2M B gh/( M A + M B )]1/ 2 = [2(4.0 kg)(9.8 m/s2 )(0.50 m)/(12 kg + 4.0 kg)] = 1.6 m / s
Assess: This is a reasonable speed for this situation.
P10.70. Prepare: From conservation of energy we may write ΔK + ΔU g = W f . Knowing expressions for the
change in kinetic energy, the change in gravitational potential energy and the work done by friction, we can
determine the final speed.
Solve: Inserting expressions for ΔK , ΔU g and W f into the earlier statement of conservation of work-energy, obtain
( M A + M B )v 2 / 2 + (− M B gh) = μ M A h cos180 °
Solving for v obtain
v = [2( M B − μM A ) gh/( M A + M B )]1/ 2 = [2(4.0 kg − 0.20(12 kg))(9.8 m/s2 )(0.50 m)/(16 kg)] = 0.99 m/s
Assess: A speed of about 1.0 m/s is a reasonable value.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Energy and Work
10-37
P10.71. Prepare: We can divide this problem into two parts. First, we have an elastic collision between the
20 g ball (m) and the 100 g ball (M). Second, the 100 g ball swings up as a pendulum.
The figure shows three distinct moments of time: the time before the collision, the time after the collision but before
the two balls move, and the time the 100 g ball reaches its highest point. We place the origin of our coordinate
system on the 100 g ball when it is hanging motionless.
Solve: For a perfectly elastic collision, the ball moves forward with speed
2mm
1
(v1 ) M =
(v0 ) m =
(v0 ) m
mm + mM
3.0
In the second part, the sum of the kinetic and gravitational potential energy is conserved as the 100 g ball swings up
after the collision. That is, K 2 + U g 2 = K1 + U g1. We have
1
1
M (v2 )2M + Mgy2 = M (v1 )2M + Mgy1
2
2
Using (v2 ) M = 0 m/s, (v1 ) M =
(v0 ) m
, y1 = 0 m, and y2 = L − L cos θ , the energy equation simplifies to
3.0
1 (v0 ) 2m
g ( L − L cosθ ) =
2 9.0
⇒ (v0 ) m = 18 g L(1 − cosθ ) = 18(9.80 m/s 2 )(1.0 m)(1 − cos50°) = 7.9 m/s
Assess: Since the collision is elastic, mechanical energy is conserved during the whole process. We could apply
conservation of mechanical energy alone to solve this problem. However, solving this particular problem in two parts
using momentum conservation for the first part leads to a simpler calculation.
P10.72. Reason: Since this is a perfectly inelastic collision and there are no significant outside forces acting during
the collision, momentum is conserved and energy is not.
Solve:
(a) From conservation of momentum we may write 2mv = 3mV or V = 2v /3
The final speed (V ) of the three cars is 2/3 the speed (v) of the two cars before collision, or 1.7 m/s.
vf = 32 vi = 32 (2.5 m/s) = 1.7 m/s
(b) The initial kinetic energy is K i = (2m)v 2 / 2 = mv 2
The final kinetic energy is K f = (3M )V 2 /2 = (3m)(2V/3)2 /2 = 2mv 2 /3
Since the final kinetic energy is a fraction f of the initial kinetic energy, we may write
Kf = f Ki
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10-38
Chapter 10
Or solving for f obtain
f = Kf /Ki = (2mv 2 /3)/mv 2 = 2/3
Knowing that the final kinetic energy is 2/3 the initial kinetic energy, we conclude that 1/3 of the initial kinetic
energy is transformed to thermal energy.
Assess: Losing 1/3 the initial kinetic energy to thermal energy is reasonable
P10.73. Prepare: Initially the spring is hanging at its equilibrium position and you hang the fish on the spring, but
don’t release it. Let’s agree to call this position 1 and to make this the point for zero gravitational and elastic potential
energy. For the case where the fish is at position 1, we can write the following:
The position of the fish is y1 = 0
The speed of the fish v1 = 0
The kinetic energy of the fish K1 = 0
The gravitational potential energy U g = 0
1
The elastic potential energy of the spring U S = 0
1
Total energy at this point is E1 = K1 + U g + U S
1
1
When the fish is released it falls to position 2, stretching the spring a maximum amount. For the case where the fish
is at position 2, we can write the following:
The position of the fish is y2 = −h
The speed of the fish v2 = 0
The kinetic energy of the fish K 2 = 0
The gravitational potential energy U g = −mgh
2
The elastic potential energy of the spring U S1 = kh 2 /2
Total energy at this point is E2 = K 2 + U g2 + U S2
Knowing that energy is conserved ( E1 = E2 ), we can determine the maximum distance the fish falls.
Solve: Inserting values for the kinetic energy, gravitational potential energy and the elastic potential energy into
E1 = E2
obtain
0 = − mgh + kh 2 /2
or
h = 2mg / k = 2(5.0 kg)(9.8 m/s 2 ) / 200 N/m = 0.49 m
Assess: This is a reasonable amount for the maximum stretch of the spring.
P10.74. Prepare: We can find the kinetic energies directly from the runner’s masses and final speeds. The power
is the rate at which each runner’s internal chemical energy is converted into kinetic energy, so the average power is
Δ K / Δ t.
1
1
1
1
Solve:(a) We have KS = mSvS2 = (70 kg)(10 m/s)2 = 3500 J, while KG = mGvG2 = (30 kg)(20 m/s)2 = 6000 J (here,
2
2
2
2
S stands for sprinter and G for greyhound).
(b) We have PS = Δ KS / Δ t = (3500 J)/(3.0 s) = 1200 W, and PG = Δ K G / Δ t = (6000 J)/(3.0 s) = 2000 W.
Assess: Although the greyhound has less than half the mass of the human, its final speed is twice as great. And
since kinetic energy depends on the square of the speed, the higher speed of the greyhound means that its kinetic
energy is greater than the human’s. Because 1 hp = 760 W, a human can output about 2 hp and a greyhound about
3 hp in very short bursts.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Energy and Work
10-39
P10.75. Prepare: This is the case of a perfectly inelastic collision. Momentum is conserved because no external
force acts on the system (clay block). Mechanical energy is not conserved during perfectly inelastic collisions.
Solve: (a) The conservation of momentum equation pf x = pi x is
( m1 + m2 )vfx = m1 (vix )1 + m2 (vix ) 2
Using (vix )1 = v0 and (vix ) 2 = 0, we get
m1
0.050 kg
(vix )1 =
(6.5 m/s) = 0.31 m/s
m1 + m2
(1.0 kg + 0.050 kg)
(b) The initial and final kinetic energies are given by
1
1
1
1
2
K i = m1 (vix )12 + m2 (vix ) 22 = (0.050 kg)(6.5 m/s) 2 + (1 kg) ( 0 m/s) = 1.06 J
2
2
2
2
1
1
2
2
K f = ( m1 + m2 )vfx = (1 kg + 0.050 kg)(0.31 m/s) = 0.0504 J
2
2
⎛ Ki − K f ⎞
⎛ 0.0505 J ⎞
The percent of energy lost = ⎜
× 100% = 95%
⎟ × 100% = ⎜1 −
K
1.06 J ⎟⎠
⎝
i
⎝
⎠
The energy goes into the permanent deformation of the ball of clay and into thermal energy.
Assess: Mechanical energy is never conserved during inelastic collisions.
vfx =
P10.76. Prepare: Momentum is conserved in both inelastic and elastic collisions. Mechanical energy is conserved
only in an elastic collision. We will break this problem into separate parts, using conservation of momentum and/or
conservation of energy where appropriate.
Solve: Consider the motion of the first package down the chute. Mechanical energy is conserved during its slide
since the chute is frictionless. For a package with mass m the conservation of energy equation is
1
1
K1 + U g1 = K0 + U g0 ⇒ m(v1 )2m + mgy1 = m(v0 )m2 + mgy0
2
2
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10-40
Chapter 10
Using (v0 ) m = 0 m/s and y1 = 0 m,
1
m(v1 )2m = mgy0 ⇒ (v1 )m = 2 gy0 = 2(9.80 m/s2 )(3.0 m) = 7.67 m/s
2
We will keep an additional significant figure in this intermediate calculation.
(a) For the perfectly inelastic collision the conservation of momentum equation is
pfx = pix ⇒ ( m + 2m)(v2 )3m = m(v1 ) m + (2m)(v1 ) 2m
Using (v1 ) 2m = 0 m/s, we get
(v2 )3m = (v1 ) m / 3 = 2.6 m/s
(b) For the elastic collision, the mass m package rebounds with velocity
m − 2m
1
(v3 )m =
(v1 )m = − (7.67 m/s) = −2.56 m/s
m + 2m
3
Where we keep an additional significant figure in this intermediate result for use later.
The negative sign with (v3 ) m shows that the package with mass m rebounds. It will travel back up the chute and goes
to the position y4 . During this part of its motion, mechanical energy is conserved. We can determine y4 by applying
the conservation of energy equation as follows. For a package of mass m:
1
1
Kf + U gf = Ki + U gi ⇒ m(v4 )2m + mgy4 = m(v3 )m2 + mgy3
2
2
Using (v3 ) m = − 2.56 m/s, y3 = 0 m, and (v4 ) m = 0 m/s, we get
1
mgy4 = m(−2.56 m/s)2 ⇒ y4 = 33 cm
2
Assess: The first mass (m) rebounds up the ramp to a height of 33 cm. This is reasonable since the first package (m)
has given a lot of its energy to the second mass (2m) Note that for part (b), energy is conserved during the whole
process.
P10.77. Prepare: We will use the constant-acceleration kinematic equations and the definition of power in terms
of work, Equation 10.22.
Solve: Refer to the diagram.
(a) We can find the acceleration from the kinematic equations and the horizontal force from Newton’s second law.
We have
1
1
x2 = xo + v0 (t2 − t0 ) + a(t2 − t0 ) 2 ⇒ 50 m = 0 m + 0 m + a(7.0 s − 0 s)2 ⇒ a = 2.04 m/s 2
2
2
⇒ F = ma = (50 kg)(2.04 m/s 2 ) = 102 N
which should be reported as 1.0 × 102 N to two significant figures.
(b) We obtain the sprinter’s power output by using P = WΔ t , where W is the work done by the sprinter. After
t = 2.0 s, the sprinter has moved a distance of d = 12 (2.04 m/s 2 )(2.0 s) 2 = 4.08 m. The work done by the sprinter is
then W = Fd = (102 N)(4.08 m) = 416 J. His power output is then P =
W
Δt
=
416 J
2.0 s
= 208 W, or 210 W to two
significant figures.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Energy and Work
10-41
(c) During the final two seconds of his run, the distance he has moved is given by
1
1
d = x2 − x1 = v1 (t2 − t1 ) + a(t2 − t1 )2 = v1 (2.0 s)+ (2.04 m/s2 )(2.0 s)2
2
2
We need his velocity at 5.0 s after he starts running, which is given by
v1 = v0 + aΔ t = 0 m/s + (2.04 m/s 2 )(5.0 s) = 10.2 m/s
So
1
d = (10.2 m/s)(2.0 s) + (2.04 m/s2 )(2.0 s)2 = 24.5 m
2
The work done by the sprinter is W = Fd = (102 N)(24.5 m) = 2.50 kJ. His power output is
W 2.50 kJ
=
= 1.2 kW
Δt
2.0 s
Assess: Note the power output required for the last two seconds of the sprint is much larger than during the first two
seconds. This is because the sprinter travels a much larger distance during the last two seconds of his trip because he
has accelerated to a high velocity by that time. The force is the same during both time intervals.
P=
P10.78. Prepare: The rock starts from rest. We can use the kinematic equations to find the force. Apply
conservation of energy to calculate the work.
Solve: Refer to the diagram.
(a) The force is constant, so we can use constant acceleration kinematics. Using vf 2 = vi2 + 2aΔx with vi = 0 m/s,
a=
vf 2
(30 m/s) 2
=
= 4.5 × 102 m/s 2
2Δ x (2)(1.0 m)
The force is then F = ma = (0.5 kg)(4.5 × 10 2 m/s 2 ) = 2.3 × 10 2 N
(b) The work done by Bob on the rock is
WBob = Fd = (2.3 × 102 N)(1.0 m) = 2.3 × 102 J
Assess: The numerical value of the force and the work are the same. This is because the force acts over a distance of
1.0 m.
P10.79. Prepare: To go at constant speed the net force must be zero, so the force of the motor must equal the
weight of the elevator (and passengers).
Solve: Use Equation 10.23.
P
15 000 W
P = Fv ⇒ v = =
= 1.3 m/s
F (1200 kg)(9.8 m/s 2 )
Assess: This seems like a reasonable speed for an elevator.
P10.80. Prepare: The heart provides the pressure to move blood through the body and therefore does work on the
blood. We assume all the work goes into pushing the blood through the body.
Solve: (a) Using the hint, W = PAd = PV = (1.3 × 104 N/m2 )(6.0 × 10−3 m3 ) = 78 J (in this equation P represents pressure,
not power).
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10-42
Chapter 10
W 78 J
=
= 1.3 W.
Δt 60 s
Assess: This seems like a reasonable answer, as it is a small fraction of the power required for most human
activities.
(b) P =
P10.81. Prepare: Energy is conserved during the fall of the ball since we are ignoring air resistance. We assume
the ball is dropped from rest.
Solve: Apply conservation of energy. The ball’s initial kinetic energy is 0 J since it’s dropped from rest. Taking the
reference for gravitational potential energy to be the ground, the ball’s final potential energy is 0 J. The conservation
of energy equation,
Kf + (Ug )f = Ki + (Ug )i
becomes
1
mvf 2 = mgyi ⇒ vf = 2 gyi = 2(9.80 m/s2 )(2.5 m) = 7.0 m/s
2
The answer is choice B.
Assess: All the ball’s initial potential energy is converted to kinetic energy here.
P10.82. Prepare: We will need to consider the acceleration after the ball first hits the floor, starts to compress up
to the point at which the ball reaches maximum compression, and is about to rebound.
Solve: During rebound, the ball compresses 6 mm and slows to a speed of 0 m/s 2 . Knowing the initial and final
speed of the ball and the distance over which this compression takes place, we can determine the acceleration. Since
the acceleration is constant, we can use the kinematics we learned in a previous chapter
vf 2 = vi 2 + 2aΔy so
vi 2 = −2aΔy ⇒ a =
−(7.0 m/s) 2
vi 2
=
= 4 × 103 m/s 2
2Δy (2(−0.006 m))
The closest choice is D.
Assess: This is a very large acceleration compared to the acceleration due to gravity. This is due to the large change
in velocity over a very short distance.
P10.83. Prepare: The collision is not elastic. We assume the ball recovers its exact shape from before the collision
after the rebound. Since the ball is dropped, we assume it is not rotating right after release.
Solve: As the ball compresses, its kinetic energy is converted to elastic potential energy. However, when the ball
decompresses this elastic potential energy is recovered since the ball regains its initial shape. During the compression
the ball’s gravitational energy also changes by a slight amount, but this is also recovered as the ball regains its initial
shape. If the ball is dropped straight down, there is no torque available to change the rotational state of the ball when
it hits the ground, so its rotational kinetic energy also remains the same. However, during the collision with the floor,
the thermal energy of the ball (and the floor) increase. This energy is lost from the system as heat.
The answer is C.
Assess: In inelastic collisions, these are generally the type of energy transformations that may occur. In all cases
some energy is converted to thermal energy, which is eventually lost to the environment.
P10.84. Prepare: We will need the results of Problem 10.81 and the ball’s speed after rebounding.
Solve: From Problem 10.81, the ball’s speed before hitting the ground is 7.0 m/s. Its kinetic energy is then
1
1
Kbefore = m(vbefore )2 = (0.0575 kg)(7.0 m/s)2 = 1.4 J
2
2
To find the ball’s speed after rebounding, use the fact that it rebounds to a height of 1.4 m and use conservation of
energy, as in Problem 10.81.
K after = mghafter = (0.0575 kg)(9.80 m/s 2 )(1.4 m) = 0.79 J
The percentage change in energy is given by
K after − K before 0.79 J − 1.4 J
=
= − 0.44 = − 44%
1.4 J
K before
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Energy and Work
10-43
The choice corresponding to this is B.
Assess: It is reasonable that the ball loses nearly half of its energy in rebounding from the floor. The height it
rebounds to is much less than the initial height it was dropped from.
P10.85. Prepare: Since we know that 30% of the initial kinetic energy is lost, we can calculate the final kinetic
energy and then the final velocity of the ball after being hit by a racket.
Solve: We have that 30% of the initial kinetic energy is converted to thermal energy. In equation form this is
K f = K i − (0.30) K i = (0.70) K i
1
1
mvf 2 = (0.70) mvi 2 ⇒ vf = ( 0.70)vi = (0.84)(10 m/s) = 8.4 m/s
2
2
The closest choice is A.
Assess: Note that this situation is a bounce from a stationary racket. Normally the ball collides with a moving racket
and therefore gains kinetic energy.
P10.86. Prepare: Equation 10.23 gives the power in terms of force and velocity.
Solve: Since the cyclist is moving with constant speed, the drag force must equal the force exerted by the cyclist,
assuming no friction from other forces.
Using Equation 10.23, W = Fd = (10 N)(1000 m) = 10 kJ.
The correct choice is B.
Assess: This is a reasonable amount of energy for a person to expend.
P10.87. Prepare: We can use the definition of power in the form of Equation 10.23.
Solve: The drag force on the cyclist is 10 N, and her velocity is 5 m/s. The power she is using to overcome drag is
P = Fv = (10 N)(5 m/s) = 50 W.
The correct choice is B.
Assess: This amount of power would just light a 60 W bulb and could be supplied by a 1/15 hp motor. This helps
one appreciate the efficiency of a bicycle.
P10.88. Prepare: Since the cyclist has sped up, the drag force has increased.
Solve: The drag force is proportional to the square of the velocity. Use k to denote the constant of proportionality.
Then F = kv 2 .
Before we can use this equation to calculate the new drag force, we need to calculate k. We can use the fact that at
v = 5 m/s the force is 10 N. Then
F
10 N
N
k= 2 =
= 0.40
2
v
(5 m/s)
(m/s) 2
With the new velocity,
⎛
N ⎞
F = kv 2 = ⎜ 0.40
(10 m/s) 2 = 40 N
2 ⎟
(m/s)
⎝
⎠
Then the work done is W = Fd = (40.0 N)(1000 m) = 40 kJ.
The correct choice is B.
Assess: Since force is proportional to the square of velocity, as the velocity goes up by a factor of two, and the force
goes up by a factor of four. So we would expect the work to also increase by a factor of four, which is what explicit
calculation shows.
P10.89. Prepare: We can calculate the power using the result of Problem 10.88 and the definition of power given
by Equation 10.23.
Solve: She is cycling at 10 m/s, with a drag force of 40 N. The power expended is then
P = Fv = (40 N)(10 m/s)=400 W
The correct choice is C.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10-44
Chapter 10
Assess: Comparing this to Problem 10.87, we see that when the speed increases by a factor or two, the force
increases by a factor of four and the power (which is the product of the speed and force) increases by a factor of
eight.
P10.90. Prepare: The speed of the wind relative to the rider has changed, so the drag force will change.
Solve: The wind speed is now 5 m/s relative to the ground, and the rider is moving at 5 m/s relative to the ground
and into the wind. The wind velocity relative to the rider is then 10 m/s.
The force due to a relative wind velocity of 10 m/s was calculated in Problem 10.82, and the result is F = 40 N.
Her velocity is now only 5 m/s, so the power she is exerting is p = Fv = (40 N)(5 m/s) = 200 W. The correct choice
is B.
Assess: The force is the same as in Problem 10.89 but the speed is half, as a result we expect the power to be half
that in Problem 10.89 and that is the case.
© Copyright 2015 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.