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ENERGY AND WORK
10
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.
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
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Chapter 10
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. In as much as the earth itself is not an
isolated system (heat can leave the earth) we really should consider the universe the system to consider the
system completely isolated.
Assess: In order to have an isolated system no heat can leave or enter the system and all forces must be internal.
Problems
P10.23. 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:
Eth  WMark  FMark d cos0  (110 N)(150 m)cos0   470 J
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
P10.25. 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  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  (U g )  U g  Mgh and W f  Ff L cos180    Ff 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 10 2 N
The minus reminds us that the force of friction opposes the motion of the object.
Assess: A 100 N force of friction for a child sliding down a playground slid is a reasonable number.
P10.33. 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.
Energy and Work 10-3
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: (a) The energy conservation equation for the rock Kf  U gf  Ki  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.
(b) 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.
P10.35. 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 energy of the slide and seat of the child’s pants. Use the ground as reference for calculating gravitational
potential energy.
Ki  K 0 
W 0 J
1 2
mv0  0 J U i  U g0  mgy0  (20 kg)(9.80 m/s 2 )(3.0 m)  590 J
2
1
1
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.
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(b) The energy conservation equation becomes K  U g  Eth  0. With U g  590 J and K  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.