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Work and Energy
Relevant Equations
Work
W  F prallel  d  Fd cos 
Work-energy theorem
Wnet  KE
Force of a spring
F  kx
Kinetic Energy
KE 
Gravitational potential energy
PE g  mgh
Elastic potential energy of a spring
PE sp 
Power
P
1 2
mv
2
1 2
kx
2
W
 Fv
t
Work
The meaning of the word work in physics is different from the common meaning. In order for work to
be done on an object, a force must cause displacement of the object. Furthermore, the force, or a
component of the force, must be acting parallel to the displacement. Hence, the formula
W  Fparallel  d , which is sometimes written as W  Fd cos , where θ is the angle between the
force and displacement. The parallel component of F is calculated as F cos θ. When multiple forces act
on an object, the net work done on the object is a product of the net force and displacement of the
object.
Wnet  Fnet d cos
Imagine that a crate is pulled by a net force of 10 N, 30⁰ above horizontal for a distance of 10 meters, as
illustrated below.
The work done on the crate is the parallel component of F multiplied by displacement.
W  F paralleld  Fd cos   10 N 10 m  cos 30  86 N  m
David Scott, Honors Physics, Rev. 2014
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In a circumstance in which force is perpendicular to displacement, the force does no work on the object.
Wnet  Fnet d cos 90  Fd  0  0 N  m
Units of Work
Notice in the example above that the unit of work was the newton∙meter (N∙m), known as the joule (J).
The joule is named after Prescott Joule. The joule is also the unit of energy. So, work involves a transfer
of energy. In this sense, then, energy can be defined as the capacity to do work. The Engineering unit
for work is pound∙foot (lb∙ft), where 1 J = 0.7376 lb∙ft)
Sign of Work
Work is a scalar quantity. The sign of work indicates the relative directions of force and displacement.
When the direction of force and displacement are the same, work is positive. When the directions are
opposite, work is negative. When an objects speed changes due to a net force, the sign of work
indicates whether speed is increasing or decreasing. When work is positive, work is being done to the
object. When work is negative, work is being done by the object.
Force Displacement Work Work is done…
+
+
+
To
To
+
By
+
By
Graphical Representation of Work
Work can be found by analyzing a plot of force
and displacement. A horizontal force of 5.0 N
displacing an object a distance of 5.0 meters is
easily calculated.
W  Fll  d  5 N  5 m  25 J
From the graph at the right you can see that the
product F x d is equal to the area underneath the
force-displacement curve. In the example, each
square equals one joule.
This is true even when force is not constant.
Consider the next graph. An increasingly greater
force is applied to the object during its
displacement. This calculation would be more
difficult. However, analysis of the forcedisplacement curve permits a reasonable
estimate of the work done on the object.
David Scott, Honors Physics, Rev. 2014
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Energy
Energy is often defined as the ability to do work. Clearly, work and energy are two closely related
concepts.
Kinetic Energy
Kinetic energy (KE, Ek) is the energy of an object due to its translational motion. This is apparent from its
formula. When the object is considered to be a particle,
KE 
1 2
mv
2
The kinetic energy of a 10-kg mass traveling at 1 m/s is
1
kg  m 2
 m
KE  10kg 1   5
 5N  m  5 J
2
s2
 s
2
The unit of energy is the joule, as mentioned previously.
Work-Kinetic Energy Theorem
The work done on an object is equal to the change in kinetic energy of the object. This is known as the
work-kinetic energy theorem. It is derived as follows.
1. Wnet  Fnet d
Since Fnet  ma
2. Wnet  mad
We have a kinetmatic equation involving a and d (displacem ent)
3. v f  vi  2ad Solving for ad ,
2
2
5. Wnet
6. Wnet


1 2
2
v f  vi
Substituti ng ad into equation 2 and distributi ng,
2
1
1
2
2
 mv f  mvi  KE f  KEi  KE
2
2
 KE
4. ad 
The W-KE theorem is very useful. For example, how much work is done to accelerate a 2000-kg truck
from 8.3 m/s to 11.1 m/s over a distance of 50.0 m?
What would be the speed at the moment of impact if a 9.1 kg turkey that fell off a table 0.95 m tall?
David Scott, Honors Physics, Rev. 2014
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Potential Energy
Potential energy is energy associated with an objects position, shape, or condition. It could be thought
of as “stored” energy, such as the energy in a compressed spring. It is called potential energy because
the object has the “potential” to move. Potential energy differs from kinetic in that it is due to the
objects relationship with its environment. In this chapter, we are concerned with two forms of potential
energy, gravitational and elastic PE.
Gravitational Potential Energy
Gravitational potential energy is the energy associated with a gravitational fields of two objects.
Imagine that a brick of mass m is lifted a vertical distance h. The work done to raise the brick would be
W = F∙d cos θ, where θ is 0°. The force required to lift the brick at a constant speed would be equal to its
weight, mg. Substituting the variables, then the work done on the brick is W = mgh. If the brick was
allowed to fall, then it would be capable of doing work in the amount mgh. When the brick is raised it
has the potential to do work when released. Thus, the brick has gravitational potential energy equal to
the quantity mgh.
PEg = mgh
Gravitational PE depends upon the vertical height of an object relative to some arbitrary reference level.
In most examples, the surface of the earth will be the reference level. But this must not always be so.
Elastic Potential Energy
We are specifically concerned with the potential energy of a helical (i.e. coiled) spring. When a spring is
compressed as shown in the figure, a
restoring force causes the spring to
return to its equilibrium position (i.e.
resting length). The force generated by
the spring is directly proportional to its
change in length, x, a spring constant, k.
Fspring   kx
The spring constant is peculiar to a given spring. A stiffer spring will have a higher spring constant. The
greater the spring constant, the more resistant the spring is to changing length, i.e. it takes more force
to change its length. The negative sign indicates that the force in the spring always acts in a direction
opposite of the change in length.
When a spring is compressed or stretched, it stores potential energy and is determined by the equation
1
PEspring   kx2
2
For now, we are only concerned a single compression or stretching of a spring. Later we will learn about
the behavior of an oscillating spring, one which moves back and forth or up and down.
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Conservation of Energy
Conserved Quantities
Conserved quantities are quantities that remain the same, though they may change form. For example,
mass is a conserved quantity. In a chemical reaction, the total mass at the beginning of a reaction
equals the total mass at the end of the reaction. The mass is the same, even though it changed form,
e.g. reactants have been changed into products.
Mechanical Energy
Energy takes many forms, but is a conserved quantity. We are concerned specifically with mechanical
energy. Mechanical energy includes kinetic and potential energy as we have thus far defined them.
The motion of objects commonly involves both kinetic and potential energy. Furthermore, kinetic and
potential energy are often inter-converted.
When a stone is held above the ground, it contains a certain gravitational potential energy equal to
mgh. When the stone drops, h continually decreases until it hits the ground, where h = 0. What does
this say about potential energy during the fall of the stone? Write your answer below.
Again, when the stone is held above the ground, it has no velocity. When the stone is dropped, what
happens to the velocity of the stone? What does this say about kinetic energy during the fall? Write
your answer below.
In the absence of friction, mechanical energy (ME) is a conserved quantity. What might you conclude
about the changes in PE and KE during the fall of the stone?
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Predict what you think a graph of PE and KE would like during the fall of the stone.
Conservation of Energy
Energy is a conserved quantity, like mass. This means that the total amount of energy in a closed system
remains constant. The initial energy must equal the final energy. Energy conservation is relatively
simple as long as you keep this in mind. Begin by adding all the energy terms at the beginning of the
problem and set them equal to all the energy terms at the end of the problem.
This is a more efficient, simpler approach to many kinematic problems. Consider a familiar problem
such as follows.
A block having mass 10.0 kg is held at rest on a frictionless plane inclined at 30⁰. The block slides down
to the bottom of the ramp a distance of 5.00 m. Find the velocity of the block at the end of the
displacement.
This problem can be solved using kinematic equations, but it is much simpler using a conservation of
energy approach. All that is necessary is to set the initial energy of the block equal to the final energy of
the block. Velocity is then easily determined.
Mechanical Energy
Mechanical energy includes both kinetic and potential energy. Thus, it is the sum of kinetic and various
forms of potential energies.
ME  KE  PE
So far we have only discussed gravitational and elastic potential energies. Keep in mind that both of
these could be present in a problem.
David Scott, Honors Physics, Rev. 2014
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Conservation of Mechanical Energy
Mechanical energy of a system is conserved in the absence of friction. We will return to friction later,
but for now it will be disregarded.
MEi  ME f
KEi  PEi  KE f  PE f
Let’s return to the problem above.
A block having mass 10.0 kg is held at rest on a frictionless plane inclined at 30⁰. The block slides down
to the bottom of the ramp a distance of 5.00 m. Find the velocity of the block at the end of the
displacement.
KEi  PEi  KE f  PE f
Initially, the block has no KEi, but it does have PEi. At the instant the block reaches the bottom of the
ramp, it has KEf, but no PEf. So,
0  PEi  KE f  0
1
2
mv f
2
2
2 ghi  v f
mghi 


2 ghi  v f  2 9.81m / s 2 5 cos 30
v f  7m / s
Notice that the mass term cancelled. Meaning that mass does not affect the answer.
Friction and the Conservation of Mechanical Energy
Friction does not invalidate the conservation of energy, but it must be accounted for. Remember that
the work done on an object equals the change in kinetic energy of the object, the W-KE theorem. Since
friction decreases the KE of an object, it must be included in the initial side of the equation, represented
by Wfr.
KEi  PEi  W fr  KE f  PE f
Let’s return the problem above, but include friction this time. How will the presence of friction affect
the velocity of the block.
A block having mass 10.0 kg is held at rest on a frictionless plane inclined at 30⁰. The coefficient of
friction between the block and the plane is 0.20. The block slides down to the bottom of the ramp a
distance of 5.00 m. Find the velocity of the block at the end of the displacement.
David Scott, Honors Physics, Rev. 2014
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KEi  PEi  W fr  KE f  PE f
PEi  W fr  KE f
Since friction opposes motion of the block, the term becomes negative.
PEi  F fr d  KE f
PEi  Fn d  KEf
mghi  mg cos30  
1
2
mv f
2
1 2
vf
2
2
2 g hi   cos 30   v f
ghi  g cos(30) 
The mass term m cancels.
2
Isolate v f and collect like terms.
2 g hi   cos 30   v f
v f  5.7 m / s
Power
Power is the rate at which work is done, or the rate at which energy is transformed.
P
W
t
From the above equation, the SI unit of power would be J/s. This is defined as the watt (W)
1 W  1 J/s
In the engineering system of units, a common unit is the ft∙lb/s. Another common unit for power is
horsepower (hp). The relationship of these units is
1 hp = 550 ftl∙b/s = 746 W
The difference between energy and power
A person is limited in the work they can do by both the total amount of energy they can transform and
the rate at which they can do so. The rate of energy transformation is power. For example, a person
may be able to walk up many flights of stairs before they have to stop because their energy has been
depleted. On the other hand, they same person who runs up the stairs may have to stop after only a
few flights. This is because the energy has been used at a greater rate. In this sense the person is
limited by power, the rate in which the body can transform chemical energy into mechanical energy.1
1
Giancoli, Physics(2005)158.
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Consider 60-kg jogger runs up a flight of stairs in 4.0 s. The vertical height of the stairs is 4.5 m. The
powr output of the jogger. The work done by the jogger would be work done against gravity. So,
P


W mgh 60 kg  9.81 m/s 2 4.5 m


 660 W
t
t
4.0 s
The energy required by the jogger’s body is equivalent to the work done by the jogger, which would be
the numerator in the above solution.
An alternative equation for power can be developed, also.
P
W Fd
d

 Fv , Since is velocity.
t
t
t
For example, how much power would a 1400-kg car need to travel at a steady rate of 80 km/h (28.8
m/s)? Assume retarding forces acting against the motion of the car total 700 N.
Since the car is moving at a constant speed, force equilibrium exists in the x-direction. Therefore, the
force needed to propel the car forward is equal to the retarding forces (friction, air resistance)
P  Fv  700N28.8m/s   20.2kW
David Scott, Honors Physics, Rev. 2014
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Name ______________________________________________ Date _______________ Period _____
Work done by a constant force
Problem Set 5A
1. Calculate the amount of work done by a crane as it lifts a 265 kg pile driver 2.80 m. How much
work is done by gravity when a 265-kg pile driver falls 2.80 m? (7.30 kJ)
2. A crate weighs 1.30 x 103 N. How much work is done in lifting the crate 4.00 m vertically? (5.2
kJ)
3. A worker lifts an 80.0-lb bag of cement from the ground to his shoulder, a distance of 5.00 ft.
The worker then carries the bag 75.0 ft. Determine the work done by the worker on the bag.
You may use Engineering units (lb∙ft) to solve this problem. (4.00 x 102 lb∙ft)
4. When catching a baseball, a catcher’s glove moves 10.0 cm along the line of motion of the ball.
If the ball exerts a force of 475 N on the glove, how much work is done by the ball? (47.5 J)
5. If a neighbor pushes a lawnmower four times as far as you do but exerts only half the force, who
does the most work and by how much? (Neighbor, twice as much work)
6. A shopper pushes a shopping cart with a force of 35 N at an angle of 25° downward from the
horizontal. Calculate the work done on the shopping cart by the shopper if the cart is pushed
50.0 m. (1.59 kJ)
David Scott, Honors Physics, Rev. 2014
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7. A traveler pulls her 70.0 N flight bag a distance of 253 m along a level airport floor at a constant
velocity. The force exerted is 40.0 N at an angle of 52.0° above the horizontal.
a. How much work was done by the attendant on the flight bag? (6.23 kJ)
b. What was the work done by the force of friction on the flight bag? (-6.23 kJ)
c. Calculate the coefficient of kinetic friction between the flight bag and the floor. (0.639)
8. A painter holding a 5.00-gal bucket of paint carries at a constant speed for it 10.0 ft. How much
work does the painter do to the bucket of paint? Can you answer this question with the
information given? If so, how much work was done to the bucket of paint by the painter? If
not, what information would you need?
9. A person pulls a 50-kg crate 50 m along a level floor by a constant force FP = 100 N, which acts at
an angle of 37⁰ to the horizontal. The floor is rough and exerts a friction force Fk = 50 N.
Determine
a. The work done by each force acting on the crate (Wp = 4000 J; Wk = -2500 J)
b. The net work done on the crate (Wnet = 1500 J)
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10. A variable force was applied to an object. As a result, the object was displaced 17 m. A forcedisplacement plot of the event is shown below.
a. Give a reasonable estimate the work done on the object by the force.
b. Explain how you arrived at your answer. Why is your method valid?
11. As a pendulum swings back and forth, does the tension force (FT)in the string do work on the
bob? Does the force of gravity do work on the bob? Explain your answers.
12. When a punter kicks a football, is he doing work on the ball while his foot is in contact with it? Is
he doing work on the ball after the ball loses contact with his foot? Are any forces doing work
on the ball while the ball is in flight? Explain your answers.
David Scott, Honors Physics, Rev. 2014
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Name ______________________________________________ Date _______________ Period _____
Problem Set 5B2,3
Kinetic Energy
1. Calculate the speed of an 8.0 x 104 kg airliner with a kinetic energy of 1.1 x 109 J. (1.7 x 102 m/s)
2. Two bullets have masses of 3.0 g and 6.0 g, respectively. Both are fired with a speed of 40.0
m/s. Which bullet has more kinetic energy? What is the ratio of their kinetic energies? (greater
mass; 2:1)
3. Two 3.0 g bullets are fired with speeds of 40.0 m/s and 80.0 m/s, respectively. What are their
kinetic energies? Which bullet has more kinetic energy? What is the ratio of their kinetic
energies? (2.4 J; 9.6 J; 1:4)
4. Calculate the kinetic energy of a proton (mass = 1.673 x 10-27 kg) traveling at one-tenth the
speed of light (c = 3.00 x 108 m/s). What speed would an electron (me = 9.109 x 10-31 kg) have to
obtain to equal the KE of the proton? If the speed of light is the speed limit of the known
universe, is this possible? (KEp = 7.53 x 10-13 J; ve = 1.29 x 109 m/s)
5. At room temperature, an oxygen molecule, with a mass of 5.31 x 10-26 kg, typically has a KE of
about 6.21 x 10-21 J. How fast is the molecule moving? How long would it take the molecule to
travel 755 nm in straight line (484 m/s; 1.56 ns)
6. If the KE of an object is doubled, by what factor has its speed increased? If the speed of an
object is tripled, by what factor is the KE increased? If mass is reduced by half and speed is
doubled, how is KE affected?
2
3
Problems 1-3 from Holt Physics (2000)174
Problems 5-6 from Giancoli, Physics (2005)162
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Name ______________________________________________ Date _______________ Period _____
Problem Set 5C4
Work-Kinetic Energy Theorem
1. A catcher catches a baseball (m = 145 g) traveling at a speed of 42.5 m/s. How much work was
done on the baseball to bring it to rest? (131 J)
2. How much work is required to slow a 1.50 x 103 kg car from 30.0 m/s to 20.0 m/s? (375 kJ)
3. A car traveling at 60.0 km/h can brake to a stop in a distance of 20.0 m. If the car is going twice
as fast, what is its stopping distance? Assume maximum braking force is being applied, that it
does not vary with speed, and that all other factors (e.g. μk) are unchanged. (Hint: begin by
determining the relationship between stopping distance and velocity.) (80.0 m)
4. A softball (m = 0.25 kg) is pitched at 95 km/h. When it reaches the plate, it has slowed by 10%.
Neglecting gravity, estimate the average force of air resistance affecting the ball. The distance
between the pitcher and the plate is 15 m. (1.1 N)
5. A 285-kg load is lifted 22.0 m vertically with an acceleration of a = 0.160g by a single cable.
Determine…
a. The tension on the cable (3.24 x 103 N)
b. The net work done on the load (9.83 x 103 J)
c. The work done by the cable on the load (7.13 x 104 J)
d. The work done by gravity on the load (-6.14 x 104 J)
e. The final speed of the load, assuming it started from rest (8.31 m/s)
4
Problems 3-5 from Giancoli, Physics (2005)143, 063
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Name ______________________________________________ Date _______________ Period _____
Problem Set 05D5
Potential Energy
1. A hiker (m = 55.0 kg) begins to climb from an elevation of 1600 m to 3300 m.
a. What is the hikers change in potential energy? (9.2 x 105 J)
b. What is the minimum work required of the hiker?
c. Can the actual work be more than this? Explain your answer
2. A crane lifts a 375-kg load from the ground to a height of 20.0 m.
a. Calculate the total work done on the load by the crane.
b. Calculate the change in gravitational potential energy of the load.
c. Reflect on your calculations and make a statement regarding the relationship of work
and potential energy.
3. A 5.00-kg cat jumps from the floor onto a kitchen table that is 0.85 m tall. Calculate the
gravitational PE of the cat relative to the floor.
4. A 1000-kg roller coaster car moves from point 1 to point 2
and then to point 3, as shown in the figure.
a. What is the PEg at points 2 and 3 relative to point 1?
b. What is the change in PEg when the car goes from
point 2 to point 3?
5
Poblems 1, 4 from Giancoli, Physics (2005)
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5. Use the formula for the potential energy of a spring to show the units of the spring constant.
6. How much must a spring be compressed to have 25 J of potential energy, if the spring constant
is 440 N/m?
7. A spring 10.0 cm in length with a spring constant of 50.0 N/m hangs vertically. How long will the
spring be when a 0.500 kg mass is attached to the spring? (19.8 cm)
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Name ______________________________________________ Date _______________ Period _____
Problem Set 05E
Conservation of Energy6
1. A skier, starting from rest, slides down a frictionless incline of 30.0⁰. If the skier begins at a
vertical height of 175 m, what is the skier’s speed at the bottom of the hill? (58.6 m/s)
2. A sled is given an initial push up frictionless inclined plane of 25.0⁰. The push resulted in a
velocity of 6.00 m/s.
a. What is the maximum vertical height obtained by the sled? (1.83 m)
b. What is the speed of the sled at this height?
c. What distance will the sled travel up the slope?
d. What is the speed of the sled when it is halfway down the slope? (4.24 m/s)
e. Calculate the speed of the sled when it returns to the bottom of the incline?
3. A 65.0 kg trampoline performer stands on a platform 3.0 m above the trampoline as shown in
the figure. If the performer jumps with an initial vertical velocity of 5.0 m/s,
a. Find his speed as he impacts the trampoline (9.2 m/s)
b. If the trampoline behaves like a spring with a spring constant of
6.2 x 104 N/m, how far will the trampoline be compressed?
(-0.30 m)
6
Problems 1-6 adapted from Giancoli, 163.
David Scott, Honors Physics, Rev. 2014
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4. A projectile is fired at an upward angle of 45.0 from the top of a 265 m cliff with a speed of 185
m/s. What will be the speed of the projectile when it strikes the ground below? (199 m/s)
5. A rollercoaster is pulled up to point 1, where it is released from rest. Assuming no friction,
calculate its speed at points 2, 3, and 4.
a. Speed at point 2 (26 m/s)
b. Speed at point 3 (12 m/s)
c. Speed at point 4 (20 m/s)
6. A pendulum with a length L = 0.75 m and a bob m = 0.25kg swings back and forth. At what
point(s) in its path will it have
a. The greatest potential energy?
b. The least potential energy?
c. The greatest kinetic energy?
d. The least kinetic energy?
e. Where will the kinetic and potential energy be equal?
Why?
f.
If the pendulum string makes an angle of 30⁰ from
vertical, what is the PEg of the bob?
David Scott, Honors Physics, Rev. 2014
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Name ______________________________________________ Date _______________ Period _____
Problem Set 05F7
Power
1. How long does it take a 19 kW steam engine to do 6.8 x 107 J of work? (3.6 x 103 seconds)
2. A rain cloud contains 2.66 x 107 kg of water vapor. How long would it take for a 2.00 kW pump
to raise the same amount of water to the cloud’s altitude of 2.00 km? ((2.61 x 10 8 s)
3. If an automobile engine delivers 50.0 hp of power, how much time will it take for the engine to
do 6.40 x 105 J of work? (1 hp = 746 W) (17.2 s)
4. Water flows over a section of Niagara Falls at the rate of 1.2 x 106 kg/s and falls 50.0 m. How
much power is generated by the falling water? (Hint: calculate the weight of the water) (5.9 x
108 W = 590 MW)
5. A 1.50 x 103 kg car accelerates uniformly from rest to 10.0 m/s in 3.00 seconds.
a. What is the work done on the car in this time interval? (7.50 x 104 J)
b. What is the power delivered by the engine in this time interval? (2.50 x 104 W)
6. A 1.0 x 103 kg elevator carries a maximum load of 800.0 kg. A constant frictional force of 4.0 x
103 N retards the elevator’s motion upward. What minimum power, in kilowatts, must the
motor deliver to lift the fully loaded elevator at a constant speed of 3.00 m/s? (65 kW)
7. A car with a mass of 1.50 x 103 kg starts from rest and accelerates to a speed of 18.0 m/s in 12.0
seconds. Assume that the force of resistance remains constant at 400.0 N during this time.
What is the average power developed by the car’s engine? (23.8 kW)
7
Holt Physics (2009)181, 186
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