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Physics 11 Chapter 6 Assignment: Work, Energy & Power
1. Define the following terms:
 Kinetic energy - the energy of an object
due to its motion
 Gravitational potential energy - the
potential energy an object has because of
its location in a gravitational field; objects at
higher altitudes have greater gravitational
potential energy than objects at lower
altitudes
 Elastic potential energy - a form of energy
that accumulates when an elastic object is
bent, stretched, or compressed
 Mechanical energy - the sum of the kinetic
energy and potential energy
 Conservative force - a force that does
work on an object in such a way that the
amount of work done is independent of the
path taken; work done by conservative
forces is zero for motions in which the
object returns to the starting point; force of
gravity is an example of a conservative
force
 Non-conservative force - a force that does
work on an object in such a way that the
amount of work done is dependent on the
path taken; work done by non-conservative
forces is nonzero for motions in which the
object returns to the starting point; friction
and air resistance are examples of nonconservative forces
 Work - the transfer of mechanical energy;
calculated as the product of the force and
the displacement when the force and
displacement vectors are in the same
direction
 Joule – unit of energy; equivalent to
applying one newton of force on an object
over a distance of one metre
 Work-kinetic energy theorem - the
relationship between the work done on an
object and the resulting change in kinetic
energy, W = ΔEk
 Law of conservation of energy - the total
energy of an isolated system, including all
forms of energy, always remains constant;
energy can neither be created nor
destroyed, but it may be converted from
one form to another or transferred from one
object to another
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
KEY
Hooke’s law - states that the applied force
is directly proportional to the amount of
extension or compression of a spring
Restoring force - the force exerted by a
spring on an object in the direction opposite
to the direction that the spring is stretched
or compressed
Spring constant - the amount of force a
spring can exert per unit distance of
extension or compression
Power - the rate at which work is done,
measured in watts (W), or joules per
second; also defined as the rate at which
energy is transferred
Efficiency - the ratio of useful energy (or
work output) to the total energy (or work
input); describes how well a machine or
device converts input energy or work into
output energy or work
2. Can the normal force on an object ever do work? Explain.
The normal force can do work on an object if the normal force has a component in the
direction of displacement of an object. If someone were to jump up in the air, then the floor
pushing upward on the person (the normal force) would do positive work and increase the
person’s kinetic energy. Likewise when they hit the floor coming back down, the force of
the floor pushing upwards (the normal force) would do negative work and decrease the
person’s kinetic energy.
3. A woman swimming upstream is not moving with respect to the shore. Is she doing any work? If
she stops swimming and merely floats, is work done on her?
The woman does work by moving the water with her hands and feet, because she must
exert a force to move the water some distance. As she stops swimming and begins to float
in the current, the current does work on her because she gains kinetic energy. Once she is
floating the same speed as the water, her kinetic energy does not change, and so no net
work is being done on her.
4. Why is it tiring to push hard against a solid wall even though no work in done?
While it is true that no work is being done on the wall by you, there is work being done
inside your arm muscles. Exerting a force via a muscle causes small continual motions in
your muscles, which is work, and which causes you to tire. An example of this is holding a
heavy load at arm’s length. While at first you may hold the load steady, after a time your
arm will begin to shake, which indicates the motion of muscles in your arm.
5. By approximately how much does your gravitational potential energy change when you jump as
high as you can?
Your gravitational Ep will change according to Δ Ep = mgΔy . If we choose some typical
values like m = 80 kg and y = 0.75 m , then Δ Ep = (80 kg)(9.81 m/s2)(0.75 m) = 589 J.
6. A pendulum is launched from a point that is a height h above its
lowest point in two different ways (see figure to the right). During
both launches, the bob is given an initial speed of 3.0 m/s. On the
first launch, the initial velocity of the bob is directed upward along
the trajectory, and on the second launch, it is directed downward
along the trajectory. Which launch will cause the pendulum to
swing the largest angle for the equilibrium position?
The two launches will result in the same largest angle.
Applying conservation of energy between the launching point
and the highest point, we have E1 = E2. Thus, ½ mv2 + mgh = mghmax. The direction of the
launching velocity does not matter, and so the same maximum height (and hence
maximum angle) will results from both launches. Also, for the first launch, the ball will rise
to some maximum height and then come back to the launch point with the same speed as
when launched. That then exactly duplicates the second launch.
7. A coil spring of mass m rests upright on a table. If you compress the spring by pressing down with
your hand and then release it, can the spring actually leave the table? Explain using the law of
conservation of energy.
The spring can leave the table if it is compressed enough. If the spring is compressed an
amount 0 x , then the gain in elastic Ep is ½ kxo2 . As the spring is compressed, it is lowered
by some amount. For example, if the spring is uniform, and it is lowered by ½ x o, then the
amount of decrease in gravitational Ep is ½ mgxo . If the gain in elastic Ep is more than the
loss in gravitational Ep, so that ½ kxo2 > ½ mgxo or xo > mg/k, then the released spring
should rise up off of the table, because there is more than enough elastic E p to restore the
spring to its original position. That extra elastic energy will enable the spring to “jump” off
the table – it can raise its center of mass to a higher point and thus rise up off the table.
Where does that “extra” energy come from? From the work you did in compressing the
spring.
8. Analyze the motion of a simple swinging pendulum in terms of energy, (a) ignoring friction, and (b)
taking it into account. Explain why a grandfather clock has to be wound up.
(a) Consider that there is no friction to dissipate any energy. Start the pendulum at the top
of a swing, and define the lowest point of the swing as the zero location for gravitational
Ep. The pendulum has maximum gravitational Ep at the top of a swing. Then as it falls, the
gravitational Ep is changed to kinetic energy. At the bottom of the swing, the energy is all
kinetic energy. Then the pendulum starts to rise, and kinetic energy is changed to
gravitational Ep. Since there is no dissipation, all of the original gravitational Ep is
converted to kinetic energy, and all of the kinetic energy is converted to gravitational Ep.
The pendulum rises to the same height on both sides of every swing, and reaches the
same maximum speed at the bottom on every swing.
(b) If there is friction to dissipate the energy, then on each downward swing, the pendulum
will have less kinetic energy at the bottom than it had gravitational Ep at the top. And then
on each swing up, the pendulum will not rise as high as the previous swing, because
energy is being lost to frictional dissipation any time the pendulum is moving. So each
time it swings, it has a smaller maximum displacement. When a grandfather clock is wound
up, a weight is elevated so that it has some Ep. That weight then falls at the proper rate to
put energy back in to the pendulum to replace the energy that was lost to dissipation.
9. Suppose you lift a suitcase from the floor to a table. Does the work you do on the suitcase
depend on (a) whether you lift it straight up or along a more complicated path, (b) the time it takes,
(c) the height of the table, and (d) the weight of the suitcase?
The work done to lift the suitcase is equal to the change in PE of the suitcase, which is the
weight of the suitcase times the change in height (the height of the table).
(a) Work does NOT depend on the path, as long as there are no non-conservative forces
doing work.
(b) Work does NOT depend on the time taken.
(c) Work DOES depend on the height of the table – the higher the table, the more work it
takes to lift the suitcase.
(d) Work DOES depend on the weight of the suitcase – the more the suitcase weighs, the
more work it takes to lift the suitcase.
10. Answer the previous question for the power needed rather than the work.
The power needed to lift the suitcase is the work required to lift the suitcase, divided by the
time that it takes.
(a) Since work does NOT depend on the path, the power will not depend on the path either,
assuming the time is the same for all paths.
(b) The power DOES depend on the time taken. The more time taken, the lower the power
needed.
(c) The power needed DOES depend on the height of the table. A higher table requires
more work to lift the suitcase. If we assume that the time to lift the suitcase is the same in
both cases, then to lift to the higher table takes more power.
(d) The power DOES depend on the weight of the suitcase. A heavier suitcase requires
more force to lift, and so requires more work. Thus the heavier the suitcase, the more
power is needed to lift it (in the same amount of time).
11. A 75.0-kg firefighter climbs a flight of stairs 10.0 m high. How much work in required?
Because there is no acceleration, the contact force must have the same magnitude as the
weight. The displacement in the direction of this force is the vertical displacement. Thus,
W  Fg  d  mg  d  (75.0 kg)( 9.81 m/s 2 )(10.0 m)  7.36 103 J
12. A 9.0 x 102-N crate rests on the floor. How much work is required to move it at constant speed (a)
6.0 m along the floor against a friction force of 180 N, and (b) 6.0 m vertically.
(a) Because there is no acceleration, the horizontal applied force must have the same
magnitude as the friction force. Thus,
W  F f  d  (180 N)(6.0 m)  1.08  103 J
(b) Because there is no acceleration, the vertical applied force must have the same
magnitude as the weight. Thus,
W  Fg  y  mg  y  (900 N)(6.0 m)  5.4 103 J
13. How much work did the movers do (horizontally) pushing a 150-kg
crate 12.3 m across a rough floor without acceleration, if the
effective coefficient of friction was 0.70?
Because there is no acceleration, from the force diagram we
see that:
FN  mg and F  Ff   k mg
Thus,
W  F  x   k mg  x  (0.70)(150 kg)(9.81 m/s 2 )(12.3 m)  1.27  10 4 J
14. The horizontal x component of the force on an object varies as
shown in the graph below. Determine the work done by this force
to move the object (a) from x = 0.0 m to x = 10.0 m, and (b) from
x = 0.0 m to x = 15.0 m.
The work done in moving the object is the area under the
F vs. x graph.
(a) For the motion from 0.0 m to 10.0 m, we find the area of two triangles and one rectangle:
W
1
1
(400 N)(3.0 m - 0.0 m)  (400 N)(7.0 m - 3.0 m)  (400 N)(10.0 m - 7.0 m)  2.8  10 3 J
2
2
(b) For the motion from 0.0 m t 15.0 m, we add the negative area of two triangles and one
rectangle:
1
1
W  2.8  10 3 J  (200 N)(12.0 m - 10.0 m)  (200 N)(13.5.0 m - 12.0 m)  (200 N)(15.0 m - 13.5 m)  2.1  10 3 J
2
2
15. A spring has a spring constant of k = 88 N/m. Use a graph to determine the work needed to
stretch it from x = 3.8 cm to x = 5.8 cm, where x is the displacement from its unstretched length.
We obtain the forces at the beginning and end of the motion:
At x1 = 0.038 m, F1 = kx1 = (88 N/m)(0.038 m) = 3.34 N;
At x2 = 0.058 m, F2 = kx2 = (88 N/m)(0.058 m) = 5.10 N
From the graph to the right, the work done in stretching the
object is the area under the F vs. x graph:
W
1
1
(F1  F2 )(x 1  x 2 )  (3.34 N  5.10 N)(0.058 m  0.038 m)  0.084 J
2
2
16. How much work is required to stop an electron (m = 9.11 x 10-31 kg) which is moving with a speed
of 1.90 x 106 m/s?
The work done on the electron decreases its kinetic energy:
1
1
1
2
2
W  E k  mv f  mvi  0  (9.11 x 10 -31 kg)(1.90 x 10 6 m/s) 2  1.64 x 10 -18 J
2
2
2
17. A spring has a spring constant k = 440 N/m. How much must this spring be stretched to store 25
J of elastic potential energy?
1
kx 2
2
2E
 x 

k
E p 
2(25 J)
 0.34 m
440 N/m
18. A roller coaster, as shown below, is pulled to a point 1 where it is released from rest. Assuming
no friction, calculate the speeds at points 2, 3 & 4.
We choose y = 0 at point 2. With no friction, energy is conserved.
At point 2, we have:
E k1  E p1  E k 2  E p2
1
1
2
2
mv1  mgh1  mv2  mgh2
2
2
1
2
0  mgh1  mv2  0
2
 v 2  2 ghi  2(9.81 m/s 2 )( 25 m)  22.1 m/s
At point 3, we have:
Ek2  E p2  Ek3  E p3
1
1
2
2
mv2  mgh2  mv3  mgh3
2
2
1
1
2
2
mv2  0  mv3  mgh3
2
2
1 2

2
 v3  2 v2  gh3   v2  2 gh3  (22.1 m/s) 2  2(9.81 m/s 2 )( 20 m)  9.90 m/s
2

At point 4, we have:
Ek3  E p3  Ek4  E p4
1
1
2
2
mv3  mgh3  mv4  mgh4
2
2
 v4  v3  2 g (h3  h4 )  (9.90 m/s) 2  2(9.81 m/s 2 )(5 m)  14.0 m/s
2
19. How long will it take a 1750-W motor to lift a 285-kg piano to a sixth-story window 16.0 m above?
Assuming the lifting is done at a constant velocity (i.e. the change in potential energy is the
entire amount of energy transformed) then we can use the formula for power:
W mg h

t
t
mg h (285 kg)(9.81 m/s 2 )(16.0 m)
t 

 25.6 s
P
1750 W
P =
20. An engine produces 8200 J of heat while performing 3200 J of useful work. What is the efficiency
of this engine?
We know that:
 Woutput 
  100%
Efficiency = 
W

input


where Winput is work put into running the engine and Woutput is the useful work performed by
the engine. So Winput = 8200 J +3200 J = 11400 J and Woutput = 3200 and:
 3200 J 
Efficiency = 
  100%  28%
 11400 J 