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Unit 6 – Work, Power, Efficiency, & Work-Energy Theorem
Grading: Show all work, keeping it neat and organized. Show equations used and include all units.
Vocabulary
Work: The product of the component of the force exerted on an object (in the direction of
displacement) and the magnitude of the displacement.
work = (force)·(displacement) or
W = Fd
The SI unit for work is the joule (J), which equals one newton-meter (N·m).
For maximum work to be done, the object must move in the direction of the force. If the object is
moving at an angle to the force, determine the component of the force in the direction of motion.
Remember, if the object does not move, or moves perpendicular to the direction of the force, no work
has been done.
Vocabulary
Power: The rate at which work is done.
power =
work
elapsed time
or
P=
W
t
The SI unit for power is the watt (W), which equals one joule per second (J/s). One person is more
powerful than another if he or she can do more work in a given amount of time, or can do the same
amount of work in less time.
Vocabulary
Efficiency: The ratio of the work output to the work input.
efficiency =
work output
work input
=
Wout
Win
Efficiency has no units and is usually expressed as a percent.
Solved Examples
Example 1:
After finishing her physics homework, Susan pulls her 50.0-kg body out of the living room chair and
climbs up the 5.0-m-high flight of stairs to her bedroom. How much work does Susan do in ascending
the stairs?
Solution: First find Susan’s weight. Her muscles exert a force to carry her weight up the stairs.
Given:
Unknown:
Original Equation:
Solve:
m = 50.0 kg
g = 10.0 N/kg
W=?
W = mg
W = mg = (50.0 kg)(10.0 N/kg) = 500 N
Now you have Susan’s weight (or force) to determine the amount of work done. It is important to note
that when you are solving for the work done, you need to know only the displacement of the body
moved. The number of stairs climbed or their steepness is irrelevant. All that is important is the change
in position in the direction of the applied force (in this case, 5.0 m up against gravity which pulls down).
Given:
Unknown:
Original Equation:
Solve:
F = 500 N
d = 5.0 m
W=?
W = Fd
W = Fd = (500 N)(5.0 m) = 2500 J
Don’t get confused here by the two W’s you see in this example. The W in W = mg means weight while
the W in W = Fd means work. There are many ways to tell them apart, the most important of which is
to understand how they are used in the context of the exercise. Also, the units used for each are quite
different: weight is measured in newtons, and work is measured in joules. Last of all, weight is a vector
and work is a scalar.
Example 2:
In the previous example, Susan slowly ascends the stairs, taking 10.0 s to go from bottom to top. The
next evening, in a rush to catch her favorite TV show, she runs up the stairs in 3.0 s. a) On which night
does Susan do more work? b) On which night does Susan generate more power?
a)
Solution:
Susan does the same amount of work on both nights because the force she exerts and her displacement
are the same each time.
b)
Susan’s power output varies because the time taken to do the same amount of work varies.
First night:
Given:
Unknown:
Original Equation:
W = 2500 J
t = 10.0 s
P=?
W
P= t
Solve:
P=
Second night:
Given:
W
t
=
2500 J
=
10.0 s
Unknown:
Original Equation:
W = 2500 J
t = 3.0 s
P=?
W
P= t
Solve:
P=
W
t
=
2500 J
=
3.0 s
250 W
830 W
Susan generates more power on the second night.
Example 3:
Clyde, a stubborn 3500-N mule, refuses to walk into the barn, so Farmer MacDonald must drag him up a
5.0-m ramp to his stall, which stands 0.50-m above ground level. If Farmer MacDonald needs to exert a
450-N force on the mule to drag him up the ramp with a constant speed, what is the efficiency of the
ramp?
Solution: We need to consider how much work is done on the mule (Work Output) and compare this to
how much work Farmer MacDonald does (Work Input).
Given:
Vocabulary
Unknown:
Original Equation:
Fin = 450 N
Δdin = 5.0 m
Fout = 3500 N
Δdout = 0.50 m
Efficiency = ?
W
Efficiency = Wout
Solve:
Efficiency =
in
Wout
Win
=
Fout ∆dout
Fin ∆din
=
(3500 N)(0.50 m)
(450 N)(5.0 m)
= 0.78 = 78%
Energy: The ability to do work.
There are many different types of energy. This chapter will focus on only mechanical energy, or the
energy related to position (potential energy) and motion (kinetic energy).
Vocabulary
Potential Energy: Energy of position, or stored energy.
An object gains gravitational potential energy when it is lifted from one level to a higher level.
Therefore, we generally refer to the change in gravitational potential energy or ΔGPE, which is
proportional to the change in height, Δh.
Δ gravitational potential energy = (mass)(acceleration due to gravity)(Δ height)
or
ΔGPE = mgΔh
It is important to remember that gravitational potential energy relies only upon the vertical change in
height, Δh, and not upon the path taken.
Vocabulary
Kinetic Energy: Energy of motion.
The kinetic energy of an object varies with the square of the speed.
Kinetic energy = (1/2)(mass)(speed)2
or
1
2
KE = mv2
The SI unit for energy is the joule. Notice that this is the same unit used for work.
Vocabulary
Work-Energy Theorem: When work is done on an object, energy is transformed from one form to
another. The sum of the changes in potential, kinetic, and heat energy is equal to the work done on the
object.
Work = Δ Energy
or
W= ΔE = (Ef – Ei)
In certain circumstances, work is done to change only the gravitational potential energy of an object,
and sometimes the change in gravitational potential energy of an object is directly related to work being
done. In other circumstances, work is done to change only the kinetic energy of an object, and
sometimes the change in kinetic energy of an object is directly related to work being done.
W = Δ GPE
or
W = Δ KE
Mechanical energy is transformed into heat energy when work is done to overcome friction.
Example 4:
Legend has it that Isaac Newton “discovered” gravity when an apple fell from a tree and hit him on the
head. If a 0.20 kg apple fell 7.0 m before hitting Newton, what was its change in GPE during the fall?
Solution: For a given object, the change in GPE depends only upon the change in height. The apple
does not need to fall all the way to the ground to experience an energy change.
Given: m = 0.20 kg
g = 10.0 m/s2
Δh = 7.0 m
Unknown: ΔGPE = ?
Original equation: ΔGPE = mgΔh
Solve: ΔGPE = mgΔh = (0.20 kg)(10 m/s2)(7.0 m) = 14 J
Example 5:
A greyhound at a race track can run at a speed of 16.0 m/s. What is the KE of a 20.0 kg greyhound as it
crosses the finish line?
Given: m = 20.0 kg
v = 16.0 m/s
Unknown: KE = ?
1
Original Equation: KE = 2 mv2
1
2
Solve: KE = (20.0 kg)(16.0 m/s)2 = 2560 J
Example 6:
A 2500 kg vehicle is traveling 32 m/s down the freeway when a deer jumps out onto the road. If the
average braking force of the vehicle is 17,250 N, how far will the vehicle travel before it comes to a
complete stop?
Solution: The braking force does work on the vehicle as it comes to rest, changing its kinetic energy.
The braking force is negative, because it is slowing the vehicle down.
Since the final velocity of the vehicle is 0 m/s, the final KE of the vehicle 0 J.
Given: m = 2500 kg
vi = 32 m/s
KEf = 0 J
Fbr = -17,250 N
Unknown: Δd = ?
Original Equation: W = Δ KE
Solve: W = Δ KE
but
FΔd = KEf - KEi
1
FΔd = (0 J) - 2mv2
Δd =
- mv2
2F
=
W = FΔd
- (2500 kg)(32 m/s)2
2(-17,250 N)
and
= 74.2 m
Δ KE = KEf - KEi
So…
Exercises
1) On his way off to college, Mitchell drags his suitcase 25.0 m from the door of his house to the car at
a constant speed with a horizontal force of 95.0 N. a) How much work does Mitchell do to overcome
the force of friction? b) If the floor has just been waxed, does he have to do more work or less work
to move the suitcase? Explain. c) Although work is done by Mitchell, no work is done on the
suitcase. Explain.
2) Clarissa does 5.2 J of work to lower the window shade in her bedroom a distance of 0.9 m. How
much force must Clarissa exert on the window shade?
3) A librarian lifts a 2.2 kg book from the floor to a height of 1.35 m, carries the book 8.0 m to the
stacks, and places the book on a shelf 1.75 m above the floor. How much work is done on the book?
4) Sau-Lan has a mass of 52 kg and rides up the world’s longest escalator at Ocean Park in Hong Kong.
If the escalator has a length of 227 m at an angle of 31° from the horizontal, calculate the work done
by the escalator to lift Sau-Lan.
5) Popeye & Brutus, two carnival sideshow strong men, each lift 250.-kg barbells 2.00 m off the
ground. Popeye lifts his barbell in 2.00 s and Brutus lifts his in 3.00 s. a) Which strong man does
more work? Explain. b) Calculate which man is more powerful.
6) The Dempseys are moving to a new town to start over, so they have called in the ACME moving
company to take care of their furniture. Jillian slides the 3300-N china cabinet up a 7.0 m-long ramp
to the moving van, which stands 0.9 m off the ground. If Jillian must exert a 600.-N force to move
the china cabinet up the ramp with a constant speed, what is the efficiency of the ramp?
7) Mary drops a 56 g super ball down a staircase. If the super ball hits the ground 5.5 m below, what is
the change in its gravitational potential energy?
8) Toni has a mass of 45 kg and is racing her bicycle downhill with a speed of 10 m/s. a) What is Toni’s
kinetic energy? b) If Toni's speed is doubled, what will happen to her kinetic energy?
9) An experimental train with a mass of 2.5 x 104 kg is powered by a jet engine with a thrust force of
5.0 x 105 N over a track length of 509 m. a) What is the work done on the train? b) What is the final
velocity of the train (assuming there is no friction)?
10) A 20 kg rock falls off of the edge of a tall, vertical cliff. When the rock hits the ground, it exerts a
force of 5960 N that drives the rock 70 cm into the ground. a) What is the height of the cliff? b)
Assuming the rock’s gravitational potential energy is transformed into kinetic energy as it falls, what
is the speed of the rock right before it hits the ground?