*
AP PHYSICS B
Work and Energy
Teacher Packet
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involved in the production of this material.
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Work & Energy
Objective
To review the student on the concepts, processes and problem solving strategies necessary to
successfully answer questions on work and energy.
Standards
Work and energy are addressed in the topic outline of the College Board AP* Physics Course
Description Guide as described below.
I. Newtonian Mechanics
C. Work, energy, and power
1. Work and work-energy theorem
2. Conservative forces and potential energy
3. Conservation of energy
4. Power
F. Oscillations and Gravitation
1. Simple harmonic motion (dynamics and energy relationships)
2. Mass on a spring
3. Pendulum and other oscillations
AP Physics Exam Connections
Topics relating to work and energy are tested every year on the multiple choice and in most years
on the free response portion of the exam. The list below identifies free response questions that
have been previously asked over work and energy, often in conjunction with momentum. These
questions are available from the College Board and can be downloaded free of charge from AP
Central. http://apcentral.collegeboard.com.
2005
2002
2001
1999
Free Response Questions
Question 2
2006 Form B Question 2
Question 1
2005 Form B Question 2
Question 2
2002 Form B Question 1
Question 1 (ex a)
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Work, Energy, & Power
What I Absolutely Have to Know to Survive the AP* Exam
Work is the defined quantity from which the entire theory of energy begins. It is the scalar product of the
force acting on an object and the displacement through which it acts. Power is the rate at which work is
done. Objects are said to have energy if they have the ability to do work, either due to the fact that they are
moving (kinetic energy) or due to their position in a force field (potential energy). Conservative forces are
forces that do work that is path independent. Conserved quantities are quantities that do not change with
time. If the mechanical energy of a system is conserved, then the sum of the system’s kinetic and potential
energies at any given time will always add to the same number. Work done by a nonconservative force
generally cannot be recovered as usable energy.
Key Formulas and Relationships
Work: W = F ⋅ d = Fd cos θ
W
P = F⋅v
Power: P =
t
1
Kinetic Energy: K = mv 2
2
Gravitational Potential Energy: U g = mgh
1 2
kx
2
Conservation of Energy: K + U = K ' + U '
Potential Energy stored in spring: U s =
unit = Joule = 1 N·m
unit = Watt = 1 Joule/second
unit = Joule
unit = Joule
unit = Joule
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involved in the production of this material.
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Work, Energy, & Power
Important Concepts
•
•
•
•
Work is done by a force parallel to the displacement of the object: perpendicular
forces (centripetal forces for example) do no work.
If the force is at an angle to the displacement, you must resolve it into components.
Work is a scalar quantity but can be negative
Work-Energy Theorem: Work causes a change in energy: it is the method by which
energy is transferred (W = ΔE).
F
Fy
Only the x component of
the force does work.
θ
Fx
d
•
•
•
•
•
•
•
Work is positive when the force and displacement are in the same direction (object
gains energy): work is negative when the force and displacement are in opposite
directions (object looses energy.)
Energy is defined as the ability to do work. It is also a scalar and cannot be negative.
Kinetic energy – energy through motion.
Potential energy – energy through position (position in a gravitational or electric field
or position next to a stretched or compressed spring).
Conservative force – a force where the work done in moving an object between two
positions is the same regardless of the path taken
o Three conservative forces you should know: Gravitational, Elastic (springs),
Electric
o Example: the work done against gravity in lifting a box on top of a shelf is the
same regardless of whether you lift it straight up or push it up a ramp.
When only conservative forces act on an object, then the total energy (kinetic plus
potential energy) remains constant.
Non-conservative force – a force where the work done in moving an object between
two points does depend on the path taken
o Examples: friction, drag
o Work done by a non-conservative force is equal to the change in energy of the
object (remember the work-energy theorem)
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Work, Energy, & Power
B
More work is needed to push an object
against friction along path 2 than path
1. Friction is a non-conservative
force.
1
A
2
•
•
Power is the rate at which work is
done and is measured in Watts.
Work done by a variable force is
equal to the area under a graph of
force vs. displacement.
F (N)
Work = area
under Force vs.
Displacement
Graph
x (m)
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Work, Energy, & Power
Free Response
Question 1 (15 pts)
1. A block of mass m is attached to an ideal spring of spring constant k, the other end of
which is fixed. The block is on a level, frictionless surface as shown in the diagram. At
time t0, the block is set into simple harmonic motion of period T by an external force
pushing it to the right, giving the block initial velocity v0. Express all answers in terms of
the given quantities and fundamental constants.
v0
m
A. Determine the amplitude of the block’s motion.
(4 points max)
Kinitial = Us,max
1
2
mv0 =
2
1
1 point For applying conservation of
energy
1 point For equating the initial kinetic
energy of the block with the maximum
elastic potential energy
2
kxmax
2
1 point For correctly substituting the
definitions of kinetic and elastic potential
energy
2
xmax =
mv0
k
1 point For the correct answer in terms
of the given quantities and fundamental
constants
B. On the axis below, plot the kinetic energy of the block as a function of time for two
periods. Label the vertical axis appropriately.
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Work, Energy, & Power
K (J)
t (s)
T
T
2
(4 points max)
2T
3T
2
1 point for drawing any sinusoidal
graph
sinusoidal graph, positive, with max
KE at 0, T/2, T, 3T/2, and 2T
1 point for a graph that never shows
negative kinetic energy (graph is
always positive)
1 point for showing the maximum
kinetic energy is
1
2
mv0 and the
2
minimum kinetic energy is zero
1 point for correctly showing that the
maximum kinetic energy occurs at t =
0, T/2, T, 3T/2, and 2T and that the
minimum kinetic energy occurs at t =
T/4, 3T/4, 5T/4, and 7T/4.
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Work, Energy, & Power
C. The block is stopped and a second identical block is glued on top of the first. The
blocks are returned to simple harmonic motion with the same initial velocity as before.
i. How does the amplitude of the motion of the two blocks together compare to
the amplitude found in part a)?
Greater than __________
Less than ________
Equal ________
ii. Justify your answer.
(3 points max)
1 point For correctly indicating the
amplitude increases when the mass
increases
Greater than __x__
the increased mass increases the kinetic
energy when the block is set in motion,
meaning the spring will be compressed and
stretch more
1 point for any reasonable justification
using the equation from part A, as the mass
increases so does the amplitude
2
xmax =
mv0
k
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Work, Energy, & Power
Question 2 (20 pts)
A roller coaster car of mass 650 kg is moving with a velocity of 15.0 m/s at the top of the
first hill of a roller coaster track as shown in the diagram. The cart rolls without friction
down the hill and through a vertical circular loop of radius 12 m.
15.0 m/s
55 m
12 m
30 m
A. Calculate the maximum velocity of the roller coaster.
1 point for a correct application of
conservation of energy
(4 points max)
1
2
mv0 + mgh0 =
2
vmax =
2
2
2
mvmax
⎡ 1 v 2 + gh ⎤
0
⎢⎣ 2 0
⎥⎦
(
⎡ 1 15 m
s
⎢⎣ 2
m
= 36
s
vmax =
vmax
2
1
) + (9.8 m s ) ( 55m )⎤⎥⎦
2
2
1 point for recognizing that the
maximum kinetic energy occurs
when the roller coaster is at the
bottom of the first hill (zero
potential energy)
1 point for correctly adding the
initial kinetic and potential energy
to find the total energy
1 point for the correct answer
including correct units and
reasonable number of significant
digits
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Work, Energy, & Power
The roller coaster travels without friction through the circular loop.
i. On the diagram below, draw and label all of the forces acting on the roller coaster
when it is upside down at the top of the loop.
1 point for a correct force vector for
the Normal force
(3 points max)
1 point for a correct force vector for
the weight force
FN
1 point for no extraneous vectors
Fg
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Work, Energy, & Power
ii. Calculate the magnitude of the normal force exerted on the roller coaster when it is
upside down at the top of the loop.
(6 points max)
∑F = F
+ Fg = Fcp
N
FN =
mv
1 point for recognizing that the sum
of the forces provide a centripetal
force on the car
1 point for recognizing that the sum
of the forces is the normal force
minus the weight force
2
− mg
r
K top = Etotal − U g ,top
(
K = 423, 475 J − ( 650kg ) 9.8 m
s
2
1 point for correctly solving for the
normal force and substituting the
expression for the centripetal force
) ( 30m )
K = 232, 375 J
K=
v=
1
2
mv
2K
1 point for using conservation of
energy to find the velocity
2
2 ( 232, 375 J )
=
650kg
m
v = 26.74 m
FN =
FN =
mv
1 point for the correctly solving for
the velocity
s
2
− mg
r
( 650kg ) ( 26.74 m s )
12 m
2
(
− ( 650kg ) 9.8 m
s
2
)
1 point for the correct answer
including correct units and
reasonable number of significant
digits
FN = 32, 000 N
iii. The safety engineer determines that the acceleration of the riders is too great while
they are passing through the loop. Describe a way the engineer can modify the ride to
safely reduce the acceleration of the passengers as they go through the loop. Justify
your answer.
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Work, Energy, & Power
(2 points max)
increase the radius, increase height of loop,
decrease initial velocity, introduce drag or
friction
1 point for any reasonable
modification that would reduce the
acceleration of the riders through the
loop
1 point for an appropriate justification
increasing the height of the loop would
require the roller coaster to gain more
potential energy and thus it would have less
kinetic energy and would be traveling
slower
C. At the end of the ride, the roller coaster car returns to its initial height of 55 m before
being brought to a stop by friction. Determine the amount of work that must be done in
stopping the roller coaster at this height.
(5 points max)
Etop = Eend + W f
1 point for any indication of
conservation of energy
W = ΔK
K=
1
2
mv =
2
1
2
( 650kg ) (
15 m
s
)
1 point for indicating that the work
done against friction is equal to the
change in energy as the roller coaster
comes to a stop
2
K = 73,125 J
ΔK = W = −73,125 J
1 point for indicating that at the
beginning, the roller coaster has both
potential and kinetic energy
Alternate solution
U g = mgh
(
U g = ( 650kg ) 9.8 m
s
2
) (55m )
1 point for indicating that at the end,
the roller coaster has only potential
energy
U g = 350, 350 J
K = 423, 475 J − 350, 250 J
1 point for the correct answer
including correct units and reasonable
number of significant digits
K = 73,125 J
ΔK = W = −73,125 J
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Work, Energy, & Power
Question 3 (15 pts)
An 80 kg object is to be pulled to the top of a 6.0 m ramp by a rope, the other end of
which is pulled up by a 2400 W electric winch. The ramp forms a 30° angle with the
horizontal. The coefficient of kinetic friction between the ramp and the object is 0.6 and
the coefficient of static friction is 0.8. The object moves up the ramp at a constant
velocity.
6m
30 °
A. On the diagram of the object
showing all the forces on the object. Label each one.
(4 points max)
below, draw vectors
1 point for a correct force vector for the
Normal force
FN
FT
1 point for a correct force vector for the
tension
Ff
1 point for a correct force vector for the
frictional force
Fg
1 point for a correct force vector for the
weight force
1 point deducted for any extraneous
vectors with a maximum of four points.
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Work, Energy, & Power
B. Determine the maximum constant speed at which the winch can pull the object up the
ramp.
1 point for any statement indicating that
the net force is zero since the speed is
constant
(7 points max)
∑F
x
= Fg , x + Ff ,k − FT = 0
FT = Fg , x + Ff ,k
1 point for correctly summing the forces
to find the tension
FT = mg sin θ + μ k mg cos θ
(
FT = ( 80kg ) 9.8 m
(
+.6 ( 80kg ) 9.8 m
s
s
2
2
1 point for correctly substituting
expressions for the x-component of
weight and the force of friction
) sin ( 30°)
) cos ( 30° )
1 point for using the coefficient of
kinetic friction and not static friction
FT = 799 N
1 point for a statement that the maximum
constant speed at which the winch can
pull the object up the ramp is power or
the rate at which work is done
P = Fv
P = Fv
v=
P
=
F
v = 3.0
2400 J
1 point for a correct application of P =
Fv to find the speed
s
799 N
m
s
1 point for the correct answer including
correct units and reasonable number of
significant digits
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Work, Energy, & Power
C. Determine the total amount of work that the winch must do in pulling the object up the
ramp.
(4 points max)
1 point for an indication that the net
work is the net force times the distance
∑W = ∑ F ⋅ d
d =
6m
sin ( 30° )
1 point for indicating the distance the
block travels along the ramp
= 12 m
1 point for using the net force from
part B
∑ W = ( 799 N )(12m )
1 point for the correct answer
including correct units and reasonable
number of significant digits
W = 9790 J
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Work, Energy, & Power
Multiple Choice
1. A hydraulic car lift is capable of lifting a 400 kg car to a height of 2.5 m in .8 s. At
what rate does the car do work in lifting the car?
a) 500 W
b) 1,250 W
c) 5,000 W
d) 8,000 W
e) 12,500 W
Power is the rate at
which work is done
P=
W
=
F ⋅d
t
P=
t
( 400kg )
=
mgd
t
(10 m s ) ( 2.5m ) = 12, 500W
E
2
.8 s
F = 64 N
60°
Ff = 14 N
2. An object of mass 0.5 kg is initially at rest. A constant force of 64 N is exerted on the
object at a 60° relative to the horizontal as shown in the diagram. An opposing frictional
force of 14 N pulls the object in the opposite direction. After being displaced 0.5 m to
the right, the speed of the object is most nearly
a) 0 m/s
b) 6.0 m/s
c) 8.5 m/s
d) 9.0 m/s
e) 10.0 m/s
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Work, Energy, & Power
Work is equal to change
in energy
Work is done by forces
parallel to displacement
Forces parallel to displacement are the horizontal
component of F (Fx) and friction (Ff)
Fx = Fx − Ff = ( 64 N ) cos 60 − 14 N = 18 N
∑
W = F ⋅ d = (18 N ) ( .5m ) = 9 J
K =
v=
1
2
mv
2K
B
2
=
m
2 (9J )
.5kg
= 6m
s
3. While testing a new polymer fishing line, a student ties a mass of 0.25 kg to the end of
a 0.5 m long piece of fishing line and swings it in a horizontal circle such that the mass
completes ten revolutions each second. While swinging the mass in this manner, the
work done by the fishing line on the mass during one revolution is most nearly
a) 0 J
b) 0.125 J
c) 25 J
d) 50π3 J
e) 100π3 J
Work is done by forces
parallel to displacement
The force of tension is centripetal (perpendicular to
displacement) and thus does no work
A
4. In which of the following situations would a force be exerted on an object and no
work be done on the object?
I. a centripetal force is exerted on a moving object
II. a force is exerted in the opposite direction as the object is moving
III. a force is exerted on an object that remains at rest
a) I only
b) I and II
c) I and III
d) II and III
e) I, II, and III
Work is done by forces
parallel to displacement
In I the force is perpendicular to displacement and in III
there is no displacement.
C
Questions 5 – 6
An object of mass 0.8 kg is initially at rest on a horizontal frictionless surface. The
object is acted upon by a horizontal force F, the magnitude of which varies as function of
the displacement of the object d as shown in the graph.
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Work, Energy, & Power
F (N)
20
0
0
d (m)
0.2
5. Which of the following would be most likely to produce the force shown in the graph?
a) a string tied to a falling mass
b) a stretched spring
c) a magnet placed in the direction of the displacement
d) kinetic friction
e) static friction
The force is decreasing
as the object is displaced
The graph can only be produced by a variable force that
gets weaker as the object moves – a stretched spring that
pulls the object as it relaxes is the only one capable of
producing such a graph
B
6. The amount of work done by the force F in displacing the object 0.2 m is most nearly
a) 0 J
b) 1 J
c) 2 J
d) 4 J
e) 16 J
Work done by a variable
force is equal to the area
under the curve of F/d
graph
The graph forms a triangle
W =
1
2
(.2m )( 20 N ) = 2 J
®
⎡ A = 1 bh ⎤ so the work is
2 ⎥⎦
⎣⎢
C
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Work, Energy, & Power
7. A helicopter ascends vertically upwards with constant velocity v1. When the
helicopter is at height h, a passenger throws a baseball of mass m out of the window of
the helicopter horizontally with velocity v2. Just before striking the ground, the kinetic
energy of the baseball is
a) K = mgh
1
b) K = mgh + mv12
2
1
c) K = mgh + mv22
2
1
d) K = m ( v12 + v22 )
2
1
e) K = mgh + m ( v12 + v22 )
2
Conservation of Energy
– the total energy of the
ball remains constant
The baseball begins with two sources of energy: its vertical
velocity (kinetic) and its height (potential). When thrown,
additional kinetic energy is given to the ball. As it falls to
the ground, potential energy is converted to kinetic energy.
Thus, the sum of the three give the total kinetic energy just
before it reaches the ground.
K = K vertical + K horizontal + U g
K=
1
2
(
E
)
m v1 + v2 + mgh
2
2
8. A 30 kg sled is sliding on a frictionless sheet of ice at a velocity of 4 m/s. The sled
encounters a rough patch of ice and begins to slow down. After traveling on the rough
patch of ice for 3 m, the sled’s velocity is 2 m/s. Determine the coefficient of friction
between the rough ice and the sled.
a) 0.07
b) 0.12
c) 0.20
d) 0.27
e) 0.60
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Work, Energy, & Power
Work is equal to change
in energy
Find change in energy, which is equal to the work done:
2
2
1
⎤ = −180J
W = ΔK = K f − K i = m ⎡ 2 m
− 4m
s
s ⎦
2 ⎣
Friction is the only force on the sled:
W = Ff ⋅ d = μ mg ⋅ d
(
μ=
W
mg ⋅ d
=
) (
180 J
( 30kg )
(
10 m
s
2
)
( 3m )
)
C
= .2
Questions 9 – 10
A simple pendulum is constructed from a string of length l and a bob of mass m as shown
in the diagram. It is released from rest at point I, which is a vertical distance y from the
equilibrium position. The bob has zero potential energy at point II. Friction and drag are
negligible.
l
III.
I.
y
y
II.
9. The speed of the pendulum bob when it is in position II is most nearly
1 2
ml
2
b) 2gy
a)
c)
2g ( l − y )
d)
2g ( l 2 − y 2 )
e)
2g (l − y )
m
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Work, Energy, & Power
Conservation of Energy
At I, the bob has zero kinetic energy and at II the bob has
zero potential energy. Potential energy at I is converted to
kinetic energy at II (energy is conserved).
U g , I = K II
mgy =
v=
1
2
mv
B
2
2 gy
10. Which of the following graphs best shows the total energy of the pendulum bob as a
function of displacement?
a)
I.
b)
II.
III.
c)
I.
II.
III.
I.
II.
III.
e)
d)
I.
Conservation of Energy
II.
III.
I.
II.
III.
Since there are no non-conservative forces acting on it, the
total energy of the bob remains constant. Graph E shows
this
®
E
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Work, Energy, & Power
11. A block of mass m is pulled across a rough horizontal surface by a force F at a
constant speed v. The coefficient of kinetic friction between the block and the surface is
μ. The rate at which the force does work on the block is
a)
2mgv
μ
v
μmg
μmg
c)
v
mgv
d)
b)
μ
e) μmgv
Power is the rate at
which work is done
Since the block moves with constant velocity, the frictional
force is equal to the force F = μmg.
P = F ⋅ v = μ mgv
E
12. An archer does 38 J of work in drawing a bowstring back. When he releases the
bowstring, the arrow, which has a mass of 0.25 kg, reaches a maximum height of 12 m.
Determine the velocity of the arrow when it is at its maximum height. Assume all of the
energy from the bow was transferred to the arrow.
a) 0 m/s
b) 4.0 m/s
c) 8.0 m/s
d) 15 m/s
e) 17 m/s
Conservation of Energy
Energy is conserved in this situation, meaning that the total
energy of the arrow must always equal 38 J. Find the
potential energy of the arrow at its max height:
(
U g = mgh = ( .25kg ) 10 m
)
2 (12 m ) = 30 J
s
Since there is 30J of potential energy, the arrow must have
8 J of kinetic energy. From this, we can find the velocity
1
2
K = mv
2
v=
2K
=
m
®
2 (8J )
.25kg
= 8m
C
s
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