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Chapter 11. Work
Chapter Goal: To develop
a more complete
understanding of energy and
its conservation.
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Ch 11 Student Learning Objectives
• To introduce and use the basic energy model.
• To recognize transformations between kinetic,
potential, and thermal energy.
• To define work and use the work-kinetic
energy theorem.
• To develop a complete statement of the law of
conservation of energy.
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The Basic Energy Model
W > 0: The environment does work on the system and the
system’s energy increases.
W < 0: The system does work on the environment and the
system’s energy decreases.
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Work and Kinetic Energy
A force acts on a particle as the particle moves along the saxis from si to sf. The force component Fs parallel to the saxis causes the particle to speed up or slow down, thus
transferring energy to or from the particle. We say that the
force does work on the particle.
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Work and Kinetic Energy
A force acts on a particle as the particle moves along the saxis from si to sf. The force component Fs parallel to the saxis causes the particle to speed up or slow down, thus
transferring energy to or from the particle. We say that the
force does work on the particle.
As the particle is moved by this single force, its kinetic
energy changes as follows:
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Work-Kinetic Energy Theorem.
Starting with Newton’s 2nd Law: Fs = mas, it can be
shown that:
sf
2 – ½ mv 2 or:
=
½
mv
F
ds
f
0
 s
s0
This result is known as the Work-Kinetic Energy
Theorem. Note that there is no mention of U, or
potential energy, as of yet.
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A particle moving along the x-axis experiences the
force shown in the graph. If the particle has 2.0 J
of kinetic energy as it passes x = 0 m, what is its
kinetic energy when it reaches x = 4 m?
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A particle moving along the x-axis experiences the
force shown in the graph. If the particle has 2.0 J
of kinetic energy as it passes x = 0 m, what is its
kinetic energy when it reaches x = 4 m?
6.0 J
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Work Done by a Constant Force
A particle experiences a constant force which makes an
angle θ with respect to the particle’s displacement, ∆r. The
work done is
Both F and θ are constant, so they can be taken outside the
integral. Thus
This as the dot product of the force vector and the
displacement vector:
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Work done by a constant Force
Therefore, for a constant force:
W = |F∆r| cos θ
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Tactics: Calculating the work done by a
constant force
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Tactics: Calculating the work done by a
constant force
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Evaluate the dot products
1.
2.
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1a)15.3 1b) -4.0 1c) 0 2a) -4.1 2b) -20 2c) 10.4
1.
2.
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A crane lowers a steel girder into place at a
construction site. The girder moves with
constant speed. Consider the work Wg done
by gravity and the work WT done by the
tension in the cable. Which of the following is
correct?
A.
B.
C.
D.
E.
Wg and WT are both zero.
Wg is negative and WT is negative.
Wg is negative and WT is positive.
Wg is positive and WT is positive.
Wg is positive and WT is negative.
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A crane lowers a steel girder into place at a
construction site. The girder moves with
constant speed. Consider the work Wg done
by gravity and the work WT done by the
tension in the cable. Which of the following is
correct?
A.
B.
C.
D.
E.
Wg and WT are both zero.
Wg is negative and WT is negative.
Wg is negative and WT is positive.
Wg is positive and WT is positive.
Wg is positive and WT is negative.
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Workbook Problem #7e
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Workbook Problem #7e
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The Work Done by a Variable Force
To calculate the work done on an object by a force that
either changes in magnitude or direction as the object
moves, we use the following:
We must evaluate the integral either geometrically, by
finding the area under the curve, or by actually doing the
integration (!).
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Workbook #13
A 1 kg particle moving
along the x-axis
experiences the force
shown in the graph.
The particle’s speed is 2
m/s at x = 0 m. What is
its speed when it gets to
x = 5 m?
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Workbook #13
A 1 kg particle moving
along the x-axis
experiences the force
shown in the graph.
The particle’s speed is 2
m/s at x = 0 m. What is
its speed when it gets to
x = 5 m?
Ans: 4 m/s
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Conservation of Mechanical Energy revisited
If no energy is added or removed
from the system:
K0 + Ug0 + Us0 = K + Ug + Us
If energy is added or removed via
work, this becomes:
K0 + Ug0 + Us0 + W = K + Ug + Us
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Conservation of Mechanical Energy revisited
K0 + Ug0 + Us0 + W = K + Ug + Us
It seems this would conflict with the Work-Kinetic
Energy, W = K – K0
This would imply that perhaps potential energy is
a form of work.
It can be shown that W = - ∆U
But only for conservative forces.
A conservative force is one that is pathindependent.
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The Work-Kinetic Energy Theorem when
Nonconservative Forces Are Involved
A force for which the work is not independent of the path
is called a nonconservative force. It is not possible to
define a potential energy for a nonconservative force.
If Wc is the work done by all conservative forces, and Wnc
is the work done by all nonconservative forces, then
But the work done by the conservative forces is the
negative of the change in potential energy, so the workkinetic energy theorem becomes
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Energy Bar Charts
We may express the conservation of energy concept as an
energy equation.
We may also represent this equation graphically with an
energy par chart.
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EOC #36
A particle moves from A to D
while experiencing a force
of (6iˆ  8 ˆj ) N.
Calculate the work done for:
a. Path ABD
b. Path ACD
c. Path AD
d. Is this a conservative
force?
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EOC #36
A particle moves from A to D
while experiencing a force
of (6iˆ  8 ˆj ) N.
Calculate the work done for:
a. Path ABD: 50 J
b. Path ACD: 50 J
c. Path AD: 50 J
d. Is this a conservative
force? - Yes
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EOC #36 -Answer
• Path ABD: 50 J
• Path ACD
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Conservation of Mechanical Energy, revisited
K0 + Ug0 + Us0 + Wnc = K + Ug + Us
can be rewritten as:
∆K + ∆U = Wnc
and
∆K + ∆U = ∆Emech
Therefore mechanical energy is conserved if
there are no non-conservative forces doing
work on the system
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Finding Force from Potential Energy
so Fs = dW/ds
W = - ∆U: The work done by a conservative
force is equal to the negative change in
potential energy
Therefore: Fs = - dU/ds
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A function and its derivative
Fs = - dU/ds
The derivative of a function shows the trend the change
of that function.
• If the derivative is zero, the function is not changing
(but not necessarily 0!).
• If the derivative is constant, the function is changing
at a constant rate.
• If the derivative is increasing/decreasing the function
is changing increasingly faster/slower.
• The negative sign is a bit of a challenge.
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• The top graph is of the
gravitational force, mg
which we model as
constant. In our
coordinate system it is
negative.
• The gravitational
potential energy curve is
linear (changing at a
constant rate), with a
positive slope equal to
the magnitude of mg.
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A particle moves along the x-axis with
the potential energy shown. The force on
the particle when it is at x = 4 m is
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A particle moves along the x-axis with
the potential energy shown. The force on
the particle when it is at x = 4 m is
–2 N.
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Workbook #17 a,b,c
The graph shows the potentialenergy curve of a particle
moving along the x-axis under
the influence of a conservative
force.
a. At which intervals of x is the
force on the particle to the right
(positive)?
b. At which intervals of x is the
force on the particle to the left
(negative)?
c. At what value(s) of x is the
force on the particle maximum?
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d. At what value(s) is
the force on the
particle zero?
Workbook #17 a,b,c
d. force is
zero at: x=2m,
x=5m, x=8m
(where slope is
zero).
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Dissipative Forces – Two perspectives
Dissipative forces (e.g. friction and drag) can be
modeled in two ways:
1. One can consider the dissipative force as an
external, non-conservative force that is
always negative and takes energy away from
the system:
K0 + Ug0 + Us0 + Wnc – Wdis = K + Ug + Us
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Dissipative Forces – Two perspectives
2. One can consider the dissipative force as increasing
the thermal energy of the system, as discussed in
the text:
K0 + Ug0 + Us0 + Wnc = K + Ug + Us + ∆Eth
This method is probably the more correct way to think
about it. However ∆Eth can only be computed as
work (e.g. ∆Eth due to friction is ukn ∆s), so in the
end it doesn’t matter where you include it on the bar
chart or the equation. Pick a method and stick with
it
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Conservation of Energy
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A child at the playground slides down a
pole at constant speed. This is a situation
in which
A. U  Eth. Emech is conserved.
B. U  Eth. Emech is not conserved but Esys is.
C. U  Wext. Neither Emech nor Esys is conserved.
D. U  K. Emech is not conserved but Esys is.
E. K  Eth. Emech is not conserved but Esys is.
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
A child at the playground slides down a
pole at constant speed. This is a situation
in which
A. U  Eth. Emech is conserved.
B. U  Eth. Emech is not conserved but Esys is.
C. U  Wext. Neither Emech nor Esys is conserved.
D. U  K. Emech is not conserved but Esys is.
E. K  Eth. Emech is not conserved but Esys is.
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
A child at the playground slides down a
pole at constant speed. This is a situation
in which
A. U  Eth. Emech is conserved.
B. U  Eth. Emech is not conserved but Esys is.
C. U  Wext. Neither Emech nor Esys is conserved.
D. U  K. Emech is not conserved but Esys is.
E. K  Eth. Emech is not conserved but Esys is.
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EOC problems 26, 27,28
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EOC #26 - Answer
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EOC #27 - Answer
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EOC #28 - Answer
• Wext = -1 J. This means
energy was transferred
from the system to the
environment
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EOC #37
A 100 g particle experiences
only the one-dimensional,
conservative force shown
in the figure.
a. Draw a graph of potential
energy from x = 0m to x =
5 m. Assume U = 0 J at
0m.
b. The particle is shot toward
the right from x = 1.0 m
with a speed of 25 m/s.
What is the particle’s total
mechanical energy?
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EOC #37
a. see graph below
b. Total Emech = 51.25
Emech = (K + U)
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EOC #48
A runaway truck ramp near
Denver, CO slopes upward
at 6.0° and has a large
coefficent of rolling friction,
ur = 0.40. Find the length of
a ramp that will stop a
15,000 kg truck that enters
the ramp at 35.0 m/s.
a. Draw an energy bar chart
for this situation.
b. Use work and energy and
kinematics to solve.
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EOC #48: Answer = 124 meters
A runaway truck ramp near
Denver, CO slopes upward
at 6.0° and has a large
coefficent of rolling friction,
ur = 0.40. Find the length of
a ramp that will stop a
15,000 kg truck that enters
the ramp at 35.0 m/s.
a. Draw an energy bar chart
for this situation.
b. Use work and energy and
kinematics to solve.
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EOC #55
A 5.0 kg box slides down a 5.0 m high frictionless hill
starting from rest. It slides across a 2-m long
“rough” horizontal section with uk = 0.25. At the
end of the horizontal section the box hits a
horizontal spring with k = 500 N/m.
a. How fast is the box going before reaching the rough
surface?
b. How fast is it going just before hitting the spring?
c. How far is the spring compressed?
d. How many complete crossings can the box make
before coming to rest?
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EOC #55 - Answers
A 5.0 kg box slides down a 5.0 m high frictionless hill starting
from rest. It slides across a 2-m long “rough” horizontal
section with uk = 0.25. At the end of the horizontal section
the box hits a horizontal spring with k = 500 N/m.
a. How fast is the box going before reaching the rough
surface? 9.9 m/s
b. How fast is it going just before hitting the spring? 9.4 m/s
c. How far is the spring compressed? 0.94 m
d. How many complete crossings can the box make before
coming to rest? 10 trips
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Final Review #1 (Impulse momentum
A superball of mass 0.05 kg is dropped from a height of 10 cm above floor. It
bounces and rebounds to a three-fourths of its original height.
a. Sketch a quantitatively correct graph of the force of gravity on the ball
during the time of impact. Use an appropriate number scale for the axes:
b. Is the magnitude of the average force of the ball on the floor greater than,
less than or equal to the magnitude of the average force of the floor on the
ball? Explain your answer.
c. Is the magnitude of the average force of the floor on the ball greater than,
less than or equal to the force of gravity on the ball? Explain your answer.
d. Determine the maximum force that the floor exerts on the ball during the
collision. Assume that the impulse due to the weight is negligible.
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Final Review #2 –
Ch 8 - EOC #47
A 100 g ball on a
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