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Work and Energy
Physics Chapter 5
In Chapter 5…
 Work
 Energy
 Conservation
 Power
of Energy
Section 1 Objectives
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Recognize the difference between the scientific and
ordinary definitions of work.
Define work by relating it to force and displacement.
Identify where work is being performed in a variety of
situations.
Calculate the net work done when many forces are
applied to an object.
Below is a link to a video about Work and Energy:
http://videos.howstuffworks.com/hsw/20908-physics-work-andenergy-video.htm
Work
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Work: The product of the component of a force along
the direction of displacement and the magnitude of
the displacement.
Work is done on an object when a force causes a
displacement of the object.
W = Fd
Work is done only when components of a force are
parallel to a displacement.
Wnet = Fnetdcos θ
Work
W = Fd
Wnet = Fnetdcos θ
Units of Work Newtons times meters (N × m), or joules (J).
Work
Work is NOT being done here.
The force applied is
perpendicular to the direction of
the displacement.
Work is done only when
components of a force are
parallel to a displacement.
Sample Problem 1
Page 162 Questions 3
A shopper in a supermarket pushes a cart with a force of 35 N
directed at an angle of 25° downward from the horizontal. Find the
work done by the shopper on the cart as the shopper moves along a
50.0 m length of aisle.
Wnet = ?
Wnet = Fnetdcos θ
Fnet = 35 N
θ = 25°
Wnet = (35 N)(50.0 m)(cos 25°)
d = 50.0 m
Wnet = 1.6 ×103 J (to 2 sig figs)
Sample Problem 2
Page 162 Questions 4
If 2.0 J of work is done in raising a 180 g apple, how far is it lifted?
d=?
Wnet = Fnetd
Wnet = 2.0 J
m = 180 g = 0.18 kg
Fnet = ma
a = 9.81 m/s2
Fnet = (0.18 kg)(9.81 m/s2 ) = 1.7658 N
d = Wnet/Fnet
d = (2.0 J)/(1.7658 N)
d = 1.1 m (to 2 sig figs)
Work
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Work is a scalar quantity. It can be positive or
negative.
Work is positive when the force is in the same
direction as the displacement.
Work is negative when the force is in the direction
opposite the displacement.
The sign of the net work can tell you whether the
object’s speed is increasing or decreasing.
Work
Positive Work  Force
is in same direction as
the displacement.
Positive Net Work 
Increasing speed.
Negative Work Force
is in opposite direction
from the displacement.
Negative Net Work 
Decreasing speed.
Section 2 Objectives
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Identify several forms of energy.
Calculate kinetic energy for an object.
Apply the work-kinetic energy theorem to solve
problems.
Distinguish between kinetic and potential energy.
Classify different types of potential energy.
Calculate the potential energy associated with an
object’s position.
Energy
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Energy: The ability to do work, or cause a change.
Kinetic Energy (KE): Energy of an object due to its
motion.
KE depends on speed and mass.
KE = ½mv2
KE is a scalar quantity.
Units of KE  kilograms times meters per second
squared (kg × (m/s)2), or joules (J).
Sample Problem 3
Page 166 Question 2
What is the speed of a 0.145 kg baseball if its kinetic energy is 109 J?
v=?
KE = ½mv2
m = 0.145 kg
KE = 109 J
v = √((2KE)/m)
v = √((2 × 109 J)/(0.145 kg))
v = 38.8 m/s (to 3 sig figs)
Sample Problem 4
Page 166 Question 5
A car has a kinetic energy of 4.32 × 105 J when traveling at a speed of
23 m/s. What is its mass?
m=?
KE = ½mv2
KE = 4.32 × 105 J
v = 23 m/s
m = ((2KE)/v2)
m = (2(4.32 × 105 J)/((23 m/s) 2))
m = 1.6 × 103 kg (to 2 sig figs)
Energy
W = FΔx = maΔx
vf2 = vi2 + 2aΔx
(vf2 - vi2)/2 = aΔx
W = FΔx = maΔx
W = (vf2 - vi2)/2 = ½mvf2 - ½mvi2
FΔx = ½mvf2 - ½mvi2
 Work-Kinetic Energy Theorem  Wnet = ΔKE
 Net work done by all the forces acting on an object is
equal to the change in the object’s kinetic energy.
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Sample Problem 5
Page 168 Question 4
A 75 kg bobsled is pushed along a horizontal surface by two athletes.
After the bobsled is pushed a distance of 4.5 m starting from rest, its
speed is 6.0 m/s, Find the magnitude of the net force on the bobsled.
Fnet = ?
Wnet = ΔKE
KE = ½mv2
Wnet = Fnetd
m = 75 kg vi = 0 m/s
vf = 6.0 m/s d = 4.5 m
Wnet = Fnetd = ΔKE = ½mvf2 - ½mvi2
Sample Problem 5
Page 168 Question 4 CONTINUED
A 75 kg bobsled is pushed along a horizontal surface by two athletes.
After the bobsled is pushed a distance of 4.5 m starting from rest, its
speed is 6.0 m/s, Find the magnitude of the net force on the bobsled.
Fnetd = ½mvf2 - ½(75 kg)(0 m/s)2
Fnetd = ½mvf2 - 0
Fnet = (½mvf2)/d
Fnet = (½(75 kg)(6.0 m/s)2)/(4.5 m)
Fnet = 3.0 × 10 N2 (to 2 sig figs)
Energy
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Potential energy is stored energy.
Energy associated with an object because of its
position, shape, or condition.
Gravitational potential energy depends on height
from a zero level.
Gravitational potential energy: Potential energy
stored in the gravitational fields of interacting bodies.
PEg = mgh
Sample Problem 6
A 517 g object is placed on a table. It has a gravitational potential
energy of 72.0 J. How tall is the table?
h=?
PEg = mgh
PEg = 72.0 J m = 517 g = 0.517 kg
h = PEg/mg
h = 72 J/(0.517 kg)(9.81 m/s2)
g = 9.81 m/s2
h = 14.2 m (to 3 sig figs)
Energy
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Elastic potential energy depends on distance
compressed or stretched.
Energy available for use when a deformed elastic
object returns to its original configuration.
PEelastic = ½kx2
Spring constant (k): A parameter that is a measure of
a spring’s resistance to being compressed or
stretched.
Energy
PEelastic = ½kx2
Sample Problem 7
Page 172 Question 1
A spring with a force constant of 5.2 N/m has a relaxed length of 2.45
m. When a mass is attached to the end of the spring and allowed to
come to rest, the vertical length of the spring is 3.57 m. Calculate the
elastic potential energy stored in the spring.
PEelastic = ?
PEelastic = ½kx2
k = 5.2 N/m
Relaxed length = 2.45 m Stretched length = 3.57 m
x = 3.57 m – 2.45 m = 1.12 m
PEelastic = ½(5.2N/m)(1.12 m)2
PEelastic = 3.3 J (to 2 sig figs)
Section 3 Objectives
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Identify situations in which conservation of
mechanical energy is valid.
Recognize the forms that conserved energy can take.
Solve problems using conservation of mechanical
energy.
Below is a link to a video about Conservation of
Momentum and Energy:
http://videos.howstuffworks.com/hsw/20915-physics-conservation-ofmomentum-and-energy-video.htm
Conservation of Energy
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Mechanical Energy: Sum of kinetic energy and all
forms of potential energy.
Non-mechanical Energy: Nuclear, chemical, internal,
electrical energy.
Mechanical is conserved. ALL energy is conserved.
ME = KE + PE
MEi = MEf (no friction)
KEi + PEg i + PEelastic i = KEf + PEg f + PEelastic f
Energy is conserved even when acceleration varies.
Mechanical energy is not conserved if there is
friction.
Classification of Energy
Energy
Mechanical
Kinetic
Nonmechanical
Potential
Gravitational
Potential
Conservation of Energy
Mechanical is conserved. ALL energy is conserved.
ME = KE + PE
MEi = MEf
Sample Problem 8
Page 177 Question 2
A 755 N diver drops from a board 10.0 m above the water’s surface.
Find the diver’s speed 5.00 m above the water’s surface. Then find
the diver’s speed just before striking the water.
v = ? 5.00 m above water
MEi = MEf KEi + PEg i = KEf + PEg f
½mvi2 + mghi = ½mvf2 +mghf
vi = 0 m/s
mghi = ½mvf2 + mghf
Sample Problem 8
Page 177 Question 2 CONTINUED 1
A 755 N diver drops from a board 10.0 m above the water’s surface.
Find the diver’s speed 5.00 m above the water’s surface. Then find the
diver’s speed just before striking the water.
m = (755 N)/(9.81 m/s2) = 76.96228338 kg
hi = 10.0 m
hf =5.00 m
vf = √(2(mghi – mghf)/m) = √(2(ghi – ghf)) = √(2g(hi – hf))
v = √(2(9.81 m/s2)(10.0 m – 5.00 m)
v = 9.90 m/s (to 3 sig figs)
5.00 m above water
Sample Problem 8
Page 177 Question 2 CONTINUED 2
A 755 N diver drops from a board 10.0 m above the water’s surface.
Find the diver’s speed 5.00 m above the water’s surface. Then find
the diver’s speed just before striking the water.
v = ? before hitting water
vi = 0 m/s
m = 76.96228338 kg
hi = 10.0 m
mghi = ½mvf2
vf = √(2(mghi)/m) = √(2ghi)
v = √(2 × 9.81 m/s2 × 10.0 m)
v = 14.0 m/s (to 3 sig figs)
before hitting water
hf = 0 m
Section 4 Objectives
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Relate the concepts of energy, time, and power.
Calculate power in two different ways.
Explain the effect of machines on work and power.
Power
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Power: The rate at which work is done or energy is
converted.
P = W/Δt = Joules/seconds = Watt
P = Fd/Δt
P = mad/Δt (because F = ma)
P = Fv
1 Horsepower = 746 Watts
Sample Problem 9
Page 181 Question 1
A 1.0 × 103 kg elevator carries a maximum load of 800.0 kg. A
constant frictional force of 4.0 × 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?
Power = ? (in kilowatts)
P = Fv
Elevator with max. load = (1.0 × 103) + 800.0 kg
v = 3.00 m/s
Frictional force = 4.0 × 103 N
Sample Problem 9
Page 181 Question 1 CONTINUED
A 1.0 x 103 kg elevator carries a maximum load of 800.0 kg. A
constant frictional force of 4.0 × 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?
Elevator = 1800 kg × 9.81 m/s2 = 17658 N
Fnet = 17658 N + (4.0 × 103 N) = 21658 N
P = 21658 N × 3.00 m/s = 64974 W
P (in kW) = 64974 W/(1.0 × 103) W
P = 65 kW (to 2 sig figs)
Key Terms in Chapter 5
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Work
Kinetic Energy
Work-Kinetic Energy Theorem
Potential Energy
Gravitational Potential Energy
Elastic Potential Energy
Spring Constant
"I think I can safely say that nobody understands
Mechanical Energy
Quantum Mechanics“ - Richard P Feynman
Physics is not a religion. If it were, we'd have a
Power
much easier time raising money. - Leon Lederman
Equations in Chapter 5
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Net Work  Wnet = Fnetd or Wnet = Fnetdcos θ
Kinetic Energy  KE = ½mv2
Work-Kinetic Energy Theorem  Wnet = ΔKE
Gravitational Potential Energy  PEg = mgh
Elastic Potential Energy  PEelastic = ½kx2
Mechanical Energy  ME = KE + PE
Conservation of Mechanical Energy  MEi = MEf
Power  P = W/Δt = Fv
"If Quantum Theory is correct, it
signifies the death of Physics as a
Science“ -Albert Einstein
Sources
Chapter 5 Notes from Mr. G. Johnson’s Physics Class

Holt Physics Textbook 2006 Pages 160-183
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Pictures:
Microsoft Office Clip Art
http://www.cars-bikes.info/d/2256-2/lamborghini-reventon-001.jpg
http://firstrung.co.uk/dbimgs/slow_down.jpg
http://resources.yesican-science.ca/energy_flow/images/kinetic_energy1.png
http://citruscollege.com/pic/46/c05_05.jpg
http://www.explainthatstuff.com/yoyoenergy.gif
http://images.despair.com/products/demotivators/power.jpg
http://z.about.com/d/esl/1/0/I/2/bobsled.gif
http://www.ibiblio.org/kuphaldt/electricCircuits/AC/02095.png
http://id.mind.net/~zona/mstm/physics/mechanics/energy/work/work2.gif
http://www.nu.ac.za/physics/1M2002/Energy%20work%20and%20power_files/image002.jpg
http://www.nu.ac.za/physics/1M2002/Energy%20work%20and%20power_files/image004.jpg

Sources
"I believe there are
15,747,724,136,275,002,577,605,653,961,181,555,468,044,717,
914,527,116,709,366,231,425,076,185,631,031,296
protons in the universe and an equal number of electrons". (=2256 x 136)
- Sir Arthur Eddington, in "The Philosophy of Physical Science".
Cambridge, 1939.
Other Useful Websites:
http://www.nu.ac.za/physics/1M2002/Energy%20work%20and%20power.htm
http://www.cliffsnotes.com/WileyCDA/CliffsReviewTopic/Work-and-Energy.topicArticleId10453,articleId-10418.html
http://www.fortunecity.com/greenfield/eagles/180/work_and_energy.html
http://www.sparknotes.com/testprep/books/sat2/physics/chapter7section1.rhtml
http://theory.uwinnipeg.ca/physics/work/index.html
http://videos.howstuffworks.com/hsw/20908-physics-work-and-energy-video.htm
http://videos.howstuffworks.com/hsw/20915-physics-conservation-of-momentum-and-energyvideo.htm
Quotes By Lecturers In Classroom
“A cube is obviously not an ellipse in the normal sense.”
“This circle has too much symmetry.”
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go
into the details at this time.“
http://www.astro.psu.edu/users/burrows/quotes.html