Download Kinetic and Potential Energy

Chapter 10. Energy
Chapter Goal: To introduce
the ideas of kinetic and
potential energy and to learn
a new problem-solving
strategy based on
conservation of energy.
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Ch. 10 – Student Learning Objectives
• To begin developing a concept of energy—
what it is, how it is transformed, and how it is
transferred.
• To introduce the concepts of kinetic and
potential energy.
• To learn Hooke’s law for springs and the new
idea of a restoring force.
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Energy – The Big Picture
Two basic types of mechanical
energy:
• Kinetic energy (K) is an energy of
motion.
• Gravitational potential energy (Ug)
is an energy of position.
Under some circumstances (e.g.
freefall) these two kinds of energy
can be transformed back and forth
without loss from the system.
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Kinetic and Potential Energy
For some systems (e.g. systems in freefall):
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½ mv2f + mgyf = ½ mv20 + mgy0
where y is the height above an arbitrary zero,
(not the displacement).
This result is true for motion along any
frictionless surface, regardless of the shape.
This is a generalization of a relationship we
already use for free fall:
v2f = v20 + 2(-g)(yf – y0)
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Kinetic and Potential Energy
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A box slides along the frictionless surface shown in the
figure. It is released from rest at the position shown. Is
the highest point the box reaches on the other side at
level a, at level b, or level c?
A. At level a
B. At level b
C. At level c
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
A box slides along the frictionless surface shown in the
figure. It is released from rest at the position shown. Is
the highest point the box reaches on the other side at
level a, at level b, or level c?
A. At level a
B. At level b
C. At level c
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Energy Bar charts – a visual aid
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Workbook Problems 12-14
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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
EOC #4
a. What is the kinetic energy of a 1500 kg car
traveling at 30 m/s?
b. From what height would the car have to be
dropped (!) to have the same amount of
energy upon impact?
c. Repeat parts a and b for a car of 3000 kg. By
how much did the height change?
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EOC#4
a. 6.75 x 106 Joules
b. Solve for the amount
of initial potential
energy needed to give
the car a final kinetic
energy equal to part a
to get y = 46 m
K0 + Ug0 = Kf + Ugf
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EOC#4
a. 6.75 x 106 Joules
b. Solve for the amount
of initial potential
energy needed to give
the car a final kinetic
energy equal to part a
to get y = 46 m.
c. Same height
K0 + Ug0 = Kf + Ugf
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The Zero of Potential Energy
• You can place the origin of your coordinate system, and
thus the “zero of potential energy,” wherever you
choose and be assured of getting the correct answer to a
problem.
• The reason is that only ΔU has physical significance,
not Ug itself.
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Workbook Problem #8
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Workbook Problem #8 - Answer
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Conservation of Mechanical Energy
The sum of the kinetic energy and the potential energy of a
system is called the mechanical energy.
the kinetic energy and the potential energy can change, as
they are transformed back and forth into each other, but
their sum remains constant.
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Conservation of Mechanical Energy
• Under what conditions is Emech conserved?
• What happens to the energy when Emech is not
conserved?
• Are there potential energies other than
gravitational, and how do you calculate them?
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Hooke’s Law
If you stretch a rubber band, a force appears that tries to pull
the rubber band back to its equilibrium, or unstretched,
length. A force that restores a system to an equilibrium
position is called a restoring force. If s is the position of the
end of a spring, and se is the equilibrium position, we define
Δs = s – se. If (Fsp)s is the s-component of the restoring force,
and k is the spring constant of the spring, then Hooke’s Law
states that
The minus sign is the mathematical indication of a restoring
force.
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Hooke’s Law
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Workbook #17
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Workbook #17 - Answer
a. 13 cm
b. 8 cm
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The graph shows force
versus displacement for
three springs. Rank in
order, from largest to
smallest, the spring
constants k1, k2, and k3.
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The graph shows force
versus displacement for
three springs. Rank in
order, from largest to
smallest, the spring
constants k1, k2, and k3.
Answer: k1 > k2 > k3
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Elastic Potential Energy
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Elastic Potential Energy
• Us = ½ k (∆s)2
• ∆s = (s-se) where se is the
equilibrium position of the
end of the spring
• ∆Us = ½ k [(sf - se)2 - (s0 - se)2].
Often sf or s0 = se as the spring
begins or ends in equilibrium
position.
at s=se, the ball will
lose contact with the
spring
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Workbook problems 19, 20, 21
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EOC #21
A student places his 500 g physics book on a
frictionless table and pushes it against a spring,
compressing it by 4.0 cm. The spring constant
is 1250 N/m. He then releases the book. What
is the speed as it slides away from the spring?
Draw a before and after picture:
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EOC #21
Knowns
m=.5 kg
k = 1250 N/m
x1 = -.04 m
v1 = 0m/s
Find: v2, the speed of the book
when it is released
Draw an energy bar chart to
determine what energy
transformation takes place
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EOC #21
• The elastic potential
energy of the spring is
converted into the
kinetic energy of the
book.
• Solve by replacing each
energy term in the bar
chart with a value:
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½ mv12 + ½ k(x1 – xe) 2 = ½ mv22 + ½ k(x2 – xe) 2
This simplifies to:
2
kx
½ k(x1) 2 = ½ mv22 and v2 = m1
= 2.0 m/s
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Problem with Ug and Us (EOC #44)
A 1000 kg safe is 2.0 m above a heavy-duty
spring when the rope holding the safe breaks.
The safe hits the spring and compresses it 50
cm. What is the spring constant of the spring?
Draw a before and after picture and decide on a
zero.
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Problem with Ug and Us (EOC #44)
Here the author decided on ye
(equilibrium of the spring).
Good choice, since we have
no information on the length
of the spring. The relevant
information was given
relative to the top of the
spring.
The safe has some velocity
when it first hits the spring.
Do we need to know that?
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Kinetic energy in the middle of the problem has
no bearing and does not need to be calculated
A 1000 kg safe is 2.0 m
above a heavy-duty
spring when the rope
holding the safe breaks.
The safe hits the spring
and compresses it 50
cm. What is the spring
constant of the spring?
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
A 1000 kg safe is 2.0 m above
a heavy-duty spring when
the rope holding the safe
breaks. The safe hits the
spring and compresses it 50
cm. What is the spring
constant of the spring?
Using the non-zero terms from the
energy bar chart, we get:
mgy0 = mgy1 + ½ k (y1- ye) 2
k= 196,000 N/m
Note that mgy1 is negative for the
zero chosen.
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