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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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Conservation of Mechanical 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 any loss from the system.
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Conservation of Mechanical Energy
For some systems (e.g. systems in freefall), mechanical
energy is conserved:
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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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Conservation of 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
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Analysis of energy conservation problems: The
before and after picture
Bob uses a slingshot to
launch a 20.0 g pebble
up with an initial speed
of 25.0 m/s. How high
above his hand does the
pebble go?
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Energy Bar charts – a visual aid
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Analysis of energy conservation problems: bar
charts
Bob uses a slingshot to
launch a 20.0 g pebble
up with an initial speed
of 25.0 m/s. How high
above his hand does the
pebble go?
Now solve using the
appropriate parts of the
conservation of
mechanical energy
equation.
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Analysis of energy conservation problems: bar
charts
Bob uses a slingshot to
launch a 20.0 g pebble
up with an initial speed
of 25.0 m/s. How high
above his hand does the
pebble go?
Now solve using the
appropriate parts of the
conservation of
mechanical energy
equation: 32 meters
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Workbook Problems 5-7
5. A car runs out of gas and coasts up a hill
until finally stopping
6. A pendulum is held out at 45 degrees and
released from rest. A short time later, it
swings through the lowest point on the arc.
7. A ball starts from rest on on the top of one
hill, rolls without friction through a valley
and just barely makes it to the top of an
adjacent hill before stopping
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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?
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EOC#4
a. 6.75 x 105 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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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 #2
We see a 1 kg object that is
initially 1 m above the
ground and rises to a height
of 2 m. Anjay, Brittany and
Carlos each measure its
position, but each of them
uses a different coordinate
system. Determine Ui, Uf
and ∆U for each coordinate
system. Which of these, if
any remain constant for all 3
coordinate systems?
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Workbook Problem #2 - Answer
∆ U remains the same for all.
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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 #10
A spring has an unstretched length of 10 cm. It
exerts a restoring force F when stretched to a
legth of 11 cm.
For what length is the restoring force 3F?
A. 33 cm
B. 13 cm
C. 2 cm
D. 14 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.
A. k1, k2, k3
B. k3, k2, k1
C. 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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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/after sketch and an energy bar
chart for this situation.
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EOC #21
Knowns
m=.5 kg
k = 1250 N/m
x1 = -.04 m (origin placed at
equilibrium of spring)
x2 = 0 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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v2 = 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.
Draw an energy bar chart.
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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?
Note that mgy1 is negative for the
zero chosen.
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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:
k= 196,000 N/m
Note that mgy1 is negative for the
zero chosen.
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Problem with Ug and Us
A 20-cm-tall spring with spring
constant of 5000 N/m is
placed on the ground as
shown. A 100-N block, is
released from rest, 15 cm
above the top of the spring,
compressing it. What is the
height H of the spring at the
point of maximum
compression? (Hint:
quadratic equation).
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Problem with Ug and Us
A 20-cm-tall spring with
spring constant of 5000
N/m is placed on the
ground as shown. A
100-N block, is released
from rest, 15 cm above
the top of the spring,
compressing it. What is
the height H of the
spring at the point of
maximum compression?
y1 = -10 cm in the
coordinate system chosen;
and therefore
H = 10 cm
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Collisions revisited
• Kinetic energy of the colliding objects are changed into
microscopic elastic potential energy and then back again to
kinetic energy.
• This very seldom happens without energy being transferred
from the 2-object system to the environment (as noise or heat).
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Inelastic collision
• During an inelastic
collision, momentum is
conserved, but much of the
kinetic energy transferred
out of the system.
• During an elastic collision,
kinetic energy and
momentum are conserved.
• Most real collisions fall
somewhere between the two
extremes, but we will use
this simplified model.
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Elastic collision
Elastic Collisions
Consider a head-on, perfectly elastic collision of a ball of
mass m1 having initial velocity (vix)1, with a ball of mass m2
that is initially at rest.
The solution, worked out in the text, is
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Elastic Collisions
What is vf1 if the two balls have the same mass?
What is vf2 if the two balls have the same mass?
When will vf1 be in the opposite direction of vi1? When will it
be in the same direction?
How about vf2 ? Can it ever be in the opposite direction?
Explain
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Elastic Collisions
What happens to the speed of each as m1 >> m2 ?
What happens to the speed of each as m2 >> m1 ?
Final speeds (and directions) depend on the relationship
between the masses.
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EOC - 28
Ball 1 with a mass of 100 g and traveling at 10 m/s
collides head-on with ball 2, which has a mass of
300 g, initially at rest.
a. What is the final velocity of each ball if the collision
is perfectly elastic?
b. What is the final velocity of each ball if the collision
is perfectly inelastic?
c. What percentage of the system’s original kinetic
energy is lost in the inelastic collision?
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EOC - 28
Ball 1 with a mass of 100 g and traveling at 10 m/s collides headon with ball 2, which has a mass of 300 g, initially at rest.
a. What is the final velocity of each ball if the collision is perfectly
elastic? vf1 = -5m/s, vf2 = 5 m/s
b. What is the final velocity of each ball if the collision is perfectly
inelastic? vf1 = vf2 = 2.5 m/s
c. What percentage of the system’s original energy is lost in the
inelastic collision? 75%
Where did it go?
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Energy Diagrams – For systems with no energy loss
• A graph showing a system’s potential energy and total energy as a function
of position is called an energy diagram.
• The object’s kinetic energy is TE – Ug. The object’s potential energy is
read directly off the PE graph.
• The particle can’t have more energy than TE. Even though the PE curve
goes above the TE line, the particle cannot reach any position to the right of
the intersection of TE and PE. This is the turning point.
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Energy Diagram of a mass on a horizontal
spring
You can only change the
height of the TE line.
The PE curve is
determined by the
spring. The object
cannot reach a position
where the PE curve is
above the TE line. The
intersection of the two is
the turning point.
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An oscillating mass
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Interpreting an energy diagram of an object that
has several kinds of potential energies
• If the particle starts at
the intersection of the
TE and PE curves, it has
no K.
• As PE decrease, K must
increase.
• At the turning point, its
speed K are both zero.
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Equilibrium: ΣF = 0
• Minimums and maximums
on the PE curve are points of
equilibrium.
• A minimum is a point of
stable equilibrium (eg the
bottom of the hill).
• A maximum is a point of
unstable equilibrium.
• We’ll investigate the
relationship between force
and potential energy in Ch.
11.
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A particle with the potential energy shown in the graph
is moving to the right. It has 1 J of kinetic energy at x =
1 m. Where is the particle’s turning point?
A.
B.
C.
D.
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2m
5m
6m
Not on graph
EOC #29 - This is the PE curve for a 20-g particle that is
released from rest at x = 1 m.
a. Will the particle move to
the left or right. How do
you know?
b. What is the maximum
speed of the particle?
Where does this happen?
c. What is/are the turning
point/s of the motion?
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EOC #29 - This is the PE curve for a 20-g particle that is
released from rest at x = 1 m.
a. Will the particle move to the
left or right. How do you
know? – right, particle cannot
move left since TE = 4 J
b. What is the maximum speed
of the particle? Where does
this happen?
17.3 m/s at x = 4 m.
a. What is/are the turning
point/s of the motion?
x=1m and x=6m.
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Workbook #20
Draw the potential energy curve, for the following:
• A particle with E1 oscillates between D and E.
• A particle with E2 oscillates between C and F.
• A particle with E3 oscillates between B and G.
• A particle with E4 enters from the right, bounces
at A, then never returns.
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Workbook #20 - answer
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Tactics: Interpreting an energy diagram
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Tactics: Interpreting an energy diagram
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