Download Chapter 10

Chapter 10 Lecture
physics
FOR SCIENTISTS AND ENGINEERS
a strategic approach
THIRD EDITION
randall d. knight
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Chapter 10 Energy
Chapter Goal: To introduce the concept of energy
and the basic energy model.
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Chapter 10 Preview
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Chapter 10 Preview
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Chapter 10 Preview
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Chapter 10 Preview
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Chapter 10 Preview
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Chapter 10 Reading Quiz
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Slide 10-8
Reading Question 10.1
Kinetic energy is
A. Mass times velocity.
B. ½ mass times speed-squared.
C. The area under the force curve in a forceversus-time graph.
D. Velocity per unit mass.
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Slide 10-9
Reading Question 10.1
Kinetic energy is
A. Mass times velocity.
B. ½ Mass times speed-squared.
C. The area under the force curve in a forceversus-time graph.
D. Velocity per unit mass.
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Slide 10-10
Reading Question 10.2
A method for keeping track of transformations
between kinetic energy and gravitational potential
energy, introduced in this chapter, is
A. Credit-debit tables.
B.
C.
D.
E.
Kinetic energy-versus-time graphs.
Energy bar charts.
Energy conservation pools.
Energy spreadsheets.
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Slide 10-11
Reading Question 10.2
A method for keeping track of transformations
between kinetic energy and gravitational potential
energy, introduced in this chapter, is
A. Credit-debit tables.
B.
C.
D.
E.
Kinetic energy-versus-time graphs.
Energy bar charts.
Energy conservation pools.
Energy spreadsheets.
© 2013 Pearson Education, Inc.
Slide 10-12
Reading Question 10.3
Mechanical energy is
A. The energy due to internal moving parts.
B. The energy of motion.
C. The energy of position.
D. The sum of kinetic energy plus potential
energy.
E. The sum of kinetic, potential, thermal, and elastic
energy.
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Slide 10-13
Reading Question 10.3
Mechanical energy is
A. The energy due to internal moving parts.
B. The energy of motion.
C. The energy of position.
D. The sum of kinetic energy plus potential
energy.
E. The sum of kinetic, potential, thermal, and elastic
energy.
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Slide 10-14
Reading Question 10.4
Hooke’s law describes the force of
A. Gravity.
B. A spring.
C. Collisions.
D. Tension.
E. None of the above.
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Slide 10-15
Reading Question 10.4
Hooke’s law describes the force of
A. Gravity.
B. A spring.
C. Collisions.
D. Tension.
E. None of the above.
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Slide 10-16
Reading Question 10.5
A perfectly elastic collision is a collision
A. Between two springs.
B. That conserves thermal energy.
C. That conserves kinetic energy.
D. That conserves potential energy.
E. That conserves mechanical energy.
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Slide 10-17
Reading Question 10.5
A perfectly elastic collision is a collision
A. Between two springs.
B. That conserves thermal energy.
C. That conserves kinetic energy.
D. That conserves potential energy.
E. That conserves mechanical energy.
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Chapter 10 Content, Examples, and
QuickCheck Questions
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Slide 10-19
Kinetic Energy K
 Kinetic energy is the
energy of motion.
 All moving objects
have kinetic energy.
 The more massive an
object or the faster it
moves, the larger its
kinetic energy.
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Slide 10-20
Potential Energy U
 Potential energy is
stored energy
associated with an
object’s position.
 The roller coaster’s
gravitational potential
energy depends on its
height above the
ground.
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Slide 10-21
Thermal Energy Eth
 Thermal energy is the
sum of the microscopic
kinetic and potential
energies of all the atoms
and bonds that make up
the object.
 An object has more
thermal energy when
hot than when cold.
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Slide 10-22
The Basic Energy Model
 Within a system, energy can
be transformed from one
type to another.
 The total energy of the
system is not changed by
these transformations.
 This is the law of
conservation of energy.
 Energy can also be transferred
from one system to another.
 The mechanical transfer of energy to a system via
forces is called work.
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Slide 10-23
Kinetic Energy and Gravitational Potential
Energy
 The figure shows a before-and-after
representation of an object in
free fall.
 One of the kinematics equations
from Chapter 2, with ay = g, is:
 Rearranging:
 Multiplying both sides by ½m:
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Slide 10-24
Kinetic Energy and Gravitational Potential
Energy
Define kinetic energy as an energy of motion:
Define gravitational potential energy as an
energy of position:
The sum K + Ug is not changed when an object is in
free fall. Its initial and final values are equal:
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Slide 10-25
Kinetic Energy and Gravitational Potential
Energy
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Slide 10-26
QuickCheck 10.1
A child is on a playground swing,
motionless at the highest
point of his arc. What energy
transformation takes place as
he swings back down to the
lowest point of his motion?
A. K  Ug
B. Ug  K
C. Eth  K
D. Ug  Eth
E. K  Eth
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Slide 10-27
QuickCheck 10.1
A child is on a playground swing,
motionless at the highest
point of his arc. What energy
transformation takes place as
he swings back down to the
lowest point of his motion?
A. K  Ug
B. Ug  K
C. Eth  K
D. Ug  Eth
E. K  Eth
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QuickCheck 10.2
A skier is gliding down a gentle slope at a constant
speed. What energy transformation is taking place?
A.
K  Ug
B. Ug  K
C. Eth  K
D. Ug  Eth
E. K  Eth
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Slide 10-29
QuickCheck 10.2
A skier is gliding down a gentle slope at a constant
speed. What energy transformation is taking place?
A.
K  Ug
B. Ug  K
C. Eth  K
D. Ug  Eth
E. K  Eth
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Slide 10-30
Example 10.1 Launching a Pebble
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Example 10.1 Launching a Pebble
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Example 10.1 Launching a Pebble
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Energy Bar Charts
 A pebble is tossed up into the air.
 The simple bar charts below show how the sum of K + Ug
remains constant as the pebble rises and then falls.
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Slide 10-34
Energy Bar Charts
 The figure below shows how to make an energy bar chart
suitable for problem solving.
 The chart is a graphical representation of the energy
equation Kf + Ugf = Ki + Ugi.
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Slide 10-35
QuickCheck 10.3
Ball A has half the mass and eight times the kinetic
energy of ball B. What is the speed ratio vA/vB?
A. 16
B.
4
C.
2
D. 1/4
E. 1/16
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Slide 10-36
QuickCheck 10.3
Ball A has half the mass and eight times the kinetic
energy of ball B. What is the speed ratio vA/vB?
A. 16
B.
4
C.
2
D. 1/4
E. 1/16
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Slide 10-37
QuickCheck 10.4
Rank in order, from largest to
smallest, the gravitational
potential energies of the balls.
A. 1 > 2 = 4 > 3
B. 1 > 2 > 3 > 4
C. 3 > 2 > 4 > 1
D. 3 > 2 = 4 > 1
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Slide 10-38
QuickCheck 10.4
Rank in order, from largest to
smallest, the gravitational
potential energies of the balls.
A. 1 > 2 = 4 > 3
B. 1 > 2 > 3 > 4
C. 3 > 2 > 4 > 1
D. 3 > 2 = 4 > 1
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Slide 10-39
The Zero of Potential Energy
 Amber and Bill use
coordinate systems with
different origins to
determine the potential
energy of a rock.
 No matter where the rock
is, Amber’s value of Ug
will be equal to Bill’s
value plus 9.8 J.
 If the rock moves, both will calculate exactly the same
value for Ug.
 In problems, only Ug has physical significance, not the
value of Ug itself.
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Slide 10-40
Example 10.2 The Speed of a Falling Rock
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Example 10.2 The Speed of a Falling Rock
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Example 10.2 The Speed of a Falling Rock
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Example 10.2 The Speed of a Falling Rock
ASSESS The figure below shows energy bar charts for Amber and Bill.
despite their disagreement over the value of Ug, Amber and Bill arrive at the
same value for vf and their Kf bars are the same height. 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.
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Gravitational Potential Energy on a Frictionless
Surface – Slide 1 of 4
 Figure (a) shows an object
of mass m sliding along a
frictionless surface.
 Figure (b) shows a magnified
segment of the surface that,
over some small distance, is
a straight line.
 Define an s-axis parallel to
the direction of motion
 Newton’s second law along
the axis is:
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Slide 10-45
Gravitational Potential Energy on a Frictionless
Surface – Slide 2 of 4
 Using the chain rule, we can
write Newton’s second law as:
 It is clear from the diagram
that the net force along s is:
 So Newton’s second law is:
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Slide 10-46
Gravitational Potential Energy on a Frictionless
Surface – Slide 3 of 4
 Rearranging, we obtain:
 Note from the diagram
that sin ds = dy, so:
 Integrating this from
“before” to “after”:
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Slide 10-47
Gravitational Potential Energy on a Frictionless
Surface – Slide 4 of 4
 With K = ½ mv2 and Ug = mgy,
we find that:
 The total mechanical energy
for a particle moving along
any frictionless smooth
surface is conserved,
regardless of the shape of
the surface.
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Slide 10-48
QuickCheck 10.5
Starting from rest, a marble
first rolls down a steeper hill,
then down a less steep hill of
the same height. For which is
it going faster at the bottom?
A.
Faster at the bottom of the steeper hill.
B.
Faster at the bottom of the less steep hill.
C.
Same speed at the bottom of both hills.
D.
Can’t say without knowing the mass of the marble.
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Slide 10-49
QuickCheck 10.5
Starting from rest, a marble
first rolls down a steeper hill,
then down a less steep hill of
the same height. For which is
it going faster at the bottom?
A.
Faster at the bottom of the steeper hill.
B.
Faster at the bottom of the less steep hill.
C.
Same speed at the bottom of both hills.
D.
Can’t say without knowing the mass of the marble.
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Slide 10-50
QuickCheck 10.6
A small child slides down the four frictionless slides A–
D. Rank in order, from largest to smallest, her speeds at
the bottom.
A. vD > vA > vB > vC
B. vD > vA = vB > vC
C. vC > vA > vB > vD
D. vA = vB = vC = vD
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Slide 10-51
QuickCheck 10.6
A small child slides down the four frictionless slides A–
D. Rank in order, from largest to smallest, her speeds at
the bottom.
A. vD > vA > vB > vC
B. vD > vA = vB > vC
C. vC > vA > vB > vD
D. vA = vB = vC = vD
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Example 10.3 The Speed of a Sled
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Example 10.3 The Speed of a Sled
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Example 10.3 The Speed of a Sled
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Problem-Solving Strategy: Conservation
of Mechanical Energy
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Slide 10-56
QuickCheck 10.7
Three balls are thrown from
a cliff with the same speed
but at different angles.
Which ball has the greatest
speed just before it hits the
ground?
A.
Ball A.
B.
Ball B.
C.
Ball C.
D.
All balls have the same speed.
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Slide 10-57
QuickCheck 10.7
Three balls are thrown from
a cliff with the same speed
but at different angles.
Which ball has the greatest
speed just before it hits the
ground?
A.
Ball A.
B.
Ball B.
C.
Ball C.
D.
All balls have the same speed.
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Slide 10-58
QuickCheck 10.8
A hockey puck sliding on smooth ice at 4 m/s comes to a
1-m-high hill. Will it make it to the top of the hill?
A.
Yes.
B.
No.
C.
Can’t answer without knowing the mass of the puck.
D.
Can’t say without knowing the angle of the hill.
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Slide 10-59
QuickCheck 10.8
A hockey puck sliding on smooth ice at 4 m/s comes to a
1-m-high hill. Will it make it to the top of the hill?
A.
Yes.
B.
No.
C.
Can’t answer without knowing the mass of the puck.
D.
Can’t say without knowing the angle of the hill.
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Slide 10-60
Restoring Forces and Hooke’s Law
 The figure shows how a
hanging mass stretches
a spring of equilibrium
length L0 to a new
length L.
 The mass hangs in static
equilibrium, so the upward
spring force balances the
downward gravity force.
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Slide 10-61
Restoring Forces and Hooke’s Law
 The figure shows measured
data for the restoring force
of a real spring.
 s is the displacement
from equilibrium.
 The data fall along the
straight line:
 The proportionality constant k is called the spring
constant.
 The units of k are N/m.
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Slide 10-62
Hooke’s Law
 One end of a spring is
attached to a fixed wall.
 (Fsp)s is the force produced
by the free end of the spring.
 s = s – se is the
displacement from
equilibrium.
 The negative sign is the
mathematical indication of
a restoring force.
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Slide 10-63
QuickCheck 10.9
The restoring force of three
springs is measured as they are
stretched. Which spring has the
largest spring constant?
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Slide 10-64
QuickCheck 10.9
The restoring force of three
springs is measured as they are
stretched. Which spring has the
largest spring constant?
Steepest slope.
Takes lots of force for
a small displacement.
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Slide 10-65
Example 10.5 Pull Until It Slips
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Example 10.5 Pull Until It Slips
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Example 10.5 Pull Until It Slips
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Example 10.5 Pull Until It Slips
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Example 10.5 Pull Until It Slips
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Example 10.5 Pull Until It Slips
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Slide 10-71
Stick-Slip Motion
 Earthquakes are an example of stick-slip motion.
 Tectonic plates are attempting to slide past each other,
but friction causes the edges of the plates to stick
together.
 Large masses of rock are somewhat elastic and can be
“stretched”.
 Eventually the elastic force
of the deformed rocks
exceeds the friction force
between the plates.
 An earthquake occurs as
the plates slip and lurch
forward.
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The slip can range from a few centimeters in a relatively
small earthquake to several meters in a very large
earthquake.
Slide 10-72
Elastic Potential Energy
 Springs and rubber bands
store potential energy that
can be transformed into
kinetic energy.
 The spring force is not
constant as an object
is pushed or pulled.
 The motion of the mass is not constant-acceleration
motion, and therefore we cannot use our old
kinematics equations.
 One way to analyze motion when spring force is
involved is to look at energy before and after some
motion.
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Slide 10-73
Elastic Potential Energy
 The figure shows a beforeand-after situation in which
a spring launches a ball.
 Integrating the net force
from the spring, as given by
Hooke’s Law, shows that:
 Here K = ½ mv2 is the kinetic
energy.
 We define a new quantity:
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Slide 10-74
Elastic Potential Energy
 An object moving without friction on an ideal spring
obeys:
where
 Because s is squared, Us is
positive for a spring that is
either stretched or compressed.
 In the figure, Us has a positive
value both before and after the
motion.
© 2013 Pearson Education, Inc.
Slide 10-75
QuickCheck 10.10
A spring-loaded gun shoots a plastic ball with a launch
speed of 2.0 m/s. If the spring is compressed twice as
far, the ball’s launch speed will be
A.
B.
C.
D.
E.
1.0 m/s.
2.0 m/s.
2.8 m/s
4.0 m/s.
16.0 m/s.
© 2013 Pearson Education, Inc.
Slide 10-76
QuickCheck 10.10
A spring-loaded gun shoots a plastic ball with a launch
speed of 2.0 m/s. If the spring is compressed twice as
far, the ball’s launch speed will be
A.
B.
C.
D.
E.
1.0 m/s.
2.0 m/s.
2.8 m/s
4.0 m/s.
16.0 m/s.
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Conservation of energy:
Double x double v
Slide 10-77
QuickCheck 10.11
A spring-loaded gun shoots a plastic ball with a launch
speed of 2.0 m/s. If the spring is replaced with a new
spring having twice the spring constant (but still
compressed the same distance), the ball’s launch
speed will be
A.
B.
C.
D.
E.
1.0 m/s.
2.0 m/s.
2.8 m/s.
4.0 m/s.
16.0 m/s.
© 2013 Pearson Education, Inc.
Slide 10-78
QuickCheck 10.11
A spring-loaded gun shoots a plastic ball with a launch
speed of 2.0 m/s. If the spring is replaced with a new
spring having twice the spring constant (but still
compressed the same distance), the ball’s launch
speed will be
A.
B.
C.
D.
E.
1.0 m/s.
2.0 m/s.
2.8 m/s.
4.0 m/s.
16.0 m/s.
© 2013 Pearson Education, Inc.
Conservation of energy:
Double k  increase
v by square root of 2
Slide 10-79
Example 10.6 A Spring-Launched Plastic Ball
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Slide 10-80
Example 10.6 A Spring-Launched Plastic Ball
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Slide 10-81
Example 10.6 A Spring-Launched Plastic Ball
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Slide 10-82
Example 10.6 A Spring-Launched Plastic Ball
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Slide 10-83
Example 10.6 A Spring-Launched Plastic Ball
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Slide 10-84
Energy Diagrams
 Potential energy is a function of position.
 Functions of position are easy to represent as graphs.
 A graph showing a system’s potential energy and total
energy as a function of position is called an energy
diagram.
 Shown is the energy diagram
of a particle in free fall.
 Gravitational potential energy
is a straight line with slope
mg and zero y-intercept.
 Total energy is a horizontal
line, since mechanical
energy is conserved.
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Slide 10-85
A Four-Frame Movie of a Particle in Free Fall
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Slide 10-86
Energy Diagrams
 Shown is the energy
diagram of a mass on a
horizontal spring.
 The potential energy (PE)
is the parabola:
Us = ½k(x – xe)2
 The PE curve is determined
by the spring constant; you
can’t change it.
 You can set the total energy (TE) to any height you
wish simply by stretching the spring to the proper
length at the beginning of the motion.
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Slide 10-87
A Four-Frame Movie of a Mass Oscillating on a
Spring
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Slide 10-88
Energy Diagrams
 Shown is a more general
energy diagram.
 The particle is released
from rest at position x1.
 Since K at x1 is zero,
the total energy TE = U
at that point.
 The particle speeds up
from x1 to x2.
 Then it slows down from x2 to x3.
 The particle reaches maximum speed as it passes x4.
 When the particle reaches x5, it turns around and
reverses the motion.
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Slide 10-89
Equilibrium Positions: Stable
 Consider a particle
with the total energy
E2 shown in the figure.
 The particle can be
at rest at x2, but it
cannot move away
from x2: This is
static equilibrium.
 If you disturb the particle,
giving it a total energy slightly
larger than E2, it will oscillate very close to x2.
 An equilibrium for which small disturbances cause small
oscillations is called a point of stable equilibrium.
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Slide 10-90
Equilibrium Positions: Unstable
 Consider a particle
with the total energy
E3 shown in the figure.
 The particle can be at
rest at x3, and it does
not move away from
x3: This is static equilibrium.
 If you disturb the particle,
giving it a total energy slightly
larger than E3, it will speed
up as it moves away from x3.
 An equilibrium for which small disturbances cause the
particle to move away is called a point of unstable
equilibrium.
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Slide 10-91
Tactics: Interpreting an Energy Diagram
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Slide 10-92
Example 10.9 Balancing a Mass on a Spring
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Slide 10-93
Example 10.9 Balancing a Mass on a Spring
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Slide 10-94
Example 10.9 Balancing a Mass on a Spring
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Slide 10-95
Example 10.9 Balancing a Mass on a Spring
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Slide 10-96
Example 10.9 Balancing a Mass on a Spring
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Slide 10-97
Example 10.9 Balancing a Mass on a Spring
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Slide 10-98
QuickCheck 10.12
A particle with the potential
energy shown is moving
to the right. It has 1.0 J of
kinetic energy at x = 1.0 m.
In the region 1.0 m < x < 2.0 m,
the particle is
A.
Speeding up.
B.
Slowing down.
C.
Moving at constant speed.
D.
I have no idea.
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Slide 10-99
QuickCheck 10.12
A particle with the potential
energy shown is moving
to the right. It has 1.0 J of
kinetic energy at x = 1.0 m.
In the region 1.0 m < x < 2.0 m,
the particle is
A.
Speeding up.
B.
Slowing down.
C.
Moving at constant speed.
D.
I have no idea.
© 2013 Pearson Education, Inc.
Slide 10-100
QuickCheck 10.13
A particle with the potential
energy shown is moving to
the right. It has 1.0 J of kinetic
energy at x = 1.0 m. Where
is the particle’s turning point?
A.
1.0 m.
B.
2.0 m.
C.
5.0 m.
D.
6.0 m.
E.
It doesn’t have a turning point.
© 2013 Pearson Education, Inc.
Slide 10-101
QuickCheck 10.13
A particle with the potential
energy shown is moving to
the right. It has 1.0 J of kinetic
energy at x = 1.0 m. Where
is the particle’s turning point?
A.
1.0 m.
B.
2.0 m.
C.
5.0 m.
D.
6.0 m.
E.
It doesn’t have a turning point.
© 2013 Pearson Education, Inc.
Slide 10-102
QuickCheck 10.14
A particle with this potential
energy could be in stable
equilibrium at x =
A.
0.0 m.
B.
1.0 m.
C.
2.0 m.
D.
Either A or C.
E.
Either B or C.
© 2013 Pearson Education, Inc.
Slide 10-103
QuickCheck 10.14
A particle with this potential
energy could be in stable
equilibrium at x =
A.
0.0 m.
B.
1.0 m.
C.
2.0 m.
D.
Either A or C.
E.
Either B or C.
© 2013 Pearson Education, Inc.
Slide 10-104
Molecular Bonds
 Shown is the energy diagram
for the diatomic molecule HCl
(hydrogen chloride).
 x is the distance between
the hydrogen and the
chlorine atoms.
 The molecule has a
stable equilibrium at
an atomic separation
of xeq = 0.13 nm.
 When the total energy is E1,
the molecule is oscillating, but stable.
 If the molecule’s energy is raised to E2, we have broken
the molecular bond.
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Elastic Collisions
 During an inelastic collision of two objects, some of the
mechanical energy is dissipated inside the objects as
thermal energy.
 A collision in which mechanical energy is conserved is
called a perfectly elastic collision.
 Collisions between
two very hard objects,
such as two billiard
balls or two steel balls,
come close to being
perfectly elastic.
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A Perfectly Elastic Collision
 Consider a head-on, perfectly
elastic collision of a ball of
mass m1 and initial velocity
(vix)1, with a ball of mass m2
initially at rest.
 The balls’ velocities after the
collision are (vfx)1 and (vfx)2.
 Momentum is conserved in all isolated collisions.
 In a perfectly elastic collision in which potential energy is
not changing, the kinetic energy must also be conserved.
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A Perfectly Elastic Collision
 Simultaneously solving
the conservation of
momentum equation and
the conservation of kinetic
energy equations allows
us to find the two unknown
final velocities.
 The result is:
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A Perfectly Elastic Collision: Special Case 1
 Consider a head-on,
perfectly elastic collision of
a ball of mass m1 and initial
velocity (vix)1, with a ball of
mass m2 initially at rest.
 Case 1: m1 = m2.
 Equations 10.42 give vf1 = 0 and vf2 = vi1.
 The first ball stops and transfers all its momentum to
the second ball.
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A Perfectly Elastic Collision: Special Case 2
 Consider a head-on, perfectly
elastic collision of a ball of
mass m1 and initial velocity
(vix)1, with a ball of mass m2
initially at rest.
 Case 2: m1 >> m2.
 Equations 10.42 give vf1  vi1 and vf2  2vi1.
 The big first ball keeps going with about the same speed,
and the little second ball flies off with about twice the speed
of the first ball.
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A Perfectly Elastic Collision: Special Case 3
 Consider a head-on, perfectly
elastic collision of a ball of
mass m1 and initial velocity
(vix)1, with a ball of mass m2
initially at rest.
 Case 3: m1 << m2.
 Equations 10.42 give vf1 ≈ −vi1 and vf2 ≈ 0.
 The little first rebounds with about the same speed,
and the big second ball hardly moves at all.
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Perfectly Elastic Collisions: Using Reference
Frames
 Equations 10.42 assume ball 2 is at rest.
 What if you need to analyze a head-on collision when
both balls are moving before the collision?
 You could solve the simultaneous momentum and
energy equations, but there is an easier way.
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Using Reference Frames: Quick Example
A 200 g ball moves to the right at 2.0 m/s. It has a
head-on, perfectly elastic collision with a 100 g ball that
is moving toward it at 3.0 m/s. What are the final
velocities of both balls?
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Using Reference Frames: Quick Example
 Figure (a) shows the situation just before the collision
in the lab frame L.
 Figure (b) shows the situation just before the collision
in the frame M that is moving along with ball 2.
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Using Reference Frames: Quick Example
 We can use Equations 10.42 to find the post-collision
velocities in the moving frame M:
 Transforming back to the lab frame L:
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Chapter 10 Summary Slides
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General Principles
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General Principles
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Important Concepts
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Important Concepts
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Important Concepts
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