Transcript Ch. 5 PP - Lemon Bay High School

Chapter 5
Section 1 Work
Preview
• Objectives
• Definition of Work
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Chapter 5
Section 1 Work
Objectives
• 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.
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Chapter 5
Section 1 Work
Definition of Work
• Work is done on an object when a force causes a
displacement of the object.
• Work is done only when components of a force are
parallel to a displacement.
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Chapter 5
Section 1 Work
Definition of Work
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Chapter 5
Section 1 Work
Sign Conventions for Work
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Visual Concept
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Chapter 5
Section 2 Energy
Preview
• Objectives
• Kinetic Energy
• Sample Problem
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Chapter 5
Section 2 Energy
Objectives
• 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.
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Chapter 5
Section 2 Energy
Kinetic Energy
• Kinetic Energy
The energy of an object that is due to the object’s
motion is called kinetic energy.
• Kinetic energy depends on speed and mass.
1
KE  mv 2
2
1
2
kinetic energy =  mass   speed 
2
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Chapter 5
Section 2 Energy
Kinetic Energy
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Visual Concept
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Chapter 5
Section 2 Energy
Kinetic Energy, continued
• Work-Kinetic Energy Theorem
– The net work done by all the forces acting on an
object is equal to the change in the object’s kinetic
energy.
• The net work done on a body equals its change in
kinetic energy.
Wnet = ∆KE
net work = change in kinetic energy
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Chapter 5
Section 2 Energy
Work-Kinetic Energy Theorem
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Visual Concept
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Chapter 5
Section 2 Energy
Sample Problem
Work-Kinetic Energy Theorem
On a frozen pond, a person kicks a 10.0 kg sled,
giving it an initial speed of 2.2 m/s. How far does the
sled move if the coefficient of kinetic friction between
the sled and the ice is 0.10?
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Chapter 5
Section 2 Energy
Sample Problem, continued
Work-Kinetic Energy Theorem
1. Define
Given:
m = 10.0 kg
vi = 2.2 m/s
vf = 0 m/s
µk = 0.10
Unknown:
d=?
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Chapter 5
Section 2 Energy
Sample Problem, continued
Work-Kinetic Energy Theorem
2. Plan
Choose an equation or situation: This problem can be
solved using the definition of work and the work-kinetic
energy theorem.
Wnet = Fnetdcos
The net work done on the sled is provided by the force
of kinetic friction.
Wnet = Fkdcos = µkmgdcos
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Chapter 5
Section 2 Energy
Sample Problem, continued
Work-Kinetic Energy Theorem
2. Plan, continued
The force of kinetic friction is in the direction opposite d,
 = 180°. Because the sled comes to rest, the final
kinetic energy is zero.
Wnet = ∆KE = KEf - KEi = –(1/2)mvi2
Use the work-kinetic energy theorem, and solve for d.
1
– mv i2  k mgd cos 
2
–v i2
d
2k g cos 
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Chapter 5
Section 2 Energy
Sample Problem, continued
Work-Kinetic Energy Theorem
3. Calculate
Substitute values into the equation:
(–2.2 m/s)2
d
2(0.10)(9.81 m/s2 )(cos180)
d  2.5 m
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Chapter 5
Section 2 Energy
Sample Problem, continued
Work-Kinetic Energy Theorem
4. Evaluate
According to Newton’s second law, the acceleration
of the sled is about -1 m/s2 and the time it takes the
sled to stop is about 2 s. Thus, the distance the sled
traveled in the given amount of time should be less
than the distance it would have traveled in the
absence of friction.
2.5 m < (2.2 m/s)(2 s) = 4.4 m
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Chapter 5
Section 2 Energy
Potential Energy
• Potential Energy is the energy associated with an
object because of the position, shape, or condition of
the object.
• Gravitational potential energy is the potential
energy stored in the gravitational fields of interacting
bodies.
• Gravitational potential energy depends on height
from a zero level.
PEg = mgh
gravitational PE = mass  free-fall acceleration  height
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Section 2 Energy
Chapter 5
Potential Energy
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Chapter 5
Section 2 Energy
Potential Energy, continued
•
Elastic potential energy is the energy available for
use when a deformed elastic object returns to its
original configuration.
1 2
PEelastic  kx
2
elastic PE =
1
 spring constant  (distance compressed or stretched)
2
•
The symbol k is called the spring constant, a
parameter that measures the spring’s resistance to
being compressed or stretched.
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2
Chapter 5
Section 2 Energy
Elastic Potential Energy
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Section 2 Energy
Chapter 5
Spring Constant
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Visual Concept
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Chapter 5
Section 2 Energy
Sample Problem
Potential Energy
A 70.0 kg stuntman is attached to a bungee cord with
an unstretched length of 15.0 m. He jumps off a
bridge spanning a river from a height of 50.0 m. When
he finally stops, the cord has a stretched
length of 44.0 m. Treat the stuntman as a point mass,
and disregard the weight of the bungee cord.
Assuming the spring constant of the bungee cord is
71.8 N/m, what is the total potential energy relative to
the water when the man stops falling?
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Chapter 5
Section 2 Energy
Sample Problem, continued
Potential Energy
1. Define
Given:m = 70.0 kg
k = 71.8 N/m
g = 9.81 m/s2
h = 50.0 m – 44.0 m = 6.0 m
x = 44.0 m – 15.0 m = 29.0 m
PE = 0 J at river level
Unknown: PEtot = ?
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Chapter 5
Section 2 Energy
Sample Problem, continued
Potential Energy
2. Plan
Choose an equation or situation: The zero level for
gravitational potential energy is chosen to be at the
surface of the water. The total potential energy is the
sum of the gravitational and elastic potential energy.
PEtot  PEg  PEelastic
PEg  mgh
PEelastic 
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1 2
kx
2
Chapter 5
Section 2 Energy
Sample Problem, continued
Potential Energy
3. Calculate
Substitute the values into the equations and solve:
PEg  (70.0 kg)(9.81 m/s2 )(6.0 m) = 4.1 10 3 J
1
PEelastic  (71.8 N/m)(29.0 m)2  3.02  10 4 J
2
PEtot  4.1 103 J + 3.02  10 4 J
PEtot  3.43  10 4 J
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Chapter 5
Section 2 Energy
Sample Problem, continued
Potential Energy
4. Evaluate
One way to evaluate the answer is to make an
order-of-magnitude estimate. The gravitational
potential energy is on the order of 102 kg  10
m/s2  10 m = 104 J. The elastic potential energy
is on the order of 1  102 N/m  102 m2 = 104 J.
Thus, the total potential energy should be on the
order of 2  104 J. This number is close to the
actual answer.
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Chapter 5
Section 3 Conservation of
Energy
Preview
• Objectives
• Conserved Quantities
• Mechanical Energy
• Sample Problem
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Chapter 5
Section 3 Conservation of
Energy
Objectives
• 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.
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Chapter 5
Section 3 Conservation of
Energy
Conserved Quantities
• When we say that something is conserved, we mean
that it remains constant.
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Chapter 5
Section 3 Conservation of
Energy
Mechanical Energy
• Mechanical energy is the sum of kinetic energy and
all forms of potential energy associated with an object
or group of objects.
ME = KE + ∑PE
• Mechanical energy is often conserved.
MEi = MEf
initial mechanical energy = final mechanical energy
(in the absence of friction)
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Chapter 5
Section 3 Conservation of
Energy
Conservation of Mechanical Energy
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Chapter 5
Section 3 Conservation of
Energy
Sample Problem
Conservation of Mechanical Energy
Starting from rest, a child zooms down a frictionless
slide from an initial height of 3.00 m. What is her
speed at the bottom of the slide? Assume she has a
mass of 25.0 kg.
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Chapter 5
Section 3 Conservation of
Energy
Sample Problem, continued
Conservation of Mechanical Energy
1. Define
Given:
h = hi = 3.00 m
m = 25.0 kg
vi = 0.0 m/s
hf = 0 m
Unknown:
vf = ?
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Chapter 5
Section 3 Conservation of
Energy
Sample Problem, continued
Conservation of Mechanical Energy
2. Plan
Choose an equation or situation: The slide is
frictionless, so mechanical energy is conserved.
Kinetic energy and gravitational potential energy are
the only forms of energy present.
1
KE 
mv 2
2
PE  mgh
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Chapter 5
Section 3 Conservation of
Energy
Sample Problem, continued
Conservation of Mechanical Energy
2. Plan, continued
The zero level chosen for gravitational potential
energy is the bottom of the slide. Because the child
ends at the zero level, the final gravitational potential
energy is zero.
PEg,f = 0
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Chapter 5
Section 3 Conservation of
Energy
Sample Problem, continued
Conservation of Mechanical Energy
2. Plan, continued
The initial gravitational potential energy at the top of
the slide is
PEg,i = mghi = mgh
Because the child starts at rest, the initial kinetic
energy at the top is zero.
KEi = 0
Therefore, the final kinetic energy is as follows:
1
KEf  mv f2
2
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Chapter 5
Section 3 Conservation of
Energy
Sample Problem, continued
Conservation of Mechanical Energy
3. Calculate
Substitute values into the equations:
PEg,i = (25.0 kg)(9.81 m/s2)(3.00 m) = 736 J
KEf = (1/2)(25.0 kg)vf2
Now use the calculated quantities to evaluate the
final velocity.
MEi = MEf
PEi + KEi = PEf + KEf
736 J + 0 J = 0 J + (0.500)(25.0 kg)vf2
vf = 7.67 m/s
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Chapter 5
Section 3 Conservation of
Energy
Sample Problem, continued
Conservation of Mechanical Energy
4. Evaluate
The expression for the square of the final speed can
be written as follows:
2mgh
2
vf 
 2gh
m
Notice that the masses cancel, so the final speed
does not depend on the mass of the child. This
result makes sense because the acceleration of an
object due to gravity does not depend on the mass
of the object.
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Chapter 5
Section 3 Conservation of
Energy
Mechanical Energy, continued
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•
Mechanical Energy is
not conserved in the
presence of friction.
•
As a sanding block
slides on a piece of
wood, energy (in the
form of heat) is
dissipated into the
block and surface.
Chapter 5
Section 4 Power
Preview
• Objectives
• Rate of Energy Transfer
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Chapter 5
Section 4 Power
Objectives
• Relate the concepts of energy, time, and power.
• Calculate power in two different ways.
• Explain the effect of machines on work and power.
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Chapter 5
Section 4 Power
Rate of Energy Transfer
• Power is a quantity that measures the rate at which
work is done or energy is transformed.
P = W/∆t
power = work ÷ time interval
• An alternate equation for power in terms of force and
speed is
P = Fv
power = force  speed
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Chapter 5
Section 4 Power
Power
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Visual Concept
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