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Chapter 5
Preview
• Objectives
• Definition of Work
Section 1 Work
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.
Chapter 5
Section 1 Work
Definition of Work
• Work is done on an object when a force causes a
displacement of the object.
– No displacement = no work done
• Work is done only when components of a force are
parallel to a displacement.
– In other words, if an object moves in the +x
direction but the force is applied in the +y
direction, NO work is done. (ex: waitress
carrying a tray)
Chapter 5
Section 1 Work
Definition of Work
Work
• Work has SI units of Joules (J)
– 1 J = 1 Nm
• Work is scalar
• +W means force and displacement are in the
same direction
• -W means force and displacement are in
opposite directions (ex: friction)
Chapter 5
Section 1 Work
Sign Conventions for Work
Click below to watch the Visual Concept.
Visual Concept
Examples
1. A tugboat pulls a ship with a constant net horizontal
force of 5.00 x 103 N and causes the ship to move
through a harbor. How much work is done on the
ship it if moves a distance of 3.00 km?
2. A weightlifter lifts a set of weights a vertical
distance of 2.00 m. If a constant net force of 350 Nis
exerted on the weights, what is the net work done on
the weights?
Examples
3. A shopper in a supermarket pushes a cart with
a force of 35 N directed at an angle of 25⁰
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.
4. If 2.0 J of work is done in raising a 180 g
apple, how far is it lifted?
Chapter 5
Preview
• Objectives
• Kinetic Energy
• Sample Problem
Section 2 Energy
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.
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
• Units of energy are Joules (J)
• Kinetic energy depends on speed and mass.
1
KE  mv 2
2
1
2
kinetic energy =  mass   speed 
2
Chapter 5
Section 2 Energy
Kinetic Energy
Click below to watch the Visual Concept.
Visual Concept
Examples
1. Calculate the speed of an 8.0 x 104 kg airliner with a
kinetic energy of 1.1 x 109 J.
2. What is the speed of a 0.145 kg baseball if its kinetic
energy is 109 J?
3. Two bullets have masses of 3.0 g and 6.0 g,
respectively. Both are fired with a speed of 40.0 m/s.
Which bullet has more kinetic energy? What is the
ratio of their kinetic energies?
Examples
4. Two 3.0 g bullets are fired with speeds of 40.0 m/s
and 80.0 m/s, respectively. What are their kinetic
energies? Which bullet has more kinetic energy?
What is the ratio of their kinetic energies?
5. A car has a kinetic energy of 4.32 x 105 J when
traveling at a speed of 23 m/s. What is its mass?
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.
Wnet = ∆KE
net work = change in kinetic energy
Chapter 5
Section 2 Energy
Work-Kinetic Energy Theorem
Click below to watch the Visual Concept.
Visual Concept
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?
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=?
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 = Fnetdcosq
The net work done on the sled is provided by the force
of kinetic friction.
Wnet = Fkdcosq = µkmgdcosq
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,
q = 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 q
2
–v i2
d
2k g cos q
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
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
Examples
1. A student wearing frictionless in-line skates on a
horizontal surface is pushed by a friend with a
constant force of 45 N. How far must the student be
pushed, starting from rest, so that their final kinetic
energy is 352 J?
2. A 2.1 x 103 kg car starts from rest at the top of a
driveway that is sloped at an angle of 20.0⁰ with the
horizontal. An average friction force of 4.0 x 103 N
impedes the car’s motion so that the car’s speed at
the bottom of the driveway is 3.8 m/s. What is the
length of the driveway?
Examples
3. A 2.0 x 103 kg car accelerates from rest under the
actions of 2 forces. One is a forward force of 1140 N
provided by the traction between the wheels and the
road. The other is a 950 N resistive force due to
various frictional forces. Use the work-kinetic energy
theorem to determine how far the car must travel for
its speed to reach 2.0 m/s.
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.
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
Section 2 Energy
Chapter 5
Potential Energy
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Visual Concept
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.
2
Chapter 5
Section 2 Energy
Elastic Potential Energy
Section 2 Energy
Chapter 5
Spring Constant
Click below to watch the Visual Concept.
Visual Concept
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?
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 = ?
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 
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 103 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
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.
Examples
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.
2. The staples inside a stapler are kept in place by a
spring with a relaxed length of 0.115 m. If the spring
constant is 51.0 N/m, how much elastic potential
energy is stored in the spring when its length is
0.150 m?
Examples
3. A 40.0 kg child is in a swing that is attached to ropes
2.00 m long. Find the gravitational potential energy
associated with the child relative to the child’s lowest
position under the following conditions:
a. When the ropes are horizontal
b. When the ropes make a 30 degree angle with the
vertical
c. At the bottom of the circular arc
Chapter 5
Section 3 Conservation of
Energy
Preview
• Objectives
• Conserved Quantities
• Mechanical Energy
• Sample Problem
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.
Chapter 5
Section 3 Conservation of
Energy
Conserved Quantities
• When we say that something is conserved, we mean
that it remains constant.
– It may change forms but the total amount is
always the same
– For example, an ice cube has a mass of 11 g. It
begins melting. Now it has a mass of 8 grams but
there is 3 g of water. The overall mass stays the
same but the form changes.
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)
Conservation of Energy
• If initial mechanical energy equals final mechanical
energy then…
1 2
1 2 1 2
1 2
mv  mgh  kx  mv  mgh  kx
i 2 i 2
f
f 2 f
2 i
• What energy transfers are being made when…
– A person swings on a swing?
– A rollercoaster rolls?
– A clock pendulum swings?
Chapter 5
Section 3 Conservation of
Energy
Conservation of Mechanical Energy
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Visual Concept
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.
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 = ?
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
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
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
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
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.
Chapter 5
Section 3 Conservation of
Energy
Mechanical Energy, continued
•
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.
Examples
1. A bird is flying with a speed of 18.0 m/s over water
when it accidentally drops a 2.00 kg fish. If the
altitude of the bird is 5.40 m and friction is
disregarded, what is the speed of the fish when it
hits the water?
2. A 755 N diver drops from a board 10.0 m above the
water’s surface. Find the diver’s speed 5.0 m above
the water’s surface. Then find the diver’s speed just
before striking the water.
Examples
3. If the diver in item 2 leaves the board with an initial
upward speed of 2.00 m/s, find the diver’s speed
when striking the water.
4. An Olympic runner leaps over a hurdle. If the
runner’s initial vertical speed is 2.2 m/s, how much
will the runner’s center of mass be raised during the
jump?
5. A pendulum bob is released form some initial height
such that the speed of the bob at the bottom of the
swing is 1.9 m/s. What is the initial height of the
bob?
Chapter 5
Section 4 Power
Preview
• Objectives
• Rate of Energy Transfer
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.
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
Chapter 5
Section 4 Power
Power
Click below to watch the Visual Concept.
Visual Concept
Examples
1. A 1.0 x 103 kg elevator carries a maximum load of
800.0 kg. A constant frictional force of 4.0 x 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.
2. A car with a mass of 1.5 x 103 kg starts from rest
and accelerates to a speed of 18 m/s in 12 s.
Assume that the force of resistance remains
constant at 400 N during this time. What is the
average power developed by the car’s engine?
Examples
3. A rain cloud contains 2.66 x 107 kg of water vapor.
How long would it take for 2.00 kW pump to raise
the same amount of water to the cloud’s altitude ,
2.00 km?
4. How long does it take a 19 kW steam engines to do
6.8 x 107 J of work?
5. A 1.50 x 103 kg car accelerates uniformly from rest
to 10.0 m/s in 3.00 s
a. What is the work done on the car in this time
interval?
b. What is the power delivered by the engine in this
time interval?