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Work and Energy
Chapter 7
Conservation of Energy
Energy is a quantity that can be
converted from one form to another but
cannot be created or destroyed.
 The total amount of energy present in
any process remains the same.

Work
Has a very specific definition in physics.
 Involves a force moving a body through
a displacement.
 If the force is in the same direction as the
motion,

W  Fs
Vector equation for work

If the force is directed at an angle to
the displacement, then only the
component of the force that is parallel
to the displacement does work.
F

W  Fs cos
Vector equation for work
Important – work is a scalar quantity,
not a vector, even though it is derived
from two vectors.
 Work does not have a direction.

Work

If  is between 0° and 90°, cos  is
positive, so work done is positive.

If  is between 90° and 180°, cos  is
negative, so work done is negative.

If  is 90°, cos  is zero, so work done
is zero.
Example

Lifting, carrying, and lowering a book.
Units on work

The unit for work is a joule.
1J 1 N m
Example
During your winter break you enter a
dogsled race across a frozen lake. To
get started you pull the sled (m = 80 kg)
with a force of 180 N at 20° above the
horizontal. You pull the sled 5 m.
 Find the amount of work you do

On your own
A truck of mass 3000 kg is to be loaded
onto a ship by a crane that exerts an
upward force of 31 kN on the truck. This
force is applied over a distance of 2 m.
 Find the work done by the crane on the
truck

Work and energy with varying
forces
The equations we’ve developed so far
for work involve constant forces and
motion along a straight line.
 Now let’s look at varying forces.
 We could also look at motion along
curved lines, but we would need to use
integrals, which you don’t know how to
do yet.

The spring force

The farther you stretch a spring, the harder
it is to stretch.
 The force you must apply follows the
following equation, known as Hooke’s Law
Fx  kx

Where k is called the force constant of the
spring or the spring constant. k has units of
N/m
Work done on a spring
1 2
W  kX
2
Where X is the total elongation of the
spring.
 See page 151.

Work done on a spring
1 2 1 2
W  kx2  kx1
2
2
Where x1 is the initial position of the
spring, x2 is the final position of the
spring
 x = 0 is the equilibrium position of the
spring (when it is neither stretched nor
compressed.)

Example
A 4-kg block on a frictionless table is
attached to a horizontal spring with a
force constant of 400 N/m. The spring
is initially compressed with the block at
x1 = -5 cm.
 Find the work done on the block as the
block moves to its equilibrium position
x2 = 0

On your own

To stretch a spring 3.00 cm from its
unstretched length, 12.0 J of work must
be done. How much work must be done
to compress this spring 4.00 cm from its
unstretched length?
Work and Kinetic Energy

The total work done on a body is related
to changes in the speed of the body.
v2  v1  2as
2
2
v2  v1
a
2s
2
2
Work and Kinetic Energy
v2  v1
F  ma  m
2s
2
2
v2  v1
Fs  m
2
1
1
2
2
Fs  mv2  mv1
2
2
2
2
Work-Kinetic Energy Theorem
1
1
2
2
Wtot  mv2  mv1
2
2

The quantity ½ mv2 is called the kinetic
energy and represented by the letter K. It is
a scalar quantity and can never be negative.
It is zero when an object is not moving.
Wtot  K 2  K1  K
Units on kinetic energy

Looking at the equation, we have
2
m
kg  2
s
m
 kg  2  m
s
 N m
J
Work-kinetic energy theorem
The work-kinetic energy theorem still
works for varying forces.
 See the last paragraph on page 153 for a
mathematical explanation of this.

1
1
2
2
Wtot  mv2  mv1
2
2
Motion along a curved path

W   F  dl
P2
P1

Where dl is a distance along the path

The work-kinetic energy theorem still
holds true.
Example
During your winter break you enter a
dogsled race across a frozen lake. To
get started you pull the sled (m = 80 kg)
with a force of 180 N at 20° above the
horizontal. You pull the sled for 5 m.
 Find

The work you do
 The final speed of the sled

On your own
A truck of mass 3000 kg is to be loaded
onto a ship by a crane that exerts an
upward force of 31 kN on the truck. This
force is applied over a distance of 2 m.
 Find

The work done by the crane on the truck
 The total work done on the truck
 The upward speed of the truck after the 2 m
if it started from rest

Example
A 4-kg block on a frictionless table is
attached to a horizontal spring with a
force constant of 400 N/m. The spring
is initially compressed with the block at
x1 = -5 cm.
 Find

The work done on the block as the block
moves to its equilibrium position x2 = 0
 The speed of the block at x2=0

Where does kinetic energy come
from?
If an object falls off a cliff, gravity does
work on it as it falls, and increases its
kinetic energy.
 But where does that energy come from?

Potential Energy
Energy seems to be stored in some form
related to height.
 This energy is related to the position of a
body, not its motion.
 It is called potential energy. – measures
potential or possibility for work to be
done.
 (Some kinds of potential energy are
related to things other than height.)

Work done by gravity

Work done by gravity as a body falls
from height y1 to height y2.
Wgrav  Fs cos 0  Fs

 w y1  y2 
 mgy1  mgy2

Also works if object is rising – then y2 is
greater than y1, so the work is negative,
as it should be.
Gravitational Potential Energy

We define the gravitational potential
energy as
U  mgy

So, the work done by gravity is
Wgrav  U1  U 2  U 2  U1 
 U
If only gravity does work
Wtot  Wgrav
K  U
K2  K1  U1  U 2
K2  U 2  K1  U1
Conservation of Mechanical
Energy
E  K  U  constant

When only gravity does work, the total
mechanical energy is constant.
Where do I measure y from?
Wherever you want.
 It is only the change in potential energy
that we are interested in, not the value of
U at a particular point.
 So choose your zero potential energy
point wherever it is convenient.

Example

Standing near the edge of the roof of a 12-m
high building, you kick a ball with an initial
speed of vi = 16 m/s at an angle of 60° above
the horizontal. Neglecting air resistance, use
conservation of energy to find


How high above the height of the building the ball
rises
Its speed just before it hits the ground

9.79 m
22.2 m/s
You try

You throw a 0.200-kg ball straight up in
the air, giving it an initial upward velocity
of 20.0 m/s. Use conservation of energy
to find how high it goes.

20.4 m
Effect of other forces

If forces other than gravity are acting on
a body
Wtot  Wgrav  Wother
 K2  K1
U1  U 2  Wother  K 2  K1
K1  U1  Wother  K 2  U 2
Example
A child of mass 40 kg goes down an
8.0 m long slide inclined at 30° above
the horizontal. The coefficient of kinetic
friction between the slide and the child is
0.35. If the child starts from rest at the
top of the slide, how fast is she traveling
when she reaches the bottom?
 5.60 m/s

Curved path
The shape of the path makes no
difference for gravitational potential
potential energy.
 You can still use the same equations for
conservation of energy.

On your own

A pendulum consists of a bob of mass m
attached to a string of length L. The bob
is pulled aside so that the string make an
angle 0 with the vertical, and is released
from rest. Find an expression for its
speed as it passes through the bottom of
the arc.
0
Elastic potential energy
Think slingshot.
 The farther you stretch it, the more
kinetic energy the projectile can gain.

Springs

For springs, the elastic potential energy
is given by
1 2
U  kx
2

The work done by a spring is
1 2 1 2
Wel  kx1  kx2  U1  U 2
2
2

 U
The equations work whether the spring
is stretched or compressed.
Where is x = 0?
We don’t get to choose this time
 x = 0 is always at the equilibrium position
of the spring – when it is neither
stretched nor compressed.

Conservation of energy

The conservation of energy works the
same for springs as it does for gravity.
K1  U1  Wother  K 2  U 2
Conservation of energy

If we have both gravitational and elastic
potential energies, then
K1  U grav,1  U el ,1  Wother  K 2  U grav, 2  U el , 2
Example

A 1.20-kg piece of cheese is placed on a
vertical spring of negligible mass and
force constant k = 1800 N/m that is
compressed 15.0 cm. When the spring
is released, how high does the cheese
rise from its initial position? (The spring
and the cheese are not attached.)

1.72 m
On your own
A 2000-kg elevator with broken cables
is falling at 25 m/s when it first contacts
a cushioning spring at the bottom of the
shaft. The spring is supposed to stop
the elevator, compressing 3.00 m as it
does so. During the motion, a safety
clamp applies a constant 17,000 N
frictional force to the elevator.
 What is the force constant of the
spring?


1.41 x 105 N/m
Conservative forces

The work done by conservative forces:
Can always be expressed as the difference
between the initial and final values of a
potential energy function.
 Is reversible
 Is path-independent (It only depends on the
starting and ending points.)
 Equals zero when the starting and ending
points are the same.

Conservative forces
Gravitational (without air resistance)
 Spring (without friction)

Nonconservative forces
Depend on path
 Are not reversible
 Cannot be expressed in terms of a
potential energy function
 Do work even when the starting and
ending points are the same

Nonconservative forces
Friction
 Chemical reaction forces

Nonconservative forces
Increase or decrease the internal energy
of a system
 Frequently, this happens through heat.

U int  Wother
Law of conservation of energy
K  U  U int  0

This is always true – the most general
form of the equation.
Power
Power is not the same as energy or
force.
 Power is the time rate at which work is
done.
W

Pav 
t
Power unit

The unit of power is the watt
J
1W 1
s
The kWh
Your electric bill tells you how many kWh
of electricity you used over the last
month.
 This is a unit of energy or work, not a
unit of power

J

 1000  3600 s

6
s 
1 kWh 
  3.6 10 J or 3.6 MJ
 1 kW  1 h 




Power in terms of velocity and
force
W
Pav 
t
Pav 
Fparallels
t
P  Fparallelv
s  F
 Fparallel
parallelvav
t
P  Fv cos
Example
Force A does 5 J of work in 10 s.
 Force B does 3 J of work in 5 s.
 Which force delivers greater power?

On your own

A 5-kg box is being lifted upward at a
constant velocity of 2 m/s by a force
equal to the weight of the box.
What is the power input of the force?
 How much work is done by the force in 4 s?
