Chapter 5 Work and Energy
5.1 Work
•Term ‘work’ has special meaning in science – work is done ONLY if a
force moves an object.
•The distance an object moves ALSO must be in the same direction as
the force applied.
Equation: Work = Force x Distance
Work = F x d
(N) x (m)
SI Units of work - Newton – meter called a ‘Joule’
Units of Work = Joules (J)
Work is a Scalar quantity – does not have a direction.
1. A force of 22N is applied over a distance of 5.0m
W
= F • d
= 22N • 5.0m
= 110 J
2. A force of 250,000N is applied over a distance of 15 km.
W
= F • d
= 250,000N
= 3.75 x 109 J
• 15,000m
Is Work Being Done?
It is only the
horizontal component
of the tension force in
the chain that causes
Fido to be displaced
to the right.
Requires the same amount of work.
The amount of work done by a force on any object is given by
the equation
Work = F * d * cosine(θ)
For work to be done, the Force must be in the direction of the motion.
If it is at an angle, the work done is:
Work = Force • cos θ • displacement
W = F • cos θ • d
1. How much work do you do if you push a lawn mower 25m with a
force of 85N at an angle of 30° ?
W = F • cosθ • d
=
=
1840 J
2. How much force would it take to do the same amount of work
if the angle was 60° ?
1840 J = F x cos ϴ x d
F =
1840 J
cos ϴ x d
= 147 N
5-2 Kinetic Energy
Energy comes in many forms.
 Electrical, chemical, heat, and atomic are just a few examples.
 A good definition of Energy is the ability to do work.
 Work is a form of energy – energy is a form of work.
 The Kinetic Energy (KE) of an object is the amount of
“work” stored by that object due to its motion.
 The velocity of the object is squared and therefore the most
influential component.
Formula:
KE = ½ mv2
Units: J (Joules)
Example #1
A bowling ball ( m = 6.0 kg ) is traveling at 4.0 m/s.
Find its K.E. at this speed.
K. E. = ½ mv2
= ½ • 6.0 kg • (4.0) 2
= 48 J
More Practice 1. What is the K. E. of a chicken ( m = 1.5 kg ) that is
caught in a tornado (v = 130 m/s)?
KE = ½ m x v2
= 12675 J or 13,000 J (2 sig figs)
2. What is the K.E. of a rain drop ( m = 6.5x10-4 kg ) as
it hits your face? ( vy = 12 m/s )
KE = ½ m x v2
= 0.0468 J or 4.7 x 10 -2 J
3. What is the K.E. of an excited Irish Setter (m = 34 kg )
that leaps on you with a velocity of 6.0 m/s ?
KE = ½ m x v2
= 612 J
5-2 continuation: WORK – ENERGY THEOREM
 When work is done on an object, the net Work is converted to K.E.
Wnet = ∆ K. E.
Ex. ( p.176 )
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 her final K.E.
is 352 J?
Given F = 45 and
Find ∆X (d) = ?
Wnet = F ● ∆X
KE = 352 J
Formula: Wnet = ∆ KE
F ● ∆X = ∆ KE
∆X = ∆ KE / F
= 7.8 m
5-2 continuation: WORK – ENERGY THEOREM
 When work is done on an object, the net Work is converted to K.E.
Wnet = ∆ K. E.
Ex. ( p.176 )
2. A 2.0 x 103 kg car accelerates from rest under the action of two
forces. One is a forward force of 1140 N provided by traction
between the tires and the road. The other is a 950 N resistive force
due to various frictional forces. Use the WORK – ENERGY theorem to
determine how far the car must travel for its speed to reach 2.0 m/s.
Given: m= 2,000 kg
H.W. – GR 5.2 and do problems p. 176 # 4 & 5
 Since K.E. depends on the square of the velocity, it increases
dramatically as velocity increases.
Example: Fill out the following table for a car ( m = 1,600 kg )
at different velocities; μs = 0.9 and Fs = μs x FN and Wbreaks = Fs x d
Speed Zone
Residential
Velocity
12 m/s
(25mph)
Main road
21 m/s
(45mph)
Freeway (legal)
32 m/s
(65mph)
Freeway (speeding) 40 m/s
(85mph)
K.E. ( Joules )
Braking ∆ x
 Since K.E. depends on the square of the velocity, it increases
dramatically as velocity increases.
Example Fill out the following table for a car ( m = 1,600 kg )
at different velocities; μs = 0.9 and Fs = μs x FN and Wbreaks = Fs x d
Speed Zone
Residential
Velocity
12 m/s
(25mph)
Main road
21 m/s
(45mph)
Freeway (legal)
32 m/s
(65mph)
Freeway (speeding) 40 m/s
(85mph)
K.E. ( Joules )
Braking ∆ x
115,200 J
8.2 m
352,800 J
25 m
819,200 J
58 m
1,280,000 J
91 m
POTENTIAL ENERGY
 Stored energy that is capable of doing work.
 Energy of position
 It can be electricity in a battery, energy in a compressed spring, a
stretched rubber band, or an object that can fall.
Gravitational P.E. –
 Energy associated with an object due to its height
 The GPE of an object at any given height can be found using the
equation:
P.E. = m • g • h
= mass • gravity • height
Unit = Joules (J)
 When an object falls, the P.E. converts into K.E.
 As P.E. decreases, K.E. increases
 If there is no friction or air resistance, 100% of the P.E. will
become K.E.
Gravitational PE does not depend on the
path taken to get it there
Elastic Potential Energy
 Energy stored in a spring that is stretched or compressed.
The force on a spring is given by Hooke’s Law.
F = k • d
k = spring constant (or force constant)
units of k = N/m
The Elastic P.E. in the spring (the work it can do) is:
P.E. = Favg • d
P.E. = ½
( Favg = Ftotal /2 )
2
kd
CONSERVATION of ENERGY
 Energy cannot be created or destroyed, It can only transform from
one form to another.
∆KE = ∆ PE
Power
 In science ‘power’ means how fast work is being done.
 Power is the rate work is done (amount of work per time)
Power = Work / Time
P = W / ∆t
OR
Power = Force x distance / Time
P = F x d / ∆t
 SI Units of Power = watt (W). (named after James Watt)
 1 watt = 1 joule per second (1 J/sec)
 50 watt light bulb doing work at the rate of 50 joules per sec
 1 kilowatt (kW) = 1000 watts. Electric power in your home
is measured in kWh (kilowatt – hour). Unit of energy – not
power.
1 horsepower can raise 330 pounds of
coal 100 feet in a minute
1 hp = 750 watts
POWER
The rate of doing work:
The faster work is done, the more power it takes.
Example:
1.
An aircraft carrier catapult does 1.05 x 108 J of work in 2.2 s.
a. How much power does it generate?
b. How many horsepower is that?
(1 hp = 750 W)
2.
A dragster does 1.7 x 107 J of work in 4.0s
a. How much power does it generate?
b. How many horsepower is that?
POWER
The rate of doing work:
The faster work is done, the more power it takes.
Examples:
1.
An aircraft carrier catapult does 1.05 x 108 J of work in 2.2 s.
a. How much power does it generate?
4.8 x 107 watts
b. How many horsepower is that?
(1 hp = 750 W)
64,000 hp
2.
A dragster does 1.7 x 107 J of work in 4.0s
a. How much power does it generate?
4.3 x 10 6 watts
b. How many horsepower is that?
5,667 hp
Objectives:
1. Identify several forms of energy.
2. Calculate kinetic energy for an object.
3. Apply the work-kinetic energy theorem to solve
problems.
4. Distinguish between kinetic and potential energy.
5. Classify different types of potential energy.
6. Calculate the potential energy associated with an
object's position.