Energy
Introduction
Lesson 1
Definition of energy
When energy is transferred from one form to
another it may be transferred by doing work.
For example, when you lift an object you do work by transferring
chemical energy to kinetic energy and gravitational potential energy.
This concept of work gives us a way of defining energy
• Energy is defined as the stored ability to do
work
Work
• When a force acts upon an object to cause a
displacement of the object, it is said that work was done
upon the object.
• There are three key ingredients to work - force,
displacement, and cause.
• In order for a force to qualify as having done work on an
object, there must be a displacement and the force
must cause the displacement.
Everyday Examples
• a horse pulling a plow through
the field,
• a father pushing a grocery cart
down the aisle of a grocery
store,
• a student lifting a backpack full
of books upon her shoulder,
• a weightlifter lifting a barbell
above his head,
• an Olympian launching the
shot-put, etc.
In each case described here
there is a force exerted upon
an object to cause that object
to be displaced.
Activity (in small groups)
• Read the five statements on the cards and
determine whether or not they represent
examples of work.
• Write down the reasons for your answers
• We will then discuss the answers as a
class
A man applies a force to a wall and
becomes exhausted
• No.
• This is not an
example of work.
The wall is not
displaced. A
force must cause
a displacement
in order for work
to be done.
A book falls off a table and free falls
to the ground
• Yes.
• This is an example
of work. There is a
force (gravity) which
acts on the book
which causes it to
be displaced in a
downward direction
(i.e., "fall").
A waiter carries a tray full of meals above his
head by one arm straight across the room at
constant speed.
• No.
• This is not an example of
work. There is a force (the
waiter pushes up on the tray)
and there is a displacement
(the tray is moved
horizontally across the
room). Yet the force does not
cause the displacement. To
cause a displacement, there
must be a component of
force in the direction of the
displacement.
A rocket accelerates through
space.
• Yes.
• This is an example
of work. There is a
force (the expelled
gases push on the
rocket) which
causes the rocket
to be displaced
through space.
Mathematically, work can be expressed
by the following equation:
W = F x d x Cos θ
where F is the force, d is the
displacement, and the angle (theta) is
defined as the angle between the force
and the displacement vector.
How do we find the angle, θ?
• The angle measure is defined as the angle
between the force and the displacement.
To gather an idea of it's meaning, consider
the following three scenarios:
• A force acts rightward upon an object as
it is displaced rightward. In such an
instance, the force vector and the
displacement vector are in the same
direction. Thus, the angle between F and
d is 0 degrees.
A force acts upward on an object as
it is displaced rightward. In such
an instance, the force vector and
the displacement vector are at
right angles to each other. Thus,
the angle between F and d is 90
degrees .
A force acts leftward upon an object that is
displaced rightward. In such an instance,
the force vector and the displacement
vector are in the opposite direction. Thus,
the angle between F and d is 180 degrees.
An example of negative work
Negative Work
• Sometimes a force acts upon a moving object to hinder a
displacement. Examples might include:
– a car skidding to a stop on a roadway surface
– baseball runner sliding to a stop on the infield dirt.
• In such instances, the force acts in the direction opposite
the objects motion in order to slow it down. The force
doesn't cause the displacement but rather hinders it. The
negative refers to the numerical value that results when
values of F, d and theta are substituted into the work
equation. Since the force vector is directly opposite the
displacement vector, theta is 180 degrees. The
cosine(180 degrees) is -1 and so a negative value
results for the amount of work done upon the object.
Which path needs the most
energy?
• link to animation
Units of Work
• For work (and also energy), the standard
metric unit is the Joule (abbreviated J).
• One Joule is equivalent to one Newton of
force causing a displacement of one
meter. In other words:
1 Joule = 1 Newton * 1 meter
1J=1N*m
Worked Example
Consider an Arctic explorer dragging a sledge across a frozen lake.
The explorer attaches the rope to his waist and the force of 200 N is
applied at 30o to the horizontal. We can model this situation in the
diagram below.
How much work is done by the explorer in dragging the sledge 150
metres across the ice at a steady speed?
The difficulty here is that the force, F (200 N), and the displacement, s,
are not in the samedirection. One solution is to resolve the 200 N
force into its vertical and horizontal components as shown below.
Provided the sledge does not rise above the ice, no work is done by the
vertical force of 100 N.
Work is done only by the horizontal component of the applied
force (173.2 N).
Work done = constant force × distance moved in direction of the force
= 173.2 N × 150 m
= 25 980 J
The other way of solving this is to use the
general formula:
W = Fd cos θ
To apply this formula to the sledge example
above we would write:
W = Fd cos θ
= 200 × 150 × cos 30o
= 200 × 129.9
= 25 980 J
Check Your Understanding
• Show your understanding of the concept
and mathematics of work by answering the
questions on the worksheet
Q1 Diagram A Answer:
W = (100 N) * (5 m)* cos(0 degrees) = 500 J
The force and the displacement are given in the
problem statement. It is said (or shown or
implied) that the force and the displacement are
both rightward.
Since F and d are in the same direction, the angle
is 0 degrees.
Diagram B Answer:
W = (100 N) * (5 m) * cos(30 degrees) = 433 J
The force and the displacement are given in the
problem statement. It is said that the
displacement is rightward. It is shown that the
force is 30 degrees above the horizontal. Thus,
the angle between F and d is 30 degrees.
Diagram C Answer:
W = (147 N) * (5 m) * cos(0 degrees) = 735 J
The displacement is given in the problem
statement.
The applied force must be 147 N since the 15-kg
mass (Fgrav=147 N) is lifted at constant speed.
Since F and d are in the same direction, the
angle is 0 degrees.
Answers to Q2
a) Only Fapp does work. Fgrav and Fnorm do not do work since a
vertical force cannot cause a horizontal displacement.
Wapp= (10 N) * (5 m) *cos (0 degrees) =+50 Joules
b) Only Ffrict does work. Fgrav and Fnorm do not do work since a
vertical force cannot cause a horizontal displacement
Wfrict =(10 N) * (5 m) * cos (180 degrees) =-50 Joules
c) Fapp and Ffrict do work. Fgravand Fnorm do not do work since a
vertical force cannot cause a horizontal displacement.
Wapp = (10 N) * (5 m) * cos (0 deg) =+50 Joules
Wfrict = (10 N) * (5 m) * cos (180 deg) =-50 Joules
Answer to question 3
• Be careful!
• If the F is parallel to the incline and the d is
parallel to the incline, then the angle theta
in the work equation is 0 degrees.
• In each case, the work is
approximately1.18 x106 Joules.
• The angle does not affect the amount of
work done on the roller coaster car.
Answer Q4
The motion has two parts:
pulling vertically to displace the suitcase vertically (angle =
0 degrees) and
pushing horizontally to displace the suitcase horizontally
(angle = 0 degrees).
For the vertical part, W = (200 N) * (10 m) * cos (0 deg)
= 2000 J.
For the horizontal part, W = (50 N) * (35 m) * cos (0 deg)
= 1750 J.
The total work done is 3750 J (the sum of the two parts).
Answers
Q5:
W = F * d * cos θ
W = (50 N) * (3 m) * cos (30 degrees)
=129.9 Joules
Q6:
To lift a 15-Newton block at constant speed, 15-N
of force must be applied to it (Newton's laws).
Thus,
W = (15 N) * (3 m) * cos (0 degrees)
=45 Joules