Work-Energy Theorem
F
m
d
In this presentation you will:
 investigate quantities using the work-energy
theorem
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Introduction
Many words used in everyday conversation have
very precise meanings in physics.
When you think of work,
you may think of going to a
place of work, or working
hard, or even something
like a computer working.
When you think of energy,
you may think of how lively
or tired you are.
In physics, work and energy have very specific
meanings.
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What is work?
Consider a force (F) acting on an object while it
moves a distance (d).
The object’s velocity will
change; it will accelerate.
a = F/m
Using the equations of
motion, we can find the
change in velocity.
vf2 – vi2 = 2ad
F
m
d
vf
vi
m
m
d
Replacing a = F/m and multiplying both sides by
m/2, we get:
Fd = ½mvf2 – ½mvi2
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What is work?
Fd = ½mvf2 – ½mvi2
m
The left side of the
equation is the work
done on the system.
vf
vi
m
F
d
In physics, work has a precise definition:
Work is the product of force × distance.
Work = F × d
So:
W = ½mvf2 – ½mvi2
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Energy
W = ½mvf2 – ½mvi2
The right side of the
equation describes the
change in quantity before
and after the force acts.
vf
vi
m
m
F
d
The quantity depends on the mass and the velocity
of the object.
This quantity is known as energy and as it is the
energy of a moving body, kinetic energy.
KE = ½mv2
So: W = KEf – KEi
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Work-Energy Theorem
Work is equal to the
change in kinetic energy.
vf
vi
m
m
F
W = ΔKE
d
In physics, the Δ symbol is used to represent a
change in something.
The units of work and energy are the joule (J),
named after the physicist James Joule, who
discovered the relationship.
1 J = 1 Nm (Fd) = 1 kgm2/s2 (½mv2)
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Calculating Work
Work is calculated using the equation:
W = Fd
F
m
d
A force of 10 N is used to move an object over a
distance of 10 m. How much work is done?
W = Fd = 10 × 10 = 100 J
Work = 100 J
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Question 1
How much work is done when a force of 25 N is
used to move an object a distance of 5 m?
Give your answer as a number in J.
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Question 1
How much work is done when a force of 25 N is
used to move an object a distance of 5 m?
W = F × d = 25 × 5 = 125 J
Work = 125 J
Give your answer as a number in J.
125 (J)
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Friction Force
In the real world, friction tends to act on a moving
body to slow it down.
Ff
F
m
Now we have a friction
d
force (Ff ) acting in the
opposite direction.
Ff acts in the opposite direction,
so it does negative work.
F
m
f
W = (F – Ff )d
d
If the friction force Ff was the only force acting on
the object when it is in motion, the object would
slow down, and its kinetic energy would be reduced.
W = (0 - Ff )d = - Ff d
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Question 2
An object is in motion. A friction force of 10 N acts
on it over a distance of 10 m. How much work is
done on the object?
Give your answer as a number in J.
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Question 2
An object is in motion. A friction force of 10 N acts
on it over a distance of 10 m. How much work is
done on the object?
W = -Ff × d = -10 × 10 = -100 J
Work = -100 J
Give your answer as a number in J.
-100 (J)
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Question 3
An object is in motion. A friction force of 10 N acts
on it over a distance of 10 m. What is its change in
kinetic energy?
Give your answer as a number in J.
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Question 3
An object is in motion. A friction force of 10 N acts
on it over a distance of 10 m. What is its change in
kinetic energy?
W = -Ff × d = -10 × 10 = -100 J
Work = -100 J
Work = ΔKE
ΔKE = -100 J
Give your answer as a number in J.
-100 (J)
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Question 4
A force of 100 N is used to move an object over a
distance of 10 m. A friction force of 10 N is acting on
the object to slow it down. What is the work done?
Give your answer as a number in J.
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Question 4
A force of 100 N is used to move an object over a
distance of 10 m. A friction force of 10 N is acting on
the object to slow it down. What is the work done?
W = (F-Ff ) × d = (100 – 10) × 10 = 900 J
Work = 900 J
Give your answer as a number in J.
900 (J)
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Question 5
A moving object has a kinetic energy of 100 J.
A force of 30 N is applied to the object in the same
direction as its travel over a distance of 10 m.
What is the total kinetic energy after 10 m?
Give your answer as a number in J.
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Question 5
A moving object has a kinetic energy of 100 J.
A force of 30 N is applied to the object in the same
direction as its travel over a distance of 10 m.
What is the total kinetic energy after 10 m?
W = F × d = 30 × 10 = 300 J
Work = 300 J
Work = ΔKE
ΔKE = 300 J
Total KE = 300 + 100 = 400 J
Give your answer as a number in J.
400 (J)
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Force at an Angle
Enrichment
To calculate force applied at an angle, the force is
split into x and y components.
Only the component of
the force that acts in
the direction of the
displacement is used.
F
m
d
Fx
Fx = F cosθ
So, W = Fd cosθ
θ
Fy
F
The Fy component acts at right
angles to the direction of
displacement so no work is done.
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Force at an Angle
Enrichment
For example, if the force of 10 N is acting at an
angle of 30° over a distance of 10 m:
W = Fd cosθ
F
m
d
W = 10 × 10 × cos30° J
Fx
θ
W = 86.6 J
Fy
F
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Summary
In this presentation you have seen:
 Work is an energy transfer calculated by the
product of a force and distance.
W=F×d
 The kinetic energy of a moving body is related to
its mass and velocity.
KE = ½mv2
 The work-energy theorem relates work and
energy.
W = ΔKE
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