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

F m d

Using the equations of motion, we can find the change in velocity. v f 2 – v i 2 = 2ad

m v i d m

Replacing a = F/m and multiplying both sides by m/2, we get: Fd = ½mv f 2 – ½mv i 2

v f Next >

What is work?

Fd = ½mv f 2 – ½mv i 2 The left side of the equation is the work done on the system.

m v i F d

In physics, work has a precise definition: Work is the product of force × distance.

Work = F × d So: W = ½mv f 2 – ½mv i 2

m v f Next >

Energy

W = ½mv f 2 – ½mv i 2 The right side of the equation describes the change in quantity before and after the force acts.

m v i F d m v f

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 = ½mv 2 So: W = KE f – KE i

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Work-Energy Theorem

Work is equal to the change in kinetic energy.

W = ΔKE

m v i F m d

In physics, the Δ symbol is used to represent a change in something.

v f

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 kgm 2 /s 2 (½mv 2 )

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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 Next >

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) Next >

Friction Force

In the real world, friction tends to act on a moving body to slow it down.

F m F f

Now we have a friction force (F f ) acting in the opposite direction.

d

F f acts in the opposite direction, so it does negative work.

m F f

W = (F – F f )d

d

If the friction force F f 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 - F f )d = - F f 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 = -F f × d = -10 × 10 = -100 J Work = -100 J Give your answer as a number in J.

-100 (J) Next >

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 = -F f × d = -10 × 10 = -100 J Work = -100 J Work = ΔKE ΔKE = -100 J Give your answer as a number in J.

-100 (J) Next >

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-F f ) × d = (100 – 10) × 10 = 900 J Work = 900 J Give your answer as a number in J.

900 (J) Next >

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) Next >

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

F x = F cosθ So, W = Fd cosθ

d F x θ F

The F y 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

W = 10 × 10 × cos30° J W = 86.6 J

θ d F x F F y Next >

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 = ½mv 2  The work-energy theorem relates work and energy. W = ΔKE

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