investigate quantities using the work-energy theorem

In this presentation you will:

F

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m

Work-Energy Theorem

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Many words used in everyday conversation have very precise meanings in physics.

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Introduction

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.

WorkEnergy

Force

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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.

Using the equations of motion, we can find the change in velocity.

F

d

m

d

m

vi

m

vf

Replacing a = F/m and multiplying both sides by m/2, we get:

a = F/m

vf2 – vi

2 = 2ad

Fd = ½mvf2 – ½mvi

2

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Fd = ½mvf2 – ½mvi

2

The left side of the equation is the work done on the system.

F

d

m

vi

m

vf

So:

In physics, work has a precise definition:

Work is the product of force × distance.

Work = F × d

W = ½mvf2 – ½mvi

2

What is work?

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Energy

W = ½mvf

2 – ½mvi2

The right side of the equation describes the change in quantity before and after the force acts.

This quantity is known as energy and as it is the energy of a moving body, kinetic energy.

The quantity depends on the mass and the velocity of the object.

KE = ½mv2

So: W = KEf – KEi

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Work-Energy Theorem

Work is equal to the change in kinetic energy.

W = ΔKE

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)

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

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Calculating Work

Work is calculated using the equation:

W = Fd = 10 × 10 = 100 J

A force of 10 N is used to move an object over a distance of 10 m. How much work is done?

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m

Work = 100 J

W = Fd  

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.

Question 1

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W = F × d = 25 × 5 = 125 J

Work = 125 J

125 (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.

F

d

m Ff

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Friction Force

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

Now we have a friction force (Ff) acting in the opposite direction.Ff acts in the opposite direction, so it does negative work. W = (F – Ff)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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m Ff

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.

Question 2

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W = -Ff × d = -10 × 10 = -100 J

Work = -100 J

-100 (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?

Give your answer as a number in J.

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.

Question 3

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W = -Ff × d = -10 × 10 = -100 J

Work = -100 J

Work = ΔKE

ΔKE = -100 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.

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.

Question 4

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W = (F-Ff) × d = (100 – 10) × 10 = 900 J

Work = 900 J

900 (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.

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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W = F × d = 30 × 10 = 300 J

Work = 300 J

Work = ΔKE

ΔKE = 300 J

Total KE = 300 + 100 = 400 J

400 (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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Force at an Angle

To calculate force applied at an angle, the force is split into x and y components.

Fx = F cosθ

F

d

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θ

F

Fx

Fy

The Fy component acts at right angles to the direction of displacement so no work is done.

Only the component of the force that acts in the direction of the displacement is used.

So, W = Fd cosθ

Enrichment

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Force at an Angle

F

d

m

θ

F

Fx

Fy

For example, if the force of 10 N is acting at an angle of 30° over a distance of 10 m:

W = Fd cosθ

W = 10 × 10 × cos30° J

W = 86.6 J

Enrichment

Work is an energy transfer calculated by the product of a force and distance. W = F × d

In this presentation you have seen:

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Summary

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