According to Physics

According to physics, work is…

The transfer of energy to a body by the application of a force that causes the body to move in a direction parallel to the force.

In other words…

If a force (F) is applied to an object in the same direction as the change in position of the object (Δx), then the force does positive work on the object.

F

Δxxi xf

xFW x F

The work done by the force can be calculated by using the formula…

Δxxi xfxi Δxxi xfxi

For example…

A boy pushes a box 3.0 m across a smooth marble floor. If the boy pushes with a force of 10 N parallel to the floor how much work does he do on the box?

F

Δxxi xf

?

N 0.10

m 0.3

W

F

x

J 30

therefore

m 3.0 N 10

or

W

W

xFW

Given Calculate

F

Δxxi xf

If a force (F) is applied to an object in the opposite direction from the change in position of the object (Δx), then the force does negative work on the object.

Another possibility is…

Δxxi xf

For example…

F

A car traveling north with a velocity is brought to rest by a braking force of 2000 N. If the car travels a distance of 15 m before stopping after the brakes are applied, how much work is done by the braking force?

iv

In this problem the force and change in position are in opposite directions so we need to decide which direction is positive and which negative. Lets say that North is to the right and positive. This means that the force will be negative and the change in position is positive. Make note of this in your diagram.

Fcar iv

Δx

North+

North+

?

N 2000

m 15

W

F

x

J 000,30

therefore

m 15 N 2000

or

W

W

xFW

Given Calculate

Question…

What happens if the force is perpendicular to the change in position of the object?

F

Δxxi xf

Answer…

No work is done by that force on the object.

For example…

A 350 N rock rests on top of a box as the box slides across a smooth floor through a distance of 10 m. How much work is done on the box by the weight of the rock?

Δxxi xf

F

The answer is of course 0 J. This is due to the fact that the weight of the rock cannot accelerate the box along a horizontal surface.

Δxxi xf

F

Another Question…

What if the force isn’t parallel OR perpendicular? What if instead the force is at an angle with respect to the change in position?

F

Δxxi xf

Recall from our discussion of vectors and forces…

A force acting at an angle can be said to be acting in two dimensions at the same time.

F

Δxxi xf

For work problems you would break the force down into a component parallel to the change in position ( ).

F

Fx

xF

And a component perpendicular to the change in position ( ).

F

Fx

yF

Fy

Only the component of the force that is parallel to the change in position can be used to find the work done on the object

F

Fx

Fy

Only the component of the force that is parallel to the change in position can be used to find the work done on the object

Fx

F

Δxxi xf

Only the component of the force that is parallel to the change in position can be used to find the work done on the object

Fx

F

Δxxi xf

Only the component of the force that is parallel to the change in position can be used to find the work done on the object

Fx

Δxxi xf

For example…

A man pushing a lawn mower pushes straight along the handle

Δxxi xfF

Only the component of the force parallel to the change in position does any work.

Δxxi xf

Fx

F

xFW x

We use the subscript on the force. It helps us remember that only forces on an object that are parallel to the change in position of the object do any work.

This is why…