7. Work by a Variable Force

by M. Bourne

Don't miss
interactive spring activity
later in this section...

The work (W) done by a constant force (F) acting on a body by moving it through a distance (d) is given by:

W = F × d

Example:

An apple weighs about 1 N. If you lift the apple 1 m above a table, you have done approximately 1 Newton metre (Nm) of work.

Work done by a Variable Force

If the force varies (e.g. compressing a spring) we need to use calculus to find the work done.

If the force is given by F(x) (a function of x) then the work done by the force along the x-axis from a to b is:

Hooke's Law for Springs

The force (F) that it takes to stretch (or compress) a spring x units from its normal length is proportional to x.

F = kx

We can find the spring constant k from observing what force gives what stretch for each spring.

Flash Interactive

In this activity, you can see the forces involved, the work done and you can explore the meaning of k, the spring constant.

Things to do...

Extend or compress the spring. See the force required to do this.
When you let go, you will see the work done in compressing or extending the spring.
Change the value of k and see how much this changes the force required and the work done.
Why does the force increase as the amount of stretch (x) increases?
Does the relationship F = kx hold?
Why soes the spring slow down?

Loading Flash movie.

Notes:

This activity assumes you are looking at the spring from above and that the "ball" has zero mass. (The ball is there so you have got something to grab.)
You can see other interesting spring examples at Applications of Trigonometric Curves and in Composite Trigonometric Curves.

Example

Find the work done on a spring when you compress it from its natural length of 1 m to a length of 0.75 m if the spring constant is k = 16 N/m.

Answer: F = 16x

So

What is the work done in compressing the spring a further 30 cm?

Answer

Note: For a spring,

requires that a and b are the distance from the natural position of the spring.

Example:

A force of 1200 N compresses a spring from its natural length of 18 cm to a length of 16 cm. How much work is done in compressing it from 16 cm to 14 cm?

Answer: We must work out the spring constant first (we will work in cm).

F = kx

So 1200 = k(2)

So k = 600 N/cm

So in this case,

F = 600x

Now the work done is given by:

Check your understanding

Go back up to the spring interactive above and calculate the work done in compressing or stretching the spring for various amounts of stretch. Do your answers tally with the answer given?

Another Example of Work Done by a Variable Force

Need Graph Paper?

Download graph paper

A leaky 5N bucket is lifted into the air at a constant speed. The rope weighs 0.08 Nm^-1. The bucket starts with 2 N of water and leaks at a constant rate. It finishes draining just as it reaches the top. How much work was done:

a) lifting the water alone

Answer: a) Water only.

The force required to lift the water is the water's weight.

When the bucket is x metres off the ground, the expression for the water's weight can be found by first graphing it:

So, using y = mx + c, we see that the slope is

m = -2/20 = -1/10 and

the y-intercept is c = 2

So we can write the function for the weight at height x as:

Alternatively, we could have found the function as follows:

b) lifting the water and bucket together

Answer: b) Water and bucket

For the bucket,

W = Fd

= 5 × 20

= 100 N.m

So total work = 20 + 100 = 120 N.m

c) lifting the water, bucket and rope?

Answer: c) Water, bucket, rope

The weight of the rope at height x is:

F(x) = 0.08(-x + 20)

So the work done on the rope is:

So the total work is 20 + 100 + 16 N.m = 136 N.m

(or 136 J)