Multiple choiceQuestion 1 · Recall
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By the end you can explain why the area under a force–extension graph equals the work done, and calculate that area for straight-line and multi-stage loading.
For a constant force, work done is easy:
where is the force and the distance moved in the direction of the force.
But when you stretch a spring, the force is not constant. At the start the spring is slack and you pull with almost no force. As it extends, the force you must apply grows.
Multiplying the final force by the total extension would use the biggest force for the whole journey, so it would overestimate the work.
We need a way to add up many small amounts of work, each done over a tiny extra extension where the force is roughly constant.
Imagine extending the object in tiny steps, each of width . During one small step the force is almost constant at , so the work done in that step is
On a force–extension graph, is the area of a thin vertical strip: height , width .
Add up all the strips from the starting extension to the finishing extension and you have the total work done. The strips fill the area under the graph.
Work done by a stretching force
The work done by the force stretching an object equals the area under the force–extension graph between the starting and finishing extensions.
The area has units of newtons × metres, so work is measured in joules (J).
Worked example: Worked example: a straight-line graph
A spring is stretched from its natural length to an extension of . The force rises steadily from to , so the graph is a straight line through the origin.
How much work is done stretching the spring?