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The graph shows the force along the path of a trolley.

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

The graph shows the force along the path of a trolley. The force rises in a straight line from $0\ \text{N}$ to $10\ \text{N}$ over the first $2\ \text{m}$ , stays at $10\ \text{N}$ from $2\ \text{m}$ to $6\ \text{m}$ , and falls to $0$ at $8\ \text{m}$ .

A graph of the force on a trolley against its position. The force rises in a straight line from 0 at position 0 to 10 newtons at 2 metres, stays at 10 newtons until 6 metres, and then falls in a straight line to 0 at 8 metres. The shape under the line is a trapezoid.

How much work does the force do over the $8\ \text{m}$ ?
The trolley has a mass of $2.5\ \text{kg}$ and starts from rest on a smooth track. How fast is it moving at $8\ \text{m}$ ?

Need a reminder? The key ideaShow

The work of a force that changes along the path is the area under its force-position graph, $W = \int F\,dx$ . For a spring, $W = \tfrac12 k(x_2^2 - x_1^2)$ .

Kinetic energy is $K = \tfrac12 mv^2$ . The work-energy theorem: the net work done on a body equals its change in kinetic energy, $W_{\text{net}} = \tfrac12 mv^2 - \tfrac12 mu^2$ .

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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