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Question

A cart of mass 20 kg starts from rest and accelerates uniformly along a straight horizontal track. If it reaches a speed of 6 m/s in 3 seconds, what is the work done on the cart during this time interval?

The correct answer is

360 J

By the work–energy theorem, the net work done on a body equals the change in its kinetic energy: \(W = \frac{1}{2}mv^2 - \frac{1}{2}mu^2\).

Here \(m = 20\) kg, the initial speed \(u = 0\) (starts from rest) and the final speed \(v = 6\) m/s, so \(W = \frac{1}{2} \times 20 \times 6^2 - 0\).

So \(W = 10 \times 36 = 360\) J of work is done on the cart in the 3 s interval.

Check: the acceleration is \(a = \frac{6}{3} = 2\) m/s2, so the force is \(F = 20 \times 2 = 40\) N and the distance is \(s = \frac{1}{2} \times 2 \times 3^2 = 9\) m; then \(W = Fs = 40 \times 9 = 360\) J, the same value.

Hence, 360 J.

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