Worked example 5.1: Bucket lifted from a well

Question: A man lifts a $30$kg bucket from a well whose depth is $150$m. Assuming that the man lifts the bucket at a constant rate, how much work does he perform?

Answer: Let $m$ be the mass of the bucket and $h$ the depth of the well. The gravitational force $f'$ acting on the bucket is of magnitude $m g$ and is directed vertically downwards. Hence, $f'=-m g$ (where upward is defined to be positive). The net upward displacement of the bucket is $h$. Hence, the work $W'$ performed by the gravitational force is the product of the (constant) force and the displacement of the bucket along the line of action of that force:

$$W' = f' h= - m g h.$$

Note that $W'$ is negative, which implies that the gravitational field surrounding the bucket gains energy as the bucket is lifted. In order to lift the bucket at a constant rate, the man must exert a force $f$ on the bucket which balances (and very slightly exceeds) the force due to gravity. Hence, $f=-f'$. It follows that the work $W$ done by the man is

$$W = f h = m g h = 30\times 150\times 9.81 = 4.415\times 10^4 {\rm J}.$$

Note that the work is positive, which implies that the man expends energy whilst lifting the bucket. Of course, since $W=-W'$, the energy expended by the man equals the energy gained by the gravitational field.