Question

A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance?

Answer 1

The work done on the mouse by air resistance is equal to change in mechanical energy of a mouse:

$$W=\frac{mv^2}{2}-mgh$$

$$W=\frac{0.2\times 8.0^2}{2}-0.2\times 9.8\times 100=-190\:\rm J$$