Question

A body dropped from a height H reaches the ground with a speed of 1.2 `sqrt(gH)`. Calculate the work done by air friction.

Answer 1

The forces acting on te body are the force of gravit and the air friction. By work energy theorem, the total work done on the body is
`W=1/2m(1.2sqrt(gH))^2-0=0.72mgH`
The work done by the force of gravity is mgh. Hence, the work done by the air friction is
`0.72mgH-mgH=00.28mgH.