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.

Explanation

The forces acting on the body are
(i) force of gravity (ii) air friction.
According to work energy theorem, total work done on the body `=` Gain in K.E.
`W=(1)/(2)m upsilon^(2)=(1)/(2)m (1.2sqrt(gH))^(2)=0.72mgH`
As work done by gravity `W_(1)=mgH`
`:.` Work done by friction , `W_(2)=W-W_(1)`
`=0.72mgH-mgH`