1.

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» 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.


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