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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. |
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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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