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Question

A body dropped from a height H reaches the ground with a speed of 1.2gH. Then the work done by air-friction is ---


A
- 0.28 mgH
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B
-9.8mgH
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C
9.8mgH
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D
Zero
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Solution

The correct option is C - 0.28 mgH
The forces acting on the body are the force of gravity and air-friction. By work-energy theorem, the total work done on the body is equal to change in kinetic energy,
W=12m(1.2gH)20=0.72 mgH
The work done by the force of gravity is mgH. Hence the work done by air-friction is
0.72mgHmgH=0.28mgH.

Physics

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