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Question

A body dropped from a height H reaches the ground with a speed of 1.2 gH.Calculate the work doneby air-friction.


A

- 0.28 mgH

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B

-0.72 mgH

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C

-0.72 mgH

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D

1.2 mgH

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Solution

The correct option is A

- 0.28 mgH


The forces acting on the body are the forces of gravity and the air-friction.

By work-energy theorem, the total work done on the body is

W=12m(1.2×1.2×gH)0=0.72mgH

The work done by the force of gravity is mgH.

Hence, the work done by air-friction is

=0.72 mgH - mgH = - 0.28 mgH


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