A body of mass m dropped from a height h reaches the ground with a speed of 0 . 8 g h . The value of work done by the air-friction is:

K 1 + U 1 = K 2 + U 2 + W f W f : work done against friction 0 + m g h = 1 2 m v 2 + 0 + W f m g h = 1 2 m 0 . 64 g h + W f W f = 0.68 m g h Work done by friction W f ' = - W f = - 0.68 m g h

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PhysicsW.E.P.MiscellaneousEasy2 minQB

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A body of mass $m$ dropped from a height $h$ reaches the ground with a speed of $0.8 \sqrt{g h} .$ The value of work done by the air-friction is:

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Answer:

Solution:

$K_{1} + U_{1} = K_{2} + U_{2} + W_{f}$
$W_{f} :$ work done against friction
$0 + m g h = \frac{1}{2} m v^{2} + 0 + W_{f}$
$m g h = \frac{1}{2} m (0.64) g h + W_{f}$
$W_{f} = 0.68 m g h$
Work done by friction $W_{f}^{'} = - W_{f} = - 0.68 m g h$

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Miscellaneous

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