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Q.

A block of mass ‘m’ slides down a rough inclined plane of inclination θ  with horizontal with zero initial velocity. The coefficient of friction between the block and the plane is μ  with θ>Tan1(μ)  .Rate of work done by the  force of friction at time ‘t’ is

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a

μmg2tsinθ

b

mg2t(sinθμcosθ)

c

μmg2tcosθ(sinθμcosθ)

d

μmg2tcosθ

answer is C.

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Detailed Solution

a=g(sinθμcosθ)

 V=at=gt(sinθμcosθ)

Force of friction    f=μmgcosθ

            Rate of work done by force of friction

            =fV=μmg2tcosθ(sinθμcosθ)

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A block of mass ‘m’ slides down a rough inclined plane of inclination θ  with horizontal with zero initial velocity. The coefficient of friction between the block and the plane is μ  with θ>Tan−1(μ)  .Rate of work done by the  force of friction at time ‘t’ is