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Question

A block is placed on an inclined plane of inclination θ. The angle of inclination is such that the block slides down the plane at a constant speed. The coefficient of kinetic friction between the block and the inclined plane is equal to


A

sin θ

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B

cos θ

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C

tan θ

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D

cot θ

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Solution

The correct option is C

tan θ



As the block moves down the inclined plane at a constant velocity, its acceleration will be zero. Hence, the net force acting on it will be zero.
mg sinθ=μNmg sinθ=μmg cos θμ=tan θ
Hence, the correct choice is (c).


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