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A body of mass 2kg is sliding down on an inclined plane of inclination $53^{0}$ to the horizontal. If coefficient of kinetic friction is 0.5, time taken by a body released from the top of an inclined plane to such the ground if the length of the inclined plane is $10 (g = 10 m s^{- 2}), (\sin 53^{0} = \frac{4}{5})$

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2 s

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answer is B.

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Detailed Solution

$t = \sqrt{\frac{2 l}{g (\sin θ - μ_{k} \cos θ)}}$

$t = 2 \sec$

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