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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 $θ > T a n^{- 1} (μ)$ .Rate of work done by the force of friction at time ‘t’ is

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$μ m g^{2} t \sin θ$

$μ m g^{2} t \cos θ$

$m g^{2} t (\sin θ - μ \cos θ)$

$μ m g^{2} t \cos θ (\sin θ - μ \cos θ)$

answer is C.

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Detailed Solution

$a = g (\sin θ - μ \cos θ)$

$V = a t = g t (\sin θ - μ \cos θ)$

Force of friction $f = μ m g \cos θ$

Rate of work done by force of friction

$= f V = μ m g^{2} t \cos θ (\sin θ - μ \cos θ)$

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