A small block starts slipping down from a point B on an inclined plane A B , which is making an angle θ…

A small block starts slipping down from a point B on an inclined plane AB, which is making an angle θ with the horizontal section BC is smooth and the remaining section CA is rough with a coefficient of friction μ. It is found that the block comes to rest as it reaches the bottom (point A) of the inclined plane. If BC=2AC, the coefficient of friction is given by μ=k tanθ. The value of k is .......

   

Solution

Let AC=    BC=2    AB=3

Apply work – Energy theorem

Wt+Wmg=ΔKE

mg(3)sinθ-μmgcosθ()=0+0

μmgcosθ=3mg/sinθ

μ=3tanθ=ktanθ

k=3

Asked in: JEE Main 2020 (02 Sep Shift 1)

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