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. 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
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. 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
Asked by Shiksha User
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1 Answer
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Let AC = l, BC = 2l, AB = 3l
Apply work-Energy theorem
W? + W? = ΔKE
Mg (3l)sinθ - µmgcosθ (l) = 0+0
µmgcosθl = 3mglsinθ
µ=3tanθ = ktanθ
∴ k=3
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