Let AC=ℓ∴BC=2ℓ∴AB=3ℓ

Apply work – Energy theorem
Wt+Wmg=ΔKE
mg(3ℓ)sinθ−μmgcosθ(ℓ)=0+0
μmgcosθℓ=3mg/sinθ
μ=3tanθ=ktanθ
k=3
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=ktanθ. The value of k is .......

Held on 2 Sept 2020 · Verified 6 Jul 2026.
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