Question:


A small block starts slipping down from a point B on an inclined plane AB, which is making an angle \(\theta\) with the horizontal section BC is smooth and the remaining section CA is rough with a coefficient of friction \(\mu\). 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 \(\mu = k\tan\theta\). The value of k is ________.

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Apply the work-energy theorem from B to A: gravity acts over the whole length, friction acts only over CA.
Updated On: Nov 5, 2026
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Correct Answer: 3

Solution and Explanation

Let AC = x, so BC = 2x and AB = 3x along the incline. The block starts from rest at B and is again at rest at A.
Work done by gravity over AB \(= mg\sin\theta\times 3x\).
BC is smooth, so friction acts only on CA: work by friction \(= -\mu mg\cos\theta\times x\).
Work-energy theorem: \(3mgx\sin\theta - \mu mgx\cos\theta = 0\).
\(\mu = 3\tan\theta\), so \(k = 3\).
Answer: 3
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