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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 ___________.
Correct Answer: 3
Let $$AC = x \implies BC = 2x$$
Total length of incline $$AB$$: $$s = BC + AC = 2x + x = 3x$$
Work-energy theorem between point B and point A:
$$W_{\text{gravity}} + W_{\text{friction}} = \Delta K$$
$$(mg\sin\theta)s - (\mu mg\cos\theta)AC = K_A - K_B$$
$$mg\sin\theta(3x) - \mu mg\cos\theta(x) = 0$$
$$3mgx\sin\theta = \mu mgx\cos\theta \implies 3\sin\theta = \mu\cos\theta$$
$$\mu = 3\tan\theta$$
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