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 ______.
Numerical value type. Enter your answer.
Answer: 3
Let $AC = l$, so $AB = 3l$. The block starts and ends at rest, so by the work-energy theorem the work done by gravity equals the work done against friction:
$$mg\sin\theta\,(3l) = \mu mg\cos\theta\,(l) \;\Rightarrow\; \mu = 3\tan\theta$$
So $k = 3$.
Solution by Sreeraj P, M.Sc Physics