Solve this following


A small block starts slipping down from a point $\mathrm{B}$ on an inclined plane $\mathrm{AB}$, 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 $\mathrm{A}$ ) of the inclined plane. If $\mathrm{BC}=2 \mathrm{AC}$, the coefficient of friction is given by $\mu=\mathrm{k} \tan \theta$ ). The value of $\mathrm{k}$ is



Apply work energy theorem

$m g \sin \theta(A C+2 A C)-\mu m g \cos \theta A C=0$

$\mu=3 \tan \theta$


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