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 $\left(\text{point}A\right)$ of the inclined plane. If $BC=2AC$, the coefficient of friction is given by $\mu =k\tan \theta$. The value of $k$ is

Answer 1

If $AC=l$ then, according to question, $BC=2l$ and $AB=3l$.


According to the Work - Energy Theorem,

$W.D_{net}=\Delta K.E$

Here, the block starts from rest and comes to rest at the end.

Therefore, work done by all forces is zero.

$W_{mg}+W_{friction}=0$

$\Rightarrow mg\left(3l\right)\sin \theta -\mu mg\cos \theta \left(l\right)=0$

$\Rightarrow \mu mgl\cos \theta =3mgl\sin \theta$

$\Rightarrow \mu =3\tan \theta =k\tan \theta$

$\therefore k=3$

Hence, $3$ is the correct answer.