Question 171 of 210

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 $\mu =ktan\theta$. The value of $k$ is .......

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Answer: 3

Let $AC=\ell ∴BC=2\ell ∴AB=3\ell$

Apply work – Energy theorem
${W}_{t}+{W}_{mg}=\Delta KE$
$mg(3\ell )\sin \theta -\mu mg\cos \theta (\ell )=0+0$
$\mu mgcos\theta \ell =3mg/\sin \theta$
$\mu =3\tan \theta =ktan\theta$
$k=3$