Question

The upper half of an inclined plane of inclination $\theta$ is perfectly smooth, while the lower half is rough. A block, starting from rest at the top of the plane, again comes to rest at the bottom. The coefficient of friction between the block and lower half of the plane is -

• A. $\mu =\tan \theta$
• B. $\mu =\frac{1}{\tan \theta }$
• C. $\mu =\frac{2}{\tan \theta }$
• D. $\mu =2\tan \theta$

Answer 1

The correct option is D $\mu =2\tan \theta$



From work energy theorem $(W=\Delta KE)$

As the block starts from rest and finally comes to rest, so $\Delta KE=0$, so that we can write,
$\text{Work done by friction +work done gravity}=0$

$\left(mg\sin \theta \right)\left(2s\right)-\left(\mu mg\cos \theta \right)\left(s\right)=0$

$2mg\sin \theta s=\mu mg\cos \theta s$

$\Rightarrow \mu =2\tan \theta$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.