Question

The upper half of an inclined plane with inclination $\theta$ is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom, then the coefficient of kinetic friction for the lower half is given by

• A. $\text{2 tan}\theta$
• B. $\text{tan}\theta$
• C. $\text{2 sin}\theta$
• D. $\text{2 cos}\theta$

Answer 1

The correct option is A $\text{2 tan}\theta$


For first half,
acceleration $=g\sin \theta$
$\therefore v^2=2\left(g\sin \theta \right)l.......\left(1\right)$

For second half,
acceleration $=g(\sin \theta -{\mu }_k\cos \theta )$
$0=v^2+2g(\sin \theta -{\mu }_k\cos \theta )l....\left(2\right)$

Putting equation $\left(1\right)$ in equation $\left(2\right)$, we get:

$0=\left(2g\sin \theta \right)l+\left(2g\sin \theta \right)l-\left(2g{\mu }_k\cos \theta \right)l$
$\therefore {\mu }_k\cos \theta =2\sin \theta$
or ${\mu }_k=2\tan \theta$

Hence, the correct answer is option (a).