Q. A block moves down a smooth inclined plane of inclination $\theta$. Its velocity on reaching the bottom is $v$. If it slides down a rough inclined plane of same inclination, its velocity on reaching the bottom is $v/n$, where $n$ is a number greater than 1. The coefficient of friction is given by
A) $\mu = \tan{\theta\left( 1 - \frac{1}{n^{2}} \right)}$
B) $\mu = \cot{\theta\left( 1 - \frac{1}{n^{2}} \right)}$
C) $\mu = \tan{\theta\ \left( 1 - \frac{1}{n^{2}} \right)^{1/2}}$
D) $\mu = \cot{\theta\ \left( 1 - \frac{1}{n^{2}} \right)^{1/2}}$
Explanation
(a) For a smooth plane, $v = \sqrt{2gsin\ \theta.s}$ and for a rough plane,
$\frac{v}{n} = \sqrt{2g(\sin\theta - \mu\cos{\theta)}.s}$
$\therefore\ n = \sqrt{\frac{\sin\theta}{\sin\theta - \mu\cos\theta}}\ $ or $n^{2} = \frac{\sin\theta}{\sin\theta\ - \mu\cos\theta}$
$\Rightarrow \ \left( n^{2} - 1 \right)\sin\theta = n^{2}\mu\cos\theta$
or $\mu = \left( \frac{n^{2} - 1}{n^{2}} \right)\tan\theta = \tan{\theta\left( 1 - \frac{1}{n^{2}} \right)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.