Q. A given object takes $n$ times more time to slide down a $45{^\circ}$ rough inclined plane as it takes so slide down a perfectly smooth 45${^\circ}$ incline. The coefficient of kinetic friction between the object and the incline is
A) $\frac{1}{1 - n^{2}}$
B) $1 - \frac{1}{n^{2}}$
C) $\sqrt{1 - \frac{1}{n^{2}}}$
D) $\sqrt{\frac{1}{1 - n^{2}}}$
Explanation
(b) From $s = ut + \frac{1}{2}at^{2} = 0 + \frac{1}{2}at^{2}$
$t = \sqrt{\frac{2s}{a}}$
For smooth plane, $a = g\sin\theta$
For rough plane, $a^{'} = g\ {(\sin{\theta - \mu\cos\theta})}$
$\therefore\ \ t^{'} = \sqrt{\frac{2s}{g\ (\sin{\theta - \mu\cos\theta})}} = nt = n\sqrt{\frac{2s}{g\sin\theta}}$
$\therefore\ n^{2}g\left( \sin{\theta -}\mu\cos\theta \right) = g\sin\theta$
When $\theta = 45{^\circ},\sin{\theta\mathbf{=}\cos{\theta = 1/\sqrt{2}}}$
Solving we get $\mu = 1 - \frac{1}{n^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.