Q 11-04-025JEE MainEAMCET 2010 (Engineering)Medium
A smooth block is released from rest on a $45°$ inclined plane and it slides a distance $d$. The time taken to slide is $n$ times that on a smooth inclined plane. The coefficient of friction is
Answer: (A) $\mu_k = 1 - \dfrac{1}{n^2}$
(Read as: on a rough $45°$ incline the time is $n$ times that on a smooth one.)
For the same distance from rest, $t \propto \dfrac{1}{\sqrt{a}}$, so $a_{smooth} = n^2 a_{rough}$:
$$g\sin 45° = n^2 g(\sin 45° - \mu_k\cos 45°) \;\Rightarrow\; 1 = n^2(1 - \mu_k)$$
$$\mu_k = 1 - \frac{1}{n^2}$$
Solution by Sreeraj P, M.Sc Physics