Question

An object is placed on the surface of a smooth inclined plane of inclination $\theta$. It takes timer to reach the bottom. If the same object h allowed to slide down a rough inclined plane of inclination $\theta$, it takes time nt to reach the bottom, where n is a number greater than 1. The coefficient of friction $\mu$ is given by

• A. $\mu =tan\theta (1-\frac{1}{n^2})$
• B. $\mu =cos\theta (1-\frac{1}{n^2})$
• C. $\mu =tan\theta \sqrt{1-\frac{1}{n^2}}$
• D. $\mu =cos\theta \sqrt{1-\frac{1}{n^2}}$

Answer 1

The correct option is A $\mu =tan\theta (1-\frac{1}{n^2})$

For a body moving with constant acceleration, the kinematics equation is $s=ut+\frac{1}{2}a{t}^2$

If the initial speed is zero, then the time taken to reach a distance $s$ is $t=\sqrt{\frac{2s}{a}}$

i.e., $t\propto a^{-0.5}$

In the case of a smooth inclined plane, $a_1=g\sin \theta$

In the case of rough inclined plane, $a_2=g(\sin \theta -\mu \cos \theta )$

Time taken to travel down the smooth inclined plane is $t_1=t$

Time taken to travel down the smooth inclined plane is $t_2=nt$

$\frac{t_1}{t_2}=\sqrt{\frac{a_2}{a_1}}\Rightarrow \frac{t_1^2}{t_2^2}=\frac{a_2}{a_1}\Rightarrow \frac{1}{n^2}=\frac{\sin \theta -\mu \cos \theta }{\sin \theta }\Rightarrow \mu =\tan \theta (1-\frac{1}{n^2})$