Question

A block of mass m takes time t to slide down on a smooth inclined plane of angle of inclination $\theta$ and height h. If same block slides down on a rough inclined plane of same angle of inclination and same height and takes time n times of initial value, then coefficient friction between block and inclined plane is

• A. $[1+n^2]tan\theta$
• B. $[1-\frac{1}{n^2}]tan\theta$
• C. $[1-n^2]tan\theta$
• D. $[1+\frac{1}{n^2}]tan\theta$

Answer 1

The correct option is B $[1-\frac{1}{n^2}]tan\theta$

When an object is sliding down the incline plane of height $h$ and inclination $\theta$

then total length of the inclined plane is

$L=\frac{h}{\sin \theta }$

Also the acceleration on the smooth inclined plane will be

$a=g\sin \theta$

now we have time to slide down

$\begin{array}{c}t=\sqrt{\frac{2\times L}{a}}\\ t=\frac{1}{\sin \theta }\sqrt{\frac{2\times h}{g}}\end{array}$

now when it slide down the rough plane the acceleration will be less due to friction force

now in that case acceleration is given by

$a=\sin \theta -\mu g\cos \theta$

again by above formula

$t^{\prime }=\frac{1}{\sin \theta }\sqrt{\frac{2\times h}{g\left(1-\mu \tan \theta \right)}}$

given that the on rough plane is $n$ time more than the smooth plane.

$\begin{array}{c}{t}^{\prime }=nt\\ \frac{1}{\sin \theta }\sqrt{\frac{2\times h}{g\left(1-\mu \tan \theta \right)}}=n\times \frac{1}{\sin \theta }\sqrt{\frac{2\times h}{g}}\\ g\left(1-\mu \tan \theta \right)=\frac{g}{n^2}\end{array}$

so, friction coefficient is given by

$\mu =\frac{1}{\tan \theta }\left(1-\frac{1}{n^2}\right)$