Question

A body placed on the surface of a smooth inclined plane reaches the bottom of the plane in time $t$. When it is identically placed on a rough inclined plane, it reaches the bottom in time $nt$ where $n>1$. In both the cases, the angle of inclination $\theta$ is same. The coefficient of friction $\mu$ is equal to

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

Answer 1

Answer: (C)

$mg\sin \theta =ma$ or $a=g\sin \theta$
$s=\frac{1}{2}(g\sin \theta )t^2;t={\left(\frac{2s}{g\sin \theta }\right)}^{\frac{1}{2}}$

$N=mg\cos \theta ;mg\sin \theta -\mu mg\cos \theta =ma$
$\Rightarrow a=g\sin \theta -\mu g\cos \theta$
$s=\frac{1}{2}(g(\sin \theta -\mu \cos \theta ))(nt{)}^2\Rightarrow nt=\left[\frac{2s}{g(\sin \theta -\mu \cos \theta )}\right]$
$\Rightarrow n{\left(\frac{2s}{g\sin \theta }\right)}^{\frac{1}{2}}={\left(\frac{2s}{g(\sin \theta -\mu \cos \theta )}\right)}^{\frac{1}{2}}$
$\Rightarrow \frac{n^2}{\sin \theta }=\frac{1}{\sin \theta -\mu \cos \theta }$
$n^2\sin \theta -\mu n^2\cos \theta =\sin \theta$
$\left(n^2-1\right)\sin \theta =\mu n^2\cos \theta$
$\mu =\left(\frac{n^2-1}{n^2}\right)\cdot \tan \theta \Rightarrow \mu =\left(1-\frac{1}{n^2}\right)\cdot \tan \theta$