Question

When a body slides down from rest along a smooth inclined plane making an angle of ${30}^0$ with the horizontal, it takes time 20 s. When the same body slides down from rest along a rough inclined plane making the same distance, it takes time 20 p s, where p is some number greater than 1. The coefficient of friction between the body and the rough plane is

• A. $\mu =(1-\frac{1}{p^2})\frac{1}{\sqrt{3}}$
• B. $\mu =\sqrt{1-\frac{1}{9p^2}}$
• C. $\mu =(1-p^2)\frac{1}{\sqrt{3}}$
• D. $\mu =\sqrt{1-9p^2}$

Answer 1

The correct option is A $\mu =(1-\frac{1}{p^2})\frac{1}{\sqrt{3}}$

The acceleration of the body sliding down a smooth inclined plane is a=gsin$\theta$ ............(i)

The acceleration of the same body sliding down a rough inclined plane is

$a^{\prime }=gsin\theta -\mu cos\theta =g(sin\theta -\mu cos\theta )$ ............(ii)

As the body sildes a distance d in each case,

$\therefore d=\frac{1}{2}a{t}^2=\frac{1}{2}{a}^{\prime }{t}^{\prime 2}$ (u=0)

$\therefore \frac{a}{a^{\prime }}=\frac{t^{\prime 2}}{t^2}=p^2$ (t'=20ps, t=20s)

$\frac{gsin\theta }{g(sin\theta -\mu cos\theta )}=p^2$ (using (i) and (ii))

$sin\theta =p^{2}sin\theta -\mu p^{2}cos\theta ;\mu p^{2}cos\theta =p^{2}sin\theta -sin\theta$

$\mu cos\theta =\frac{(p^2-1)}{p^2}sin\theta ;\mu =tan\theta (1-\frac{1}{p^2})$

Here, $\theta ={30}^0.\therefore \mu =(1-\frac{1}{p^2})\frac{1}{\sqrt{3}}$