Question

A body at rest slides down a ${30}^o$ inclined plane. The time taken by it to slide down is twice the time it takes when it slides down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is

• A. $0.43$
• B. $0.37$
• C. $0.64$
• D. $0.75$

Answer 1

Answer: (A)

$0.43$

When a plane is inclined to the horizontal at an angle $\theta$, which is greater than the angle of repose, then the body placed on the inclined plane slides down with an acceleration $a$.
From figure,
$R=mg\cos \theta$ ....(i)
Net force on the body down the inclined plane
$f=mg\sin \theta -F$ ....(ii)
i.e., $f=ma=mg\sin \theta -\mu R$ $[\because \mu =\frac{F}{R}]$
$\therefore ma=mg\sin \theta -\mu mg\cos \theta$ [from equation (i)]
$=mg(\sin \theta -\mu \cos \theta )$
$\Rightarrow a=g(\sin \theta -\mu \cos \theta )$
Time taken by the body to slide down the plane
$t_1=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2s}{(g\sin \theta -\mu g\cos \theta )}}$
$t_2=\sqrt{\frac{2s}{g\sin \theta }}$ [in absence of friction]
$\because t_1=2t_2$
$\Rightarrow t_1^2=4t_2^2$ [given]
${\left|\sqrt{\frac{2s}{(g\sin \theta -\mu g\cos \theta )}}\right|}^2=4\times {\left|\sqrt{\frac{2s}{g\sin \theta }}\right|}^2$
$\Rightarrow \frac{2s}{g(\sin \theta -\mu \cos \theta )}=4\times \frac{2s}{g\sin \theta }$
$\Rightarrow \sin \theta =4\sin \theta -4\mu \cos \theta$
$\Rightarrow \mu =\frac{3}{4}\tan \theta =\frac{3}{4}\tan {30}^o=0.43$