Question

Starting from rest a body slides down a ${45}^o$ inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and inclined plane is

• A. $0.33$
• B. $0.25$
• C. $0.75$
• D. $0.80$

Answer 1

Answer: (C)

$0.75$

The free body diagram for the sliding body is shown in the attached figure.

In absence of friction, $f=0$

$ma=mg\sin \theta$

$a=g\sin \theta$

Using second equation of motion,

$s=\frac{1}{2}a{t}^2$

$s=\frac{g\sin \theta t^2}{2}...........\left(i\right)$

In presence of friction,

$f=\mu N$

$=\mu mg\cos \theta$

$m{a}^{\prime }=mg\sin \theta -f$

$m{a}^{\prime }=mg\sin \theta -\mu mg\cos \theta$

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

Using second equation of motion,

$s=\frac{1}{2}{a}^{\prime }{t}^{\prime 2}$

$s=\frac{g(\sin \theta -\mu \cos \theta )(2t{)}^2}{2}...........\left(ii\right)$

Equating (i) & (ii),

$\frac{g\sin \theta t^2}{2}=\frac{4g(\sin \theta -\mu \cos \theta )t^2}{2}$

$\mu =\frac{3\tan \theta }{4}=0.75$