Question

Starting from rest, a body slides down a $45$ degree 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 the inclined plane is:

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

Answer 1

Answer: (C)

$0.75$

When body slides down without friction ,

$mg\sin \theta =m{a}_1$

or $g\sin \theta =a_1$ .............................eq1

When body slides down with friction ,

$mg\sin \theta -f_k=m{a}_2$

but, frictional force $f_k=\mu mg\cos \theta$

Hence

$mg\sin \theta -\mu mg\cos \theta =m{a}_2$

or $g\sin \theta -\mu g\cos \theta =a_2$.............................eq2

Dividing eq1 by eq2 ,

$\frac{a_1}{a_2}=\frac{g\sin \theta }{g\sin \theta -\mu g\cos \theta }$

or $\frac{a_1}{a_2}=\frac{\sin \theta }{\sin \theta -\mu \cos \theta }$ ..........eq3

If time taken by body without friction is $t_1$ and with friction is $t_2$ , then distance covered by body is given by ,

$s=1/2a_1{t}_1^2=1/2a_2{t}_2^2$

or $\frac{a_1}{a_2}=\frac{t_2^2}{t_1^2}=2^2=4$ (because $t_2=2t_1$)

putting this value in eq3 ,

$4=\frac{\sin \theta }{\sin \theta -\mu \cos \theta }$

or $4=\frac{1/\sqrt{2}}{1/\sqrt{2}-\mu /\sqrt{2}}$ (as given $\theta ={45}^o$)

or $4(1-\mu )=1$

or $\mu =3/4=0.75$