Question

When a body slides down from rest along a smooth inclined plane making an angle of $45 ^ { \circ }$ with the horizontal, it takes time t. When the same body slides down from rest along a rough inclined plane making the same angle and through the same distance, it takes time pt, where p is same number greater than 1. The coefficient of friction between the body and the rough plane is :

• A. $\mu = 1 - \frac { 1 } { \mathrm { p } ^ { 2 } }$
• B. $\mu = \sqrt { 1 - \frac { 1 } { \mathrm { p } ^ { 2 } } }$
• C. $\mu = 1 - \mathrm { p } ^ { 2 }$
• D. $\mu = \sqrt { 1 - \mathrm { p } ^ { 2 } }$

Answer 1

Answer: (A)

$\mu = 1 - \frac { 1 } { \mathrm { p } ^ { 2 } }$

Answer 2

The acceleration of the body sliding down smooth inclined plan is

$a = g \sin \theta$ …..(i)

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

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

As the body slides a distance d in each case,

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

$\therefore \frac { a } { a ^ { \prime } } = \frac { t ^ { \prime 2 } } { t ^ { 2 } } = \frac { ( p t ) ^ { 2 } } { t ^ { 2 } } = p ^ { 2 } ( \because t = p t ($ Given $) )$ $\frac { g \sin \theta } { g ( \sin \theta - \mu \cos \theta ) } = p ^ { 2 } ( u \sin g ($ i and $($ ii $) )$ $\sin \theta = p ^ { 2 } \sin \theta - \mu p ^ { 2 } \cos \theta$ $\mu p ^ { 2 } \cos \theta = p ^ { 2 } \sin \theta - \sin \theta$

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

$\mu = \tan \theta \left( 1 - \frac { 1 } { p ^ { 2 } } \right)$

Here $\theta = 45 ^ { \circ }$

$\therefore \mu = \left( 1 - \frac { 1 } { p ^ { 2 } } \right)$