Question

The time taken by block to sliding downward from the smooth surface of an angle ${45}^o$, the time taken by block kept on same type of rough surface is 'n' times. Then what is the coefficient of friction between block and surface?

• A. ${(1-\frac{1}{n^2})}^{\frac{1}{2}}$
• B. ${\left(\frac{1}{1-n^2}\right)}^{\frac{1}{2}}$
• C. $(1-\frac{1}{n^2})$
• D. $\left(\frac{1}{1-n^2}\right)$

Answer 1

The correct option is C $(1-\frac{1}{n^2})$

Given,

Time taken by block to slide on frictionless incline plane is $T$

Time taken by block to slide on rough incline plane is $nT$

Net acceleration on smooth incline plane, $a=g\sin \theta$

Apply kinematic equation

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

$s=0+\frac{1}{2}g\sin \theta \times T^2....\left(1\right)$

Net acceleration on rough inclined plane $a=g\sin \theta -\mu \cos \theta$

Apply kinematic Equation.

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

$s=0+\frac{1}{2}(g\sin \theta -\mu g\cos \theta )\times {\left(nT\right)}^2$

Put value of $s$ from equation $\left(1\right)$

$\frac{1}{2}g\sin \theta \times T^2=\frac{1}{2}(g\sin \theta -\mu g\cos \theta ){\left(nT\right)}^2$

$1=(1-\frac{\mu }{\tan {45}^o})n^2$

$\mu =(1-\frac{1}{n^2})$

Coefficient of friction $\mu =(1-\frac{1}{n^2})$