Question

An object is placed on the surface of a smooth inclined plane of inclination $\theta$. It takes time $t$ to reach the bottom. If the same object is allowed to slide down a rough inclined plane of same inclination $\theta$, it takes times nth to reach the bottom where $n$ number greater than $1$. The coefficient of friction $\mu$ is given by:

• A. $\mu =\tan \theta (1-1/n^2)$
• B. $\mu =\cot \theta (1-1/n^2)$
• C. ${\mu =\tan \theta (1-1/n^2)}^{1/2}$
• D. ${\mu =\cot \theta (1-1/n^2)}^{1/2}$

Answer 1

Answer: (A)

$\mu =\tan \theta (1-1/n^2)$

$\frac{V_a}{n}=\sqrt{2Lg(\sin \theta -\mu \cos \theta )}⟶\left(1\right)$

$V_a=\sqrt{2Lg\sin \theta }⟶\left(2\right)$

By putting equ(2) in eq(1) we get

$\frac{\sqrt{2Lg\sin \theta }}{n}=\sqrt{2Lg(\sin \theta -\mu \cos \theta )}$

$\frac{2Lg\sin \theta }{n^2}=2Lg(\sin \theta -\mu \cos \theta )$

$\frac{\sin \theta }{n^2}=\sin \theta -\mu \cos \theta$

$\mu \cos \theta =\sin \theta -\frac{\sin \theta }{n^2}$

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

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