Question

A block is moving downwards on a smooth inclined plane of angle of inclination θ. On reaching the bottom its velocity is v. If it slips on a rough inclined plane, then its velocity at the bottom is \(\frac{v}{n}\) where n is a number greater than zero. What will be coefficient of friction µ?

Answer 1

On smooth plane,

v² = 2(g sin θ).s ......(1)

On rough surface, (inclined plane)

\((\frac{v}{n})^2\) = 2(g sin θ - µg cos θ)s .......(2)

On equating the values of v₂ from equation (1) and (2)

2g sin θ s = 2n² (g sin θ - µg cos θ)s

or g sin θ = n² g sin θ - n² µg cos θ

or n²µg cos θ = n² g sin θ - g sin θ = g sin θ(n² - 1)

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

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