Question

A given object takes \(n\) times the time to slide down \(45^\circ\) rough inclined plane as it takes the time to slide down an identical perfectly smooth \(45^\circ\) inclined plane. The coefficient of kinetic friction between the object and the surface of inclined plane is:

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

Answer 1

The Correct Option is A

For the smooth inclined plane, the acceleration is:
\[a_{\text{smooth}} = g \sin 45^\circ = \frac{g}{\sqrt{2}}.\]
For the rough inclined plane, the acceleration is:
\[a_{\text{rough}} = g (\sin 45^\circ - \mu_k \cos 45^\circ) = \frac{g}{\sqrt{2}} (1 - \mu_k).\]
The time taken is inversely proportional to the square root of acceleration:
\[t_{\text{rough}} = n \cdot t_{\text{smooth}} \implies \sqrt{\frac{a_{\text{smooth}}}{a_{\text{rough}}}} = n.\]
Substituting:
\[\sqrt{\frac{\frac{g}{\sqrt{2}}}{\frac{g}{\sqrt{2}} (1 - \mu_k)}} = n.\]
Simplify:
\[\sqrt{\frac{1}{1 - \mu_k}} = n \implies 1 - \mu_k = \frac{1}{n^2}.\]
Solving for \(\mu_k\):
\[\mu_k = 1 - \frac{1}{n^2}.\]
Thus, the coefficient of kinetic friction is:
\[\mu_k = 1 - \frac{1}{n^2}.\]