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

To find the coefficient of kinetic friction between the object and the surface of the inclined plane, we need to compare the time taken for the object to slide down a rough and a smooth inclined plane of the same angle, \(45^\circ\).

1. Time to slide down a smooth inclined plane:

• For a smooth inclined plane, there is no friction. Therefore, the only force causing the object to accelerate is the component of gravitational force along the plane.
• The acceleration \(a\) down the plane is given by \(a = g \sin \theta\), where \(g\) is the acceleration due to gravity and \(\theta = 45^\circ\).
• Thus, \(a = g \sin 45^\circ = \frac{g}{\sqrt{2}}\).
• From physics, the time \(t_1\) to slide down a distance \(d\) is given by \(t_1 = \sqrt{\frac{2d}{a}}\). Substituting for \(a\), we have:

\(t_1 = \sqrt{\frac{2d}{g/\sqrt{2}}} = \sqrt{\frac{2d\sqrt{2}}{g}}\)

2. Time to slide down a rough inclined plane:

• On the rough inclined plane, friction acts opposite to the direction of motion, reducing the net acceleration.
• The acceleration \(a'\) is given by \(a' = g \sin \theta - \mu g \cos \theta\), where \(\mu\) is the coefficient of kinetic friction.
• So, \(a' = g \sin 45^\circ - \mu g \cos 45^\circ = \frac{g}{\sqrt{2}} - \frac{\mu g}{\sqrt{2}}\).
• The time \(t_2\) to slide down the same distance \(d\) is then:

\(t_2 = \sqrt{\frac{2d}{a'}} = \sqrt{\frac{2d\sqrt{2}}{g (1 - \mu)}}\)

3. Relating the times \(t_1\) and \(t_2\):

• According to the problem, the object takes \(n\) times the time to slide down the rough inclined plane compared to the smooth one: \(t_2 = n t_1\).
• Using the expressions for \(t_1\) and \(t_2\):

\(\sqrt{\frac{2d\sqrt{2}}{g (1 - \mu)}} = n \cdot \sqrt{\frac{2d\sqrt{2}}{g}}\)

4. Solving for \(\mu\):

• Squaring both sides to remove the square root gives:

\(\frac{2d\sqrt{2}}{g (1 - \mu)} = n^2 \cdot \frac{2d\sqrt{2}}{g}\)

• Simplifying, we get: \(1 - \mu = \frac{1}{n^2}\).
• Solving for \(\mu\):

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

Therefore, the correct option is \(1 - \frac{1}{n^2}\).

Answer 2

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:
\[\mu_k = 1 - \frac{1}{n^2}.\]