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 determine the coefficient of kinetic friction (\(\mu\)) between an object and a surface, we will compare the time taken for the object to descend a rough inclined plane versus a smooth inclined plane of identical angle, \(45^\circ\).

1. Descent Time on a Smooth Inclined Plane:

• A smooth inclined plane lacks friction. Consequently, the object's acceleration is solely due to the component of gravitational force acting parallel to the plane.
• The acceleration (\(a\)) down this plane is calculated as \(a = g \sin \theta\), where \(g\) represents the acceleration due to gravity and \(\theta = 45^\circ\).
• Therefore, \(a = g \sin 45^\circ = \frac{g}{\sqrt{2}}\).
• Using kinematic equations, the time (\(t_1\)) to traverse a distance (\(d\)) is \(t_1 = \sqrt{\frac{2d}{a}}\). Substituting the value of \(a\) yields:

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

2. Descent Time on a Rough Inclined Plane:

• On a rough inclined plane, friction opposes the direction of motion, thus reducing the net acceleration.
• The acceleration (\(a'\)) on this surface is given by \(a' = g \sin \theta - \mu g \cos \theta\).
• For \(\theta = 45^\circ\), this becomes \(a' = g \sin 45^\circ - \mu g \cos 45^\circ = \frac{g}{\sqrt{2}} - \frac{\mu g}{\sqrt{2}}\).
• The time (\(t_2\)) to descend the same distance (\(d\)) is then:

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

3. Relationship Between \(t_1\) and \(t_2\):

• The problem states that the object takes \(n\) times longer to slide down the rough inclined plane than the smooth one, meaning \(t_2 = n t_1\).
• Substituting the derived expressions for \(t_1\) and \(t_2\):

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

4. Solution for \(\mu\):

• Squaring both sides of the equation eliminates the square roots:

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

• Simplifying this equation leads to \(1 - \mu = \frac{1}{n^2}\).
• Rearranging to solve for \(\mu\):

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

The coefficient of kinetic friction is determined to be \(1 - \frac{1}{n^2}\).