Question

There are two inclined surfaces of equal length inclined at an angle of \(45^\circ\) with the horizontal. One of them is rough and the other is perfectly smooth. A given body takes 2 times as much time to slide down on the rough surface than on the smooth surface. The coefficient of kinetic friction (\(\mu_k\)) between the object and the rough surface is close to :

• A. \( 0.80 \)
• B. \( 0.25 \)
• C. \( 0.75 \)
• D. \( 0.5 \)

Hint:

Use the equations of motion for constant acceleration. The acceleration down the incline depends on gravity and friction. Relate the times of descent using the given factor and solve for the coefficient of kinetic friction.

Answer 1

The Correct Option is C

To solve this problem, we first analyze the forces acting on the body sliding down each incline.

1. Smooth Surface:

Since the surface is smooth, there is no friction. The only forces are the gravitational force parallel to the incline and the normal force.

Gravitational acceleration parallel to the incline: \( g\sin\theta \) where \( \theta= 45^\circ \)

Thus, the acceleration is:

\( a_1 = g\sin\theta = \frac{g}{\sqrt{2}} \)

The formula for time of descent \( t \) on an incline of length \( l \) under constant acceleration \( a \) is given by:

\( t = \sqrt{\frac{2l}{a}} \)

Time taken on smooth surface \( t_1 = \sqrt{\frac{2l}{a_1}} = \sqrt{\frac{2l\sqrt{2}}{g}} \)

2. Rough Surface:

The forces include gravitational force, normal force, and frictional force.

The frictional force \( f_k = \mu_k N = \mu_k mg\cos\theta \)

Net force down the incline: \( mg\sin\theta - f_k = mg\sin\theta - \mu_k mg\cos\theta \)

Net acceleration:

\( a_2 = g(\sin\theta - \mu_k \cos\theta) = g\left(\frac{1}{\sqrt{2}} - \mu_k \frac{1}{\sqrt{2}}\right) \)

\( a_2 = \frac{g}{\sqrt{2}}(1 - \mu_k) \)

Time taken on the rough surface \( t_2 = \sqrt{\frac{2l}{a_2}} = \sqrt{\frac{2l\sqrt{2}}{g(1-\mu_k)}} \)

Condition Given:

Time on rough surface is twice time on smooth surface:

\( t_2 = 2t_1 \)

\( \sqrt{\frac{2l\sqrt{2}}{g(1-\mu_k)}} = 2\sqrt{\frac{2l\sqrt{2}}{g}} \)

Squaring both sides:

\( \frac{2l\sqrt{2}}{g(1-\mu_k)} = 4 \cdot \frac{2l\sqrt{2}}{g} \)

Canceling common terms and simplifying:

\( \frac{1}{1-\mu_k} = 4 \)

\( 1 = 4(1-\mu_k) \)

\( 1 = 4 - 4\mu_k \)

\( 4\mu_k = 3 \)

\( \mu_k = \frac{3}{4} = 0.75 \)

Thus, the coefficient of kinetic friction is approximately \( \mu_k = 0.75 \).