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 resolve this problem, we first evaluate the forces acting on an object descending each incline.

1. Smooth Surface:

As the surface is smooth, friction is absent. The operative forces are gravity, acting parallel to the incline, and the normal force.

The component of gravitational acceleration parallel to the incline is \( g\sin\theta \), with \( \theta= 45^\circ \).

Consequently, the acceleration is:

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

The equation for the time of descent \( t \) over an incline of length \( l \) under uniform acceleration \( a \) is:

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

The time of descent on the smooth surface is \( t_1 = \sqrt{\frac{2l}{a_1}} = \sqrt{\frac{2l\sqrt{2}}{g}} \).

2. Rough Surface:

The forces involved are gravity, the normal force, and friction.

The frictional force is calculated as \( f_k = \mu_k N = \mu_k mg\cos\theta \).

The net force acting down the incline is \( mg\sin\theta - f_k = mg\sin\theta - \mu_k mg\cos\theta \).

The net acceleration is:

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

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

The time of descent on the rough surface is \( t_2 = \sqrt{\frac{2l}{a_2}} = \sqrt{\frac{2l\sqrt{2}}{g(1-\mu_k)}} \).

Condition Provided:

The time on the rough surface is twice the time on the smooth surface:

\( t_2 = 2t_1 \)

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

Upon squaring both sides:

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

After canceling common terms and simplifying:

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

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

\( 1 = 4 - 4\mu_k \)

\( 4\mu_k = 3 \)

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

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