Question

The time taken by a block of mass \(m\) to slide down from the highest point to the lowest point on a rough inclined plane is 50 % more compared to the time taken by the same block on identical inclined smooth plane. Both inclined planes are at 45° with the horizontal. The coefficient of kinetic friction between the rough inclined surface and block is ______.

• A. 3/4
• B. 2/3
• C. 5/9
• D. 4/9

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
When a block slides down an inclined plane, its acceleration depends on gravity, the angle of inclination, and friction (if the plane is rough).
Since the distance covered is the same for both smooth and rough planes, we can use kinematic equations to relate the time of descent to the respective accelerations.
Step 2: Key Formula or Approach:
For a smooth inclined plane, the acceleration is \(a_1 = g \sin \theta\).
For a rough inclined plane, the acceleration is \(a_2 = g(\sin \theta - \mu_k \cos \theta)\).
Using the kinematic equation \(S = \frac{1}{2} a t^2\), we can express time \(t\) as \(t = \sqrt{\frac{2S}{a}}\).
This means time \(t\) is inversely proportional to the square root of acceleration: \(t \propto \frac{1}{\sqrt{a}}\).
Step 3: Detailed Explanation:
Let \(t_1\) be the time taken on the smooth plane and \(t_2\) be the time on the rough plane.
The problem states that \(t_2\) is 50% more than \(t_1\), which means:
\(t_2 = t_1 + 0.5 t_1 = 1.5 t_1 = \frac{3}{2} t_1\).
Squaring both sides gives:
\(t_2^2 = \frac{9}{4} t_1^2\).
Using the relation \(S = \frac{1}{2} a t^2\), substitute \(t^2 = \frac{2S}{a}\):
\(\frac{2S}{a_2} = \frac{9}{4} \left( \frac{2S}{a_1} \right)\).
Canceling \(2S\) from both sides gives the ratio of accelerations:
\(\frac{1}{a_2} = \frac{9}{4 a_1} \implies 4a_1 = 9a_2\).
Substitute the expressions for \(a_1\) and \(a_2\):
\(4(g \sin \theta) = 9[g(\sin \theta - \mu_k \cos \theta)]\).
Cancel \(g\) from both sides and substitute \(\theta = 45^\circ\):
\(4 \sin 45^\circ = 9(\sin 45^\circ - \mu_k \cos 45^\circ)\).
Since \(\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}\), we can divide the entire equation by \(\frac{1}{\sqrt{2}}\):
\(4(1) = 9(1 - \mu_k)\).
Step 4: Final Answer:
Solve for \(\mu_k\):
\(4 = 9 - 9\mu_k \implies 9\mu_k = 5 \implies \mu_k = \frac{5}{9}\).