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 ______.
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 ______.
Updated On: Apr 26, 2026
- 3/4
- 2/3
- 5/9
- 4/9
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
For a block sliding down an incline of length $s$, the time taken is $t = \sqrt{2s/a}$. We compare the acceleration on a smooth plane ($a_1$) versus a rough plane ($a_2$).
Step 2: Key Formula or Approach:
1. Smooth plane: $a_1 = g \sin \theta$. 2. Rough plane: $a_2 = g (\sin \theta - \mu \cos \theta)$. 3. Time relation: $t_r = t_s + 0.5t_s = 1.5 t_s = \frac{3}{2} t_s$. 4. Since $s$ is the same, $t \propto 1/\sqrt{a}$, so $\frac{t_r}{t_s} = \sqrt{\frac{a_1}{a_2}}$.
Step 3: Detailed Explanation:
1. Square the time ratio: \[ \left( \frac{3}{2} \right)^2 = \frac{a_1}{a_2} \implies \frac{9}{4} = \frac{g \sin 45^{\circ}}{g (\sin 45^{\circ} - \mu \cos 45^{\circ})} \] 2. Since $\sin 45^{\circ} = \cos 45^{\circ} = 1/\sqrt{2}$, they cancel out: \[ \frac{9}{4} = \frac{1}{1 - \mu} \] 3. Solve for $\mu$: \[ 9(1 - \mu) = 4 \implies 9 - 9\mu = 4 \] \[ 9\mu = 5 \implies \mu = 5/9 \]
Step 4: Final Answer:
The coefficient of kinetic friction is 5/9.
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