Question:medium

The angle of inclination of an inclined plane is \(45^\circ\). If the coefficient of kinetic friction between the inclined plane and a block on it is \(0.2\), then the block released from rest from the top of the inclined plane reaches the bottom of the plane in a time of \(3\) s. If the coefficient of kinetic friction between the block and the inclined plane is \(0.8\), then the time taken by the same block to reach the bottom of the plane is

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For motion from rest over the same distance, \[ \boxed{t\propto\frac1{\sqrt a}}. \] On an inclined plane, \[ \boxed{a=g(\sin\theta-\mu\cos\theta).} \]
Updated On: Jul 18, 2026
  • \(6\) s
  • \(9\) s
  • \(7.5\) s
  • \(4.5\) s
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The Correct Option is A

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