AIPMT MAINS 2011 Physics Friction MCQ Question
A conveyor belt is moving at a constant speed of 2 m/s. A box is gently dropped on it. The coefficient of friction between them is μ = 0.5. The distance that the box will move relative to belt before coming to rest on it, taking g = 10 m/s², is
Zero
0.4 m
1.2 m
0.6 m
Correct Answer
Detailed Explanation
The box will come to rest when the frictional force equals the deceleration required to stop it. Using the equation v² = u² + 2as, where u = 2 m/s, a = -μg = -5 m/s², and v = 0, we find s = 0.4 m.
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