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JEE Main Physics question

PhysicsLaws of MotionJEE Main

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 the belt before coming to rest on it, taking g = 10 ms⁻², is

Answer: (B)

Solution

The box accelerates due to friction: a = μg = 5 m/s². Time to reach belt speed (2 m/s): t = v/a = 2/5 = 0.4 s. Box's displacement in this time:

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Question ID VK-153439