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JEE Main Physics question

PhysicsLaws of MotionJEE Main

A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab. The static coefficient of friction between the block and slab is 0.60, while the kinetic coefficient is 0.40. The 10 kg block is acted upon by a horizontal force of 100 N. If g = 9.8 ms⁻², the resulting acceleration of the slab will be

Answer: (D)

Solution

Maximum static friction between block and slab = 0.6×10×9.8 = 58.8 N, less than the applied 100 N, so the block

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Question ID VK-153014