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JEE Main Physics question

PhysicsLaws of MotionJEE Main

A block of mass 2 kg is placed on the floor. The coefficient of static friction is 0.4. If a force of 2.8 N is applied on the block parallel to the floor, the force of friction between the block and the floor (taking g = 10 ms⁻²) is

Answer: (A)

Solution

Limiting friction = μmg = 0.4×2×10 = 8 N. Since the applied force (2.8 N) is less than the limiting

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Question ID VK-153131