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JEE Main Physics question

PhysicsLaws of MotionJEE Main

You are designing an elevator for a hospital. The force exerted on a passenger by the floor of the elevator is not to exceed 1.60 times the passenger's weight. The elevator accelerates upward with constant acceleration for a distance of 3.0 m and then starts to slow down. What is the maximum speed (in ms⁻¹) of the elevator?

Answer: 6

Solution

Let N be the normal force of the floor on the passenger (mass m). With N = 1.60 mg as the limiting (maximum) case: N − mg = ma ⟹ a = 0.60g = 6 ms⁻². Using v²

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Question ID VK-153135