A block rests on the horizontal floor of an elevator that is accelerating upwards at 2 m s ⁻² . The block is suddenly given a horizontal velocity of 6 m s ⁻¹ relative to the elevator floor. If the coefficient of kinetic friction between the block and the floor is 0.5 , the distance the block slides relative to the floor before coming to rest is : [Take g = 10 m s ⁻² ]

Step-by-step solution

In the non-inertial frame of the accelerating elevator, a pseudo force acts downwards on the block. The effective acceleration due to gravity is g' = g + a_e = 10 + 2 = 12 m s ⁻² . The normal force exerted by the floor on the block is N = m g' = 12m . The kinetic friction acting on the block is f_k = N = 0.5 12m = 6m . The horizontal deceleration of the block relative to the elevator is: a = f_k m = 6 m s ⁻² . Using the kinematic equation v^2 = u^2 - 2as with final velocity v = 0 relative to the elevator: 0 = (6)^2