A suitcase is gently dropped on a conveyor belt moving at a velocity of 3 m s - 1 . If the coefficient of friction between the belt and the suitcase is 0 . 5 , the displacement of the suitcase relative to conveyor belt before the slipping between the two is stopped, is ( g = 10 m s - 2 )

Step-by-step solution

Acceleration of the suitcase till the slipping continues is a = f m a x m a = μ m g m = μ g = 0.5 × 10 = 5   m ​ s - 2 Slipping will continue until its velocity also becomes 3   m   s - 1 . ∴   v = u + a t or or t = 0 . 6   s In this time, the displacement of the suitcase s 1 = 1 2 a t 2 = 1 2 × 5 × 0.6 2 = 0. 9   m and the displacement of the belt, s 2 = v t = 3 × 0.6 = 1 .8   m Displacement of the suitcase with respect to the belt s 1 -