A conveyor belt is moving at a constant speed of 2 m s - 1 . A box is gently dropped on it. The coefficient of friction between them is μ = 0.5 . The distance that the box will move relative to the belt before coming to rest on it taking g = 10 m s - 2 , is

Step-by-step solution

Frictional force on the box f = μ m g ∴       Acceleration in the box a = μ g = 5   m   s - 2 v 2 = u 2 + 2 a s ⇒       0 = 2 2 + 2 × ( 5 ) s ⇒       s = - 2 5   m (with respect to the belt) ⇒    distance = 0 .4   m