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 be

Question

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

TestHub · Accepted Answer

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

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