A conveyor belt is moving at a constant speed of 2 ms^-1. A box is gently dropped on it. The coefficient of friction between them is =0.5. The distance that the box will move relative to belt before…

Question

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

Accepted Answer

Correct answer: A. (a): Force of friction, f=μ m g a= fm= μ m gm=μ g=0.5×1 0=5 m s^−2 Using v^2−u^2=2 a S 0^2−2^2=2_(−5_)×S⇒S=0.4 m

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

Solution

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