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NEET2011PhysicsLaws of MotionActual

A conveyor belt is moving at a constant speed of 2 ~m / s . 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=10 ~ms ⁻² , is

Options

  1. A1.2 ~m
  2. B0.6 ~m
  3. Czero
  4. D0.4 ~m

Correct answer

D. 0.4 ~m

Step-by-step solution

Force, F= m g Retardation of the block on the belt a= F m = m g m = g From, aligned v^2 & =u^2+2 a s 0 & =(2)^2-2( g) s s & = 4 2 0.5 10 =0.4 ~m aligned

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