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NEET 2011
Exam
NEET
Year
2011
Subject
Physics
Chapter
Laws of Motion

NEET 2011 · Physics · Laws of Motion

A conveyor belt is moving at a constant speed of 2 ~m / s . A box...

A conveyor belt is moving at a constant speed of 2 m/s2 \mathrm{~m} / \mathrm{s}. A box is gently dropped on it. The coefficient of friction between them is μ=0.5\mu=0.5. The distance that the box will move relative to belt before coming to rest on it taking g=10 ms−2g=10 \mathrm{~ms}^{-2}, isA conveyor belt is moving at a constant speed of 2 ~m / s . A box is gently dropped on it. The coefficient of frictionA conveyor belt is moving at a constant speed of 2 ~m / s . A box is gently dropped on it. The coefficient of friction

Options

  1. A1.2 m1.2 \mathrm{~m}
  2. B0.6 m0.6 \mathrm{~m}
  3. Czero
  4. D0.4 m0.4 \mathrm{~m}

Correct answer

D. 0.4 m0.4 \mathrm{~m}

Step-by-step solution

A conveyor belt is moving at a constant speed of 2 ~m / s . A box is gently dropped on it. The coefficient of friction
Force, F=μmg\quad F=\mu m g Retardation of the block on the belt a=Fm=μmgm=μ\frac{F}{m}=\frac{\mu m g}{m}=\mug From, v2=u2+2as0=(2)2−2(μg)ss=42×0.5×10=0.4 m\begin{aligned}v² & =u²+2 a s \\ 0 & =(2)^2-2(\mu g) s \\ s & =\frac{4}{2 \times 0.5 \times 10}=0.4 \mathrm{~m} \end{aligned}
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