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Laws of Motion question

2011 · Q105
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Laws of Motion question

2011 · Q105

NEETPhysicsLaws of MotionMCQ+4 / −1
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 μ\muμ = 0.5. The distance that the box will move relativce to belt before coming to rest on it, taking g = 10 m s−-−2,, is
  1. A
    0.4 m
  2. B
    1.2 m
  3. C
    0.6 m
  4. D
    zero
View written solutionFree

Correct answer: A

Force of friction, f = μ\mu μmg

∴\therefore∴ a=fm=μmgm=μga = {f \over m} = {{\mu mg} \over m} = \mu ga=mf​=mμmg​=μg

=0.5×10=5 = 0.5 \times 10 = 5=0.5×10=5 ms-2

Using v2 – u2 = 2aS

02 – 22 = 2(–5) ×\times× S

⇒\Rightarrow⇒ S = 0.4 m

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