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  4. 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 = 0.5. The distance that the box will move relative to belt before coming to rest on it, taking g = 10 ms-2, is

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Q. 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 $\mu = 0.5$. The distance that the box will move relative to belt before coming to rest on it, taking $g = 10\, ms^{-2}$, is

AIPMTAIPMT 2011Laws of Motion

Solution:

Force of friction, $f=\mu m g$
$\therefore \quad a=\frac{f}{m}=\frac{\mu m g}{m}$
$=\mu g=0.5 \times 10=5 m s ^{-2}$
Using $v^{2}-u^{2}=2 a S$
$0^{2}-2^{2}=2(-5) \times S$
$ \Rightarrow S=0.4\, m$