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 $\mu = 0.5.$ The distance that the box will move relative to belt before coming to rest on it, taking $g = 10\, m s^{-2},$ is ........... $m$
AIPMT 2011, Medium
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Force of friction, $f=\mu m g$
$\therefore a=\frac{f}{m}=\frac{\mu m g}{m}=\mu g=0.5 \times 10=5 ms^{-2}$
Using $v^2-u^2=2 a s$
$0^2-2^2=2(-5) \times S \Rightarrow S=0.4 m$
$\therefore a=\frac{f}{m}=\frac{\mu m g}{m}=\mu g=0.5 \times 10=5 ms^{-2}$
Using $v^2-u^2=2 a s$
$0^2-2^2=2(-5) \times S \Rightarrow S=0.4 m$
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