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4-2.Friction
medium
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 ........... $m$
A
$0.4 $
B
$1.2 $
C
$0.6$
D
$0$
(AIPMT-2011)
Solution
$\begin{array}{l}
\,\,\,\,\,\,\,\,Force\,of\,friction,\,f = \mu mg\\
\therefore \,\,a = \frac{f}{m} = \frac{{\mu mg}}{m} = \mu g = 0.5 \times 10 = 5\,m{s^{ – 2}}\\
{\rm{Using}}\,{v^2} – {u^2} = 2as\\
\,\,\,\,\,\,\,\,\,\,{0^2} – {2^2} = 2\left( { – 5} \right) \times S \Rightarrow S = \,0.4\,m
\end{array}$
Standard 11
Physics
