Laws Of Motion Ques 44
44. A conveyor belt is moving at a constant speed of $2 m / s$. 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
[2011M]
(a) $1.2 m$
(b) $0.6 m$
(c) zero
(d) $0.4 m$
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Answer:
Correct Answer: 44.(d)
Solution:
- (d) Frictional force on the box $f=\mu mg$
$\therefore$ Acceleration in the box, $a=\frac{f}{m}=\frac{\mu m g}{m}$
$
\begin{matrix}
& a=\mu g=5 ms^{-2} \\
& v^{2}=u^{2}+2 as \\
\Rightarrow \quad & 0=2^{2}+2 \times(5) s \\
\Rightarrow & s=-\frac{2}{5} \text{ w.r.t. belt } \\
\Rightarrow & \text{ distance }=0.4 m
\end{matrix}
$
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