Laws of Motion - Result Question 43

43. 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: 43. (d)

Solution:

  1. (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} $