Question

The rear side of a truck is open and a box of mass $20kg$ is placed on the truck $4m$ away from the open end. The coefficient of friction between the box and the surface is $0.15$. The truck starts from rest with an acceleration of $2m{s}^{-2}$ on a straight road. The box will fall off the truck when it is at a distance from the starting point equal to

• A. $4m$
• B. $8m$
• C. $16m$
• D. $32m$

Answer 1

Answer: (C)

Maximum acceleration of box = $\mu$g
$=0.15\times 10m{s}^{-2}=1.5m{s}^{-2}$
Acceleration of truck, $a_T=2m{s}^{-2}$
Therefore, relative acceleration of the box,
$a_r=0.5m{s}^{-2}$ (backwards).
It will fall off the truck in a time,
$t=\sqrt{\frac{2S}{a_r}}=\sqrt{\frac{2\times 4}{0.5}}=4s$
Displacement of truck upto the instant is,
$S_T=\frac{1}{2}{a}_T{t}^2=\frac{1}{2}\times 2\times {\left(4\right)}^2=16m$