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  4. The rear side of a truck is open and a box of mass 20 kg is placed on the truck 4 m 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 2 m 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

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Q. The rear side of a truck is open and a box of mass $20\, kg$ is placed on the truck $4\, m$ 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 $2\, m\, s^{-2}$ on a straight road. The box will fall off the truck when it is at a distance from the starting point equal to

AIIMSAIIMS 2018Laws of Motion

Solution:

Maximum acceleration of box = $\mu$g
$= 0.15 × 10\, m \,s^{-2} = 1.5\, m\, s^{-2}$
Acceleration of truck, $a_T = 2\, m\, s^{-2}$
Therefore, relative acceleration of the box,
$a_r = 0.5\, m\, s^{-2}$ (backwards).
It will fall off the truck in a time,
$t=\sqrt{\frac{2S}{a_{r}}}=\sqrt{\frac{2\times4}{0.5}}=4\,s$
Displacement of truck upto the instant is,
$S_{T}=\frac{1}{2}a_{T}t^{2}=\frac{1}{2}\times2\times\left(4\right)^{2}=16\,m$