Question

The rear side of a truck is open and a box of mass $40 \,kg$ is placed $5 \,m$ away from the open end. The coefficient of friction between the box and the surface below it is $0.15$. The truck starts from rest with an acceleration of $2 \,m$ $s^{-2}$ on a straight road. At what distance from the starting point does the box fall off the truck?

• A. $20\,m$
• B. $30\,m$
• C. $40\,m$
• D. $50\,m$

Answer 1

Answer: (A)

Here, Mass of the box, $M$ $=40 \,kg$
Acceleration of the truck, $a$ $=2\, m$ $s^{-2}$
Distance of the box from the rear end, $d=5\,m$.
Coefficient of friction between the box and the surface below it, $\mu$ $=0.15.$
The various forces acting on the block are as shown in the figure.
As the truck moves in forward direction with the acceleration $a$ $=2 \,m$ $s^{-2}$, the box experiences a force $F$ in the backward direction and it is given by
$F=Ma$ $=\left(40\, kg\right)$ $\left(2 \,m\,s^{-2}\right)$ $=80 \,N$ in backward direction. Under the action of this force, the box will tend to move toward the rear end of the truck. As it does so, its motion will be opposed by the force of friction which acts in the forward direction and it is given by
$f=\mu \, N$ $=\mu Mg=0.15\times40\times10$ $=60\, N$
The acceleration of the box relative to the truck toward the rear end is
$a_{1}$ $=\frac{F-f}{M}$ $=\frac{80 \,N-60 \,N}{40\, kg}$ $=0.5\,m$ $s^{-2}$
Let $t$ be the time taken for the box to fall off the truck.
$d=0\times t+\frac{1}{2}$ $a_{1}$ $t^{2}$ $\quad\left(\therefore \, u=0\quad\right)$
$5=\frac{1}{2}\times0.5\times$ $t^{2}$, $t=\sqrt{\frac{2\times5}{0.5}}=\sqrt{20}\,s$
During this time, the truck covers a distance $x$.
Using $S=ut+\frac{1}{2}$ $at^{2}$
We get, $x=0\times t+\frac{1}{2}\times2\times\left(\sqrt{20}\right)^{2} $ or $x=20 \,m$