The rear side of a truck is open and a box of $40 \;kg$ mass is placed $5 \,m$ away from the open end as shown in Figure. The coeffictent of friction between the box and the surface below it is $0.15 .$ On a stratght road, the truck starts from rest and accelerates with $2\; m s ^{-2} .$ At what distance from the starting point does the box fall off the truck? (Ignore the size of the box).

Mass of the box, $m=40 \,kg$

Coefficient of friction, $\mu=0.15$

Initial velocity, $u=0$

Acceleration, $a=2 \,m / s ^{2}$

Distance of the box from the end of the truck, $s^{\prime}=5\, m$

As per Newton's second law of motion, the force on the box caused by the accelerated motion of the truck is given by:

$F=m a$

$=40 \times 2=80 \,N$

As per Newton's third law of motion, a reaction force of $80 \,N$ is acting on the box in the backward direction. The backward motion of the box is opposed by the force of friction $f,$ acting between the box and the floor of the truck. This force is given by:

$f=\mu m g$

$=0.15 \times 40 \times 10=60\, N$

$\therefore$ Net force acting on the block:

$F_{\text {net }}=80-60=20\, N$ backward

The backward acceleration produced in the box is given by:

aback $\quad=\frac{F_{\text {mat }}}{m}=\frac{20}{40}=0.5 \,m / s ^{2}$

Using the second equation of motion, time $t$ can be calculated as:

$s^{\prime}=u t+\frac{1}{2} a_{ back } t^{2}$

$5=0+\frac{1}{2} \times 0.5 \times t^{2}$

$\therefore t=\sqrt{20} \,s$

Hence, the box will fall from the truck after $\sqrt{20}$ $s$ from start.

The distance $s$, travelled by the truck in $\sqrt{20} \,s$ is given by the relation:

$s=u t+\frac{1}{2} a t^{2}$

$=0+\frac{1}{2} \times 2 \times(\sqrt{20})^{2}$

$=20\, m$