The rear side of a truck is open and a box of 40 kg mass
Question.
The rear side of a truck is open and a box of $40 \mathrm{~kg}$ mass is placed $5 \mathrm{~m}$ away from the open end as shown in Fig. $5.22$. The coefficient of friction between the box and the surface below it is $0.15$. On a straight road, the truck starts from rest and accelerates with $2 \mathrm{~m} \mathrm{~s}^{-2}$. At what distance from the starting point does the box fall off the truck? (Ignore the size of the box). solution:

Mass of the box, m = 40 kg

Initial velocity, $u=0$

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

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

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 = ma

= 40 × 2 = 80 N

As per Newton’s third law of motion, a reaction force of $80 \mathrm{~N}$ 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 × 40 × 10 = 60 N

$\therefore$ Net force acting on the block:

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

The backward acceleration produced in the box is given by:

$a_{\text {back }}=\frac{F_{\text {tet }}}{m}=\frac{20}{40}=0.5 \mathrm{~m} / \mathrm{s}^{2}$

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

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

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

$\therefore t=\sqrt{20} \mathrm{~s}$

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

The distance $s$, travelled by the truck in $\sqrt{20} \mathrm{~s}$ 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