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 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 $2 \;m s^{-2}$. At what distance from the starting point does the box fall off the truck? (Ignore the size of the box).

(C) Mass of the box,$m = 40 \;kg$.
Coefficient of friction,$\mu = 0.15$.
Initial velocity,$u = 0$.
Acceleration of the truck,$a = 2 \;m s^{-2}$.
Distance of the box from the end of the truck,$s' = 5 \;m$.
In the frame of reference of the truck,the box experiences a pseudo force $F_p = ma$ in the backward direction.
$F_p = 40 \times 2 = 80 \;N$.
The limiting friction force acting in the forward direction is $f = \mu mg = 0.15 \times 40 \times 10 = 60 \;N$.
The net force on the box in the backward direction is $F_{net} = F_p - f = 80 - 60 = 20 \;N$.
The backward acceleration of the box relative to the truck is $a_{rel} = \frac{F_{net}}{m} = \frac{20}{40} = 0.5 \;m s^{-2}$.
Using the equation of motion $s' = u t + \frac{1}{2} a_{rel} t^2$:
$5 = 0 + \frac{1}{2} \times 0.5 \times t^2$
$5 = 0.25 \times t^2$
$t^2 = 20$
$t = \sqrt{20} \;s$.
The distance $s$ traveled by the truck in time $t$ is:
$s = ut + \frac{1}{2} a t^2$
$s = 0 + \frac{1}{2} \times 2 \times (\sqrt{20})^2$
$s = 20 \;m$.