Here, mass of the box, $m = 40$ kg;acceleration of the truck, a = 2 m s$^{-2}$;distance of the box from the open end, S = 5 m;and the coefficient of friction between the box and the surface below it, $\mu = 0.15$As the truck moves in forward direction with the acceleration $a = 2\ \mathrm{m}\,\mathrm{s}^{-2}$, the box experiences a force F in the opposite (backward) direction given by $F = ma = 40 \times 2 = 80\ \mathrm{N}$Under the action of this force, the box will tend to move to the open side of the truck. As it does so, its motion will be opposed by the force of friction. The limiting friction acting between the box and the truck, $f = \mu m g$ $= 0.15 \times 40 \times 10$ $= 60\ \mathrm{N}$The net force on the box in the backward direction,$F_1 = F - f = 80 - 60 = 20\ \mathrm{N}$Therefore, acceleration produced in the motion of the box in the backward direction, $a_1 = \frac{F_1}{M} = \frac{20}{40} = 0.5\ \mathrm{m}\,\mathrm{s}^{-2}$If $t$ is the time in which the box falls off the truck, then$S = u t + \frac{1}{2} a_1 t^2$ or $5 = 0 \times t + \frac{1}{2} \times 0.5 t^2$ or $t = \sqrt{\frac{5 \times 2}{0.5}} = 4.47\ \mathrm{s}$The distance covered by the truck (a = 2 m s$^{-2}$) in this time is given by $x = u t + \frac{1}{2} a t^2$ $= 0 \times 4.47 + \frac{1}{2} \times 2 \times (4.47)^2$ $= 19.98\ \mathrm{m}$