Mass of the box, m = 40 kg

Coefficient of friction, μ = 0.15

Initial velocity, u = 0

Acceleration, a =$2 m/s^2$

Distance of the box from the end of the truck, s' = 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 = ma

= 40 × 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 = μmg

= 0.15 × 40 × 10 = 60 N

Net force acting on the block:

Fnet = 80 – 60 = 20 N backward

The backward acceleration produced in the box is given by:

${\mathrm{a}}_{\mathrm{back}} = \frac{{\mathrm{F}}_{\mathrm{net}}}{\mathrm{m}} = \frac{20}{40} = 0.5 \mathrm{m}/{\mathrm{s}}^2$

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

$s\text{'} = ut + \frac{1}{2} a_{back}{t}^2$
$5 = 0 + \frac{1}{2} \times 0.5 \times t^2$
$\Rightarrow t = \sqrt{20} \mathrm{s}$

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:

$\mathrm{s} = \mathrm{ut} + \frac{1}{2} {\mathrm{at}}^2$
$= 0 + \frac{1}{2} \times 2 \times {\left(\sqrt{20}\right)}^2 = 20 \mathrm{m}$