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**Hint**From the Newton’s Second Law, we get the expression for force which is-

$F = ma$

where, $F$ is the force, $m$ is the mass and $a$ is the acceleration.

Then, from Newton’s Third Law, we get

$f = \mu mg$(where, $\mu $ is the coefficient of friction and $g = 10$)

Calculate the net force by finding the difference between two forces.

Then, use the expression ${a_{back}} = \dfrac{{{F_{net}}}}{m}$ to calculate the backward acceleration

Using the second equation of motion, calculate the time by-

$S = ut + \dfrac{1}{2}a{t^2}$

Now, put the value of time and calculate the distance by again using the second equation of motion.

**Complete step by step answer:**

According to the question, it is given that

Coefficient of friction, $\mu = 0.15$

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

Initial Velocity, $u = 0$

Mass of the box, $m = 40kg$

Distance of box from one end, $s' = 5m$

According to the Newton’s second law of motion, the force on box due to the acceleration of truck is-

$F = ma$

Putting the values of mass and acceleration in the above equation

$F = 40 \times 2 = 80N$

Now, according to Newton’s third law, a force of $80N$ is acting on the box in a backward direction. This direction is opposed by frictional force $f$. This force acts between the box and floor of the truck. Thus, the force is-

$f = \mu mg$

Now, putting the values

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

$\therefore $The net force can now be calculated by

$

{F_{net}} = F - f \\

{F_{net}} = 80 - 60 = 20N \\

$

Let the backward acceleration produced be ${a_{back}}$ and this can be calculated by

$

{a_{back}} = \dfrac{{{F_{net}}}}{m} \\

\Rightarrow \dfrac{{20}}{{40}} \\

{a_{back}} = 0.5m/{s^2} \\

$

Let the time be $t$ so, to calculate the time we will use the second equation of motion

$s' = ut + \dfrac{1}{2}{a_{back}}{t^2}$

Putting the value in the above equation

$

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

\\

$

By further solving we get,

$t = \sqrt {20} s$

Now, we got the time when the box will fall from the truck. Therefore, we will again use second equation of motion to calculate the distance $s$ travelled by the truck-

$

s = ut + \dfrac{1}{2}a{t^2} \\

\Rightarrow s = 0 + \dfrac{1}{2} \times 2 \times {(\sqrt {20} )^2} \\

\Rightarrow s = 20m \\

$

Hence, the box will fall from the truck after reaching the $20m$ of destination.

**So, option (A) is correct.**

**Note**The Newton’s second law states that “force is equal to the product of mass and acceleration of a body”.

Newton's third law states that “every action has an equal and opposite reaction”.

As the truck starts off from the rest, take the initial velocity of the truck as zero.

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