Question

The rear side of the truck is open and a box of $4kg$ mass is placed $5m$ away from the open end. Given that, the coefficient of friction between the box and surface below it is $0.15$. The truck starts from rest and accelerates with $2m{s^{ - 2}}$. What will be the distance (in m) travelled by the truck by the time box falls off the truck.
(A) $20m$
(B) $30m$
(C) $40m$
(D) $50m$

Answer 1

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.