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 $a_{back}={\displaystyle \frac{F_{net}}{m}}$ to calculate the backward acceleration
Using the second equation of motion, calculate the time by-
$S=ut+{\displaystyle \frac{1}{2}}a{t}^2$
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, $s^{\prime }=5m$
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
$$\begin{gathered} F_{net}=F-f \\ {F}_{net}=80-60=20N \end{gathered}$$
Let the backward acceleration produced be $a_{back}$ and this can be calculated by
$$\begin{gathered} a_{back}={\displaystyle \frac{F_{net}}{m}} \\ \Rightarrow {\displaystyle \frac{20}{40}} \\ {a}_{back}=0.5m/s^2 \end{gathered}$$
Let the time be $t$ so, to calculate the time we will use the second equation of motion
$s^{\prime }=ut+{\displaystyle \frac{1}{2}}{a}_{back}{t}^2$
Putting the value in the above equation
$5=0+{\displaystyle \frac{1}{2}}\times 0.5\times t^2$
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-
$$\begin{gathered} s=ut+{\displaystyle \frac{1}{2}}a{t}^2 \\ \Rightarrow s=0+{\displaystyle \frac{1}{2}}\times 2\times (\sqrt{20}{)}^2 \\ \Rightarrow s=20m \end{gathered}$$
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.