Question

The backside of a truck is open and a box of $40kg$ is placed $5m$ away from the rear end. The coefficient of friction of the box with the surface of the truck is $\mathrm{0.15.}$ The truck starts from rest with $2m/s^2$ acceleration. The distance covered by the truck when the box falls off will be

• A. 20 m
• B. 30 m
• C. 40 m
• D. 50 m

Answer 1

Answer: (A)

Here,
Mass of the box, $M=40kg$
Acceleration of the truck, $a=2m{s}^{-2}$
Distance of the box from the rear end, $d=5m$
Coefficient of friction between the box and the surface below it,
$\mu =0.15$
The various forces acting on the block are as shown in the figure. As the truck moves in forward direction with the acceleration $a=2m{s}^{-2},$ the box experiences a force $F$ in the backward direction and it is given by
$F=Ma=(40kg)\left(2m{s}^{-2}\right)=80N$

in backward direction. Under the action of this force, the box will tend to move towards the rear end of the truck. As it does so, its motion will be opposed by the force of friction which acts in the forward direction and it is given by
$f=\mu N=\mu Mg=0.15\times 40\times 10=60N$
The acceleration of the box relative to the truck towards the rear end is
$a_1=\frac{F-f}{M}=\frac{80N-60N}{40kg}=0.5m{s}^{-1}$
Let $t$ be the time taken for the box to fall off the truck.
As $s=ut+\frac{1}{2}a{t}^2\therefore d=0\times t+\frac{1}{2}{a}_1{t}^2$ $(\because u=0)$
or $5=\frac{1}{2}\times 0.5\times t^2$ or $t=\sqrt{\frac{2\times 5}{0.5}}=\sqrt{20}s$
During this time, the truck covers a distance $x$.
$\therefore x=0\times t+\frac{1}{2}\times 2\times (\sqrt{20}{)}^2$
or $x=20m$