Question

The rear side of a truck is open and a box of $40kg$ mass is placed $5m$ away from the open end as shown in the figure. The coefficient of friction between the box and the surface below it is $0.15$. On a straight road, the truck starts from rest and accelerates with $2{ms}^{-2}$. Find the distance (in $m$) traveled by truck by the time box falls from the truck. (Ignore the size of the box)

Answer 1

Given: the rear side of a truck is open and a box of 40kg mass is placed 5m away from the open end as shown in the figure. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with $2m/s^2$

To find the distance (in m) traveled by truck by the time box falls from the truck. (Ignore the size of the box)

Solution:

Mass of the box, $m=40kg$

$\mu =0.15u=a=2m/s^{2}S=5mF=ma⟹F=40\times 2=80N$

This is due to acceleration of truck.

$f=\mu mg⟹f=0.15\times 40\times 10=60N$

Therefore, net force acting on the block

$F_{net}=80-60=20N$ backward

Backward acceleration produce is

$a_{back}=\frac{F_{net}}{m}=\frac{20}{40}=0.5m/s^2$

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

$S^{\prime }=ut+\frac{1}{2}{a}_{back}{t}^2⟹S=0+\frac{1}{2}\times 0.5\times t^2\therefore t=\sqrt{20}S$

Distance S, travelled by the truck in $\sqrt{20}S$

$S=ut+\frac{1}{2}a{t}^2⟹S=0+\frac{1}{2}\times 2\times (\sqrt{20}{)}^2⟹S=20m$

is the distance (in m) traveled by truck by the time box falls from the truck.