The rear side of a truck is open and a box of $40 kg$ mass is placed $5 m$ away from the open end. The coefficient of friction between the box and the surface is $0.15$ . On a straight road, the truck starts from rest and accelerates with $2 m/s^2$ . At what distance from the starting point does the box fall off the truck ? Ignore the size of the box.

Because of the acceleration of the truck the pseudo force on the box $=m\times a=40\times 2=80 N$ .

This force acts opposite to the acceleration of the truck.

The frictional force on the truck which acts in forward direction,

$f_k=\mu N =0.15 \times 40 g=58.8 N$ .

Since pseudo force is greater than frictional force, so block will accelerate in backward direction relative to truck with a magnitude.

$a=\frac{80-58.8}{40}=0.53 m/s^2 .$

The time taken by box to cover the distance $5 m$

                         $s=0+\frac{1}{2}{at}^2$

or                     $5=\frac{1}{2}\times 0.53\times t^2$

$\therefore$                   $t=4.34 s$ .

The distance travelled by truck in this duration

                       $s=0+\frac{1}{2}\times 2{(4.34)}^2=18.87 m$ .