### Q.36:- The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in Fig. 5.22. 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 m s^{–2}. At what distance from the starting point does the box fall off the truck? (Ignore the size of the box).

**Answer:-**Mass of the box, *m* = 40 kg

Coefficient of friction, *μ* = 0.15

Initial velocity, *u* = 0

Acceleration, *a* = 2 m/s^{2}

Distance of the box from the end of the truck, *s*‘ = 5 m

As per Newton’s second law of motion, the force on the box caused by the accelerated motion of the truck is given by:

*F* = *ma *

*= *40 × 2 = 80 N

As per Newton’s third law of motion, a reaction force of 80 N is acting on the box in the backward direction. The backward motion of the box is opposed by the force of friction *f*, acting between the box and the floor of the truck. This force is given by:

*f* = *μ**m*g

= 0.15 × 40 × 10 = 60 N

∴Net force acting on the block:

*F*_{net} = 80 – 60 = 20 N backward

The backward acceleration produced in the box is given by:

*a*_{back} = F_{net} / m = 20 / 40 = 0.5 ms^{-2}

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

s’ = ut + (1/2)a_{back}t^{2}

5 = 0 + (1/2) × 0.5 × t^{2}

∴ t = √*20* s

Hence, the box will fall from the truck after √*20* s from start.

The distance *s*, travelled by the truck in √*20* s is given by the relation:

s = ut + (1/2)at^{2}

= 0 + (1/2) × 2 × (√*20)*^{2}

= 20 m