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).

Coefficient of friction, μ = 0.15
Initial velocity, u = 0
Acceleration, a = 2 m/s²

Distance of the box from the end of the truck, s' = 5 m
According to Newton's second law of motion,
Force on the box caused by the accelerated motion, F = ma 
               = 40 × 2
                = 80 N 
According to 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 = μmg  
                   = 0.15 × 40 × 10
                   = 60 N 
Therefore,
Net force acting on the block, 
              F_net = 80 – 60 = 20 N, acting backward 
The backward acceleration produced in the box is, 
Acceleration, a_back = 
Using the second equation of motion,
Time t is, 
                   s' = ut + (1/2) a_backt² 

                   5 = 0 + (1/2) × 0.5 × t² 

∴                  t = √20 s 
Hence, the box will fall from the truck after  s from the start. 
The distance s, travelled bytruck in √20 s is given by, 
                   s = ut + (1/2)at² 

                     = 0 + (1/2) × 2 × ()² 

                     = 20 m 
Therefore, at a distance of 20 m, the box will fall off the truck.