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).
Mass of the box, m = 40 kg
Coefficient of friction, μ
Initial velocity, u
= 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
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
= 0.15 × 40 × 10
= 60 N
Net force acting on the block, Fnet
= 80 – 60 = 20 N, acting backward
The backward acceleration produced in the box is, Acceleration, aback
Using the second equation of motion,
Time t is,
s' = ut + (1/2) aback
5 = 0 + (1/2) × 0.5 × t2
∴ t = √20
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.