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. 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).

Explanation

Force on the box due to accelerated motion of the truck.
F = ma = 40 x 2 = 80 N (in forward direction)
Reaction on the box, F. = F = 80 N (in backward direction)
Force of limiting friction, `f = muR = 0.15 xx 40 xx 10 = 60 N`
Net force on the box in backward direction is P=F.- f=80 – 60 = 20 N
Backward acceleration in the box = ` a = P/m = 20/40 = 0.5 ms^(-2)`
t= time taken by the box to travel s = 5 m and falls off the truck, then from
` s = ut + 1/2 at`
`5 = 0 xx t + 1/2 xx 0.5 xx t^2`
`t = 4.47 `
If the truck travels a distance x during this time then
` x = 0 xx 4.34 + 1/2 xx 2 xx (4.471)^2`
x = 19.98 m