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).

The rear side of a truck is open and a box of 40 kg mass is placed 5 m

1.	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).
Answer» Solution :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