The rear side of a truck is open and a box of 40 kg mass is placed 5m away from the open end. 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 ?

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

1.	The rear side of a truck is open and a box of 40 kg mass is placed 5m away from the open end. 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 ?
Answer» Solution :Force on the box due to accelerated motion of the truck `F= ma =40 XX 2= 80 N` (in forward direction) Reaction on the box, F= R= 80 N (in BACKWARD direction) Force of limiting friction, `f=muR=0.15 xx 40 xx 10=60N` Net force on the box in backward direction is P =F'-f =80-60=20N Backward acceleration in the box `=a=P/m= (20)/(40)= 0·5"ms"^(-1)` t= time taken by the box to travel s= 5m and falls off the truck, then from `s=ut+1/2 at^(2)` `5=0 xx t +1/2 xx 0.5xx t^(2)` t=4.47 s 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