Question

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/s². At what distance from the starting point does the box fall off the truck ? (ignore the size of the box)

Answer 1

Force on the box due to accelerated motion of the truck 

F = ma = 40 × 2 = 80 N (in forward direction) 

Reaction on the box, F´ = F = 80 N (in backward direction) 

Force of limiting friction, f = µR = 0·15 × 40 × 10 = 60 N 

Net force on the box in backward direction is P = F´ – f 

= 80 – 60 = 20 N 

Backward acceleration in the box 

t = time taken by the box to travel s = 5 m and falls off the truck, then from

If the truck travels a distance x during this time