The rear side of a truck is open and a box of 40 kg mass is placed 5 m from the openend as shown 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 ms^(-2)` At what distance from the starting point does the box fall off the truck ? (Ignore the size of the box )
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Explanation

Here , mass of the box , ` m = 40 kg `
acceleration of truck , ` a = 2 ms^(-2) `
distance of box from open end ` s = 5 m `
Coeff . Of friction ` mu = 0. 5 `
Force on the box due to accelerated motion of the truck
` F = ma = 40 xx 2 = 80 N `
This force F is in the forward direction .
Reaction F on the box is equal to F = 80 N in the backward driction This is opposed by force of limitting
friction ` f = mu R = mu mg = 0.15 xx 40 xx 10 = 60 ` N in the forward direction
:. Net force on the box in the bacward direction is ` P = F - f = 80 - 60 = 20 N `
Backward acceleration produced in the box ` a = (p)/(m) = (20)/(40) = 0.5 ms^(-2) `
If is time taken by the box to travel s = 5 meter and fall off the truck then from
` s = ut + (1)/(2) at^(2) `
` 5 = 0 xx t + (1)/(2) xx 0.5 t^(2) `
` t = sqrt((5 xx 2)/(0 .5 ))= 4.47 s `
If the truck travels a distance x during this time , then again from
` s = ut = (1)/(2) at^(2) `
` x = 0 xx 4.34 + (1)/(2) xx 2 (4.47)^(2) = 19 . 98 M ` .