The rear side of a truck is open and a box of 40 kg mass is placed 5 m 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 ? Ignore the size of the box .

Explanation

Because of the acceleration of the truck the pseudo force on the box `m xx a = 40 xx 2 = 80N` .
This force acts opposite to the acceleration of the truck
The frictional force on the truck which acts in the forward direction `f_(k) = mu N = 0.15 xx 40 g= 58.8 N` . Since pseudo force is greater than frictional force , the block will accelerate in backward direction relative to truck with a magnitude .
`a = (80 - 58.8)/(40) = 0.53 m // s^(2)`
The time taken by box to cover the distance 5 m is given by
`s = 0 + (1)/(2) at^(2) implies t = sqrt((2s)/(a)) = 4.34` sec
The distance travelled by truck in this time is a' = `2 m s^(-2)`
`s = (1)/(2) a't^(2) = (1)/(2) xx 2 xx (4.34)^(2) = 18.87` m