The rear side of a truck is open and a box of mass 40kg is placed 5m away from the open end. The coefficient of friction between the box and the surface below it is 0.15. The truck starts from rest with an acceleration of `2ms^(-2)` on a straight road. At what distance from the starting points does not the box fall off the truck?

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 Lhe acceleration of the truck. The frictional force on the truck which acts in forward direction
`f_k = muN = 0.15 xx 40g = 58.8N`
Since pseudo force is greater than frictional force, so block will accelerate in backward direction relative to truck with a magnitude.
`a=(8) - (58.8)/40 = 0.53 m//s^(2)`
The time taken by box to cover the distance 5 m
`s = 0 + 1/2 at^(2) rArr t = sqrt((2s)/a) = 4.34`sec
The distance travelled by truck in this duration
`s=0 + 1/2 xx 2(4.34)^(2) = 18.87` m