The rear side of a truck is open. A box of `40kg` mass is placed `5m` away from the open end as shown in figure. The coefficient of friction between the box and the surface is `0.15`. On a straight road, the truck starts from rest and accel erating with `2m//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 forwed direction `f_(k) =muN =0.15 xx 40g =58.8N` Since pseudo force is greater than frictional force, the block will accelerate in backward diraction 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 `5m` is given by
`s = 0 + (1)/(2) at^(2) rArr t = sqrt((2s)/(a)) =4.3 sec`
The distance travelled by truck in this time is
`a' =2ms^(-2)`
`s' = (1)/(2) a' t^(2) = (1)/(2) xx 2 xx (4.34)^(2) = 18.87m` .