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

Force on the box due to accelerated motion of the truck `F= ma =40 xx 2= 80 N` (in forward direction)
Reaction on the box, F= R= 80 N (in backward direction)
Force of limiting friction, `f=muR=0.15 xx 40 xx 10=60N`
Net force on the box in backward direction is P =F'-f
=80-60=20N
Backward acceleration in the box `=a=P/m= (20)/(40)= 0·5"ms"^(-1)`
t= time taken by the box to travel s= 5m and falls off the truck, then from `s=ut+1/2 at^(2)`
`5=0 xx t +1/2 xx 0.5xx t^(2)`
t=4.47 s
If the truck travels a distance x during this time
then `x=0 xx 4.34 +1/2 xx 2 xx (4.471)^(2)`
x=19.98