The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in Fig. 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 of the truck ?

Explanation

Here mass of the bon m = 40 kg
Acceleration of truck `a = 2 ms^(-2)`
Distance of box from open end `S = 5 m_(1)`
Coeff of friction `u = 0.15`
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 equal to F = 80 N in the backward direction . This is opposed by force of limitating friction .
`f = mu R = mu ` mg
` = 0.15 xx 40 xx 9.8`
` = 58.8 ` N in the forward direction
` :. ` Net force on the box in the backward is `p = F. - F = 80 - 58.8 = 21.2 v`
Backward acceleration produced in the box
` a = p/m = (21.2)/40 = 0.53 ms^(-2)`
It is time taken by the box to travel S = 5 metre and ball off the trunk , ten from `S = ut + 1/2 at^(2)`
`5 = 0 xx t +1/2 xx 0.53 t^(2)`
`t = (sqrt(5xx2))/(0.53) = 4.34 S `
If the truck travels a distance `xx` during this time , then again from
`S = ut +1/2 at^(2)`
`x = 0 xx 4.34 +1/2 xx 2(4.34)^(2) = 18.84 m ` .