Question

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.

Answer 1

Force experienced by box, F = ma = 40 `xx` 2 = 80 N
Frictional force , `F_f` = `mu`mg = `0.15 xx 40 xx 10` = 60 N .
Net force = F - `F_f` = 80 - 60 = 20 N .
Backward acceleration produced in the box, a = 20 / 40(Net force / m) ltbr? `implies` a = 0.5 ms^(-2)`
If t is time taken by the box to travel s = 5 metre and fall off the truck , then from
S = ut + `1/2 ar^(2)`
S = `0 xx t + 1/2 xx 0.5 t^(2)`
t = `sqrt (5 xx2 / 0.5 )` = 4.47 s.
If the truck travels a distance x during this time, then again from
S = ut + `1/2 at^(2)`
x = `0 xx 4.47 + 1/2 xx 2(4.47)^(2)` = 19.98 m .