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 .

Options

A. 20 m
B. 30 m
C. 40 m
D. 50 m

Answer

A

Explanation

Acceleration of box = 2 - `mu g`
`= 2 - 0.15 xx 10`
`= 0.5 m//s^(2)`
For box
u = 0 , `a_(b) = 0.5 m//s^(2)`
`S = (1)/(2) a_(b) t^(2) implies 5 = (1)/(2) xx 0.5 xx t^(2)`
`implies t = sqrt((10)/(0.5)) = sqrt(20)` sec
In time t = `sqrt(20) ` sec , truck travel a distance `S_(t) = (1)/(2) a_(t) t^(2) = (1)/(2) xx 2 xx (sqrt(20))^(2)`
`S_(t) = 20`m [ (u = 0 , a = `2 m//s^(2))]`