The rear side of a truck is open A box of `40kg` mass is placed `5m` away from the open end as shown in The coefficient of friction between the box and the surface is 0.15. On a straight road, the truck starts from rest and accelerating with `2m//s^(2)`. At what dis tance from the starting point does the box distance from the starting point does the box fall from the truck? (Ignore the size of the box)
.

Options

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

Answer

A

Explanation

Here, Mass of the box, M=40kg
Acceleration of the truck `a=2ms^(-2)`
Distance of the box from the near rear, d=5m
Coefficient of friction between the box and the surface below it `mu=0.15` The various forces acting on the block are as shown in the figure. As the truck moves in forward direction with the acceleration `a=2ms^(-2)` the box expertience a force the backward direction and it is give by `F=Ma=(40kg)(2ms^(-2))=80N` is backward direction. Under the rear end of the truck. As it does so, its motion will be opposed by the force of friction which acts in the forward direction and it is given by `f=muN=muMg=0.15xx40xx60N` The acceleration of the box relative to the truck towards the near end is `a_(1)=(F-f)/(M)=(80N-60N)/(40kg)=0.5ms^(-2)`
Let t be the time taken for the box to fall of the truck. `d=0xxt+(1)/(2)a_(1)t^(2) (therefore u=0)`
`5=(1)/(2)xx0.5xxt^(2), t=sqrt((2xx5)/(0.5))=sqrt(20)s`
Using, `S=mu+(1)/(2)at^(2)`
We, get, `x=0xxt+(1)/(2)xx2xx(sqrt(20))^(2) or x=20m`