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

Here,
Mass of the box, M= 40 kg
Acceleration of the truck, `a = 2 ms^(-2)`
Distance of the box from the rear end, d = 5 m
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 = 2` ms^(-2)` , the box experiences a force F in the backward direction and it is given by
` F = Ma = (40 kg)(2 ms^(-2)`) = 80 N
in backward direction.

Under the action of this force, the box will tend to move towards 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= mu N = mu Mg = 0.15 xx 40 xx 10 =60 N`
The acceleration of the box relative to the truck towards the rear end is
` a_1 =(F -f)/(M ) = ( 80 N- 60 N) /( 40 kg) = 0.5 ms^(-1)`
Let t be the time taken for the box to fall off the truck.
As ` s = ut +1/2 xx 0.5 xx t^2 or t = sqrt((2 xx5 )/( 0.5 )) = sqrt(20 )s`

During this time, the truck covers a distance x
` therefore x=0 xx t +1/2 xx 2 xx ( sqrt(20))^2 " " ( :. u=0)`
or ` x= 20 m`