A block of mass `m` is attached to a spring of force constant `k` whose other end is fixed to a horizontal surface. Initially the spring is in its natural length and the block is released from rest. The average force acting on the surface by the spring till the instant when the block has zero acceleration for the first time is

Explanation

Due to weight of block B, it moves down and pulls the block A now as block A and B move, spring gets stretched and becomes inclined as its lower end is attached to the block A. It will break off from the ground below it when the vertical component of the spring force on block A will balance its weight mg Let it happens
when A moves by a distances as shown in figure.
At this instant let the spring be inclined at an angle with the vertical
If the stretch in the spring at this instant is x, then it is given as
`x = l sec theta - l ` or ` x = l ( sec theta - 1)` ........... (1)
If mass A breaks off from ground below it , we have
` kx cos theta = mg `
From equation (1) and (2) substituting the value of x, we get
`kl( sec theta - 1) cos theta = mg or kl(1- cos theta ) = mg or cos theta =1 -(mg)/(kl) or tan theta =([k^(2)l^(2) (kl - mg)^(2) ]^(1//2))/(kl-mg)`
At this instant the distance travelled by mass A and B is given by ` s=l tan theta (or) s=l([k^(2)l^(2)(kl-mg)^(2)]^(1//2))/(kl-mg)`