Figure shows a block A on a smooth surface attached with a spring of force constant k to the ceiling . In this state spring is in its natural length l. The block A is connected with a massless and frictionless string to another identicalmass B hanging over a light and smooth pulley . Find the distance moved by A before it leaves contact with the ground .

Figure shows a block A on a smooth surface attached with a spring of f

1.	Figure shows a block A on a smooth surface attached with a spring of force constant k to the ceiling . In this state spring is in its natural length l. The block A is connected with a massless and frictionless string to another identicalmass B hanging over a light and smooth pulley . Find the distance moved by A before it leaves contact with the ground .
Answer» Solution :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 distance as shown in figure. At this instant let the spring be inclined at an angle `theta` 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`.....(2) 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)`