When will block leave surface ?

[vissh]

hello >:D
The arrangement of my question at its initial state is as http://s1102.photobucket.com/albums/g448/vissh/?action=view&current=2BlockSpring.jpg"
The above figure shows 2 blocks A and B,each having a mass of 320g connected by a light string passing over a smooth light pulley. The horizontal surface on which the block A can slide is smooth. The block A is attached to a spring of spring constant 40N/m whose other end is fixed to a support 40cm above the horizontal surface. Initially, the spring is Vertical and unstretched when the system is released to move. Find the velocity of the block A at the instant it breaks off the surface below it. Take g=10m/s²

Homework Equations

Work done on a body by net force = Change in its K.E.
Net force (along a line) = mass*acceleration('s component in that line's direction)

The Attempt at a Solution

Hmmm.. The thing is that i can't decide how to proceed the problem. I think the block will leave contact when the spring is fully stretched and the normal rxn from below gets zero at that moment.But still not able to apply this hehe . Can I get some Hints :D .

Thanks in advance ^.^

Share knowledge, it increases on sharing :)

 

[Doc Al]

I think the block will leave contact when the spring is fully stretched and the normal rxn from below gets zero at that moment.But still not able to apply this hehe .

Sounds good to me. Why can't you apply this?

Figure out where the block will be at the instant that the normal force goes to zero.

 

[vissh]

Thx doc :D WoW got the answer . First , i found out the extension in the spring and thus got the horizontal displacement of block A. Then , applied the work energy theorem to the 2 masses and got the answer v=1.5 m/s (approx) . Thz again (^.^)

 

[Doc Al]

Excellent.