Work done with free body released

[jack1234]

I have a question here
http://tinyurl.com/2kgeun

My attempts is
mgx=(1/2)*kx^2 + (1/2)mv^2
where mgx is the potential energy of block m
and (1/2)mv^2 is kinetic energy of block 2m.
But this does not seem to work, because we don't have the value v.

May I know how to solve it?

 

[learningphysics]

I have a question here
http://tinyurl.com/2kgeun

My attempts is
mgx=(1/2)*kx^2 + (1/2)mv^2
where mgx is the potential energy of block m
and (1/2)mv^2 is kinetic energy of block 2m.
But this does not seem to work, because we don't have the value v.

May I know how to solve it?

you have the right equation... v is 0 when the object reaches maximum extension... because at this point, the hanging M mass will change from going downward to going upward.

 

[rl.bhat]

Draw the free body diagram for M and 2M. From that find the common accelaration. Now work done the masses = potential energy stored in the spring. i.e. 3Max = 1/2*kx^2

 

[jack1234]

Thanks, rl.bhat, but I still getting the correct answer(22cm) if sub v=0 in
mgx=(1/2)*kx^2 + (1/2)mv^2
as mentioned by learningphysics.