Dynamics Question (Kinetics& energy)

[gomerpyle]

Homework Statement

The two blocks are connected by a light inextensible cord, which passes around small

massless pulleys as shown below. If block B is pulled down 500 mm from the equilibrium

position and released from rest, determine its speed when it returns to the equilibrium

position.

http://s3.amazonaws.com/answer-board-image/201069134736341168802377387504353.gif

Homework Equations

T1 + U1 = T2 + U2

The Attempt at a Solution

If it's pulled down below equilibrium and held there, then T1 of the system is zero because both blocks are not moving. At the moment B passes through the equilibrium, there is no more potential energy, only kinetic, then the equation would look like:

U1 = T2

The problem I'm running into is that I get a negative value for the left side of this equation, which is impossible because then it would have to go under a square root when solving for the velocity.

for U1 I had:

mgha - mghb + 0.5kx^2

Since 'b' moves down 0.5m, a moves up 0.25 and the spring is stretched 0.25. Is this right since A is attached to the pulley and B is simply hanging? If that's the case then:

(2)(9.81)(0.25) - (10)(9.81)(0.5) + 0.5(800)(0.25)^2

Which is negative. Supposedly the answer is 2.16 m/s.

 

[PhanthomJay]

You neglected the fact that the spring is already stretched in the equilibrium position before the block B is pulled down.