- #1
gomerpyle
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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.