This problem is bugging the crap out of me. It's the last one I need to do. I swear I got the right answer. Maybe you experts can help? :) 1. The problem statement, all variables and given/known data The system shown in the figure below consists of a light, inextensible cord; light frictionless pulleys, and blocks of equal mass. It is initially held at rest so that the blocks are at the same height above the ground. The blocks are then released. Find the speed of block A at the moment when the vertical separation of the blocks is 1.00 m. http://www.webassign.net/serpop/p7-47.gif 2. Relevant equations PEi + KEi = PEf + KEf KE= (mv^2)/2 PE= mgy 3. The attempt at a solution 1. Judging from what is presented, I assume that the block A moves twice as fast as block B and therefore, it will lower at double the rate that block B will rise. 2. I figured that I would write out my equation to be: mgy[a] + mgy [initial] = mg(y-2/3)[a] + mg(y+1/3) + (mv^2)/2[a] + (m[v/2]^2)/2 3. Simplified by dividing by m and making initial y to be 0. 0 = g(-2/3) + g(1/3) + (v^2)/2 + ((v/2)^2)/2 4. Input -9.8 for g and simplify more. -3.26699 = (v^2)/2 + ((v/2)^2)/2 5. Simplify more. 2.556165 = 1.5v Answer = 1.7 m/s HOWEVER, that was wrong, and I've checked my math several times. I must be doing something horribly wrong. Kudos and much thanks to you if you can find it, like I said, this has been bugging me for a while.