PE = mgh
KE = 0.5mv^2
WD = F * s
The Attempt at a Solution
Its part i. I understand the solution when you look at the whole system. You do not need to consider the tension in the string because they cancel out, but I want to be able to do it just considering B's KE and PE.
So I said that for B, the PE at the start is mgl and at the end it is considered 0. For its kinetic energy, at the start it is 0 and at the end it is 0.5mv^2. I also realise that it is subject to resistance from the tension in the string, so I resolved the forces to 3mg - T = 3ma (For B) and T - mgSin30 = ma (for A). From this I solved for T to get T = 9/8mg. Then I calculated the work done against tension as WD = -9/8mgl.
So then using the conservation of energy I said that 3mgl = 0.5mv^2 + 9/8mgl and for v I got (√15gl)/2. (g and l also in root)
The correct answer however is v = (√5gl)/2. (g and l also in root).
I know its easier to do it considering the whole system, but I want to do it this way, so what have I done wrong?