- #1

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## Homework Statement

## Homework Equations

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 realize 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?