1. The problem statement, all variables and given/known data A 0.20-kg block and a 0.25-kg block are connected to each other by a string draped over a pulley that is a solid disc of inertia 0.50 kg and radius 0.10 m. When released, the 0.25-kg block is 0.30 m off the ground. What speed does this block have when it hits the ground? 2. Relevant equations ΔE = 0 Ei = Ef (closed system) E = K + UG + Eth + Es UiG = Ktransf + Krotf Ktrans = ½mv2 Krot = ½Iω2 ISolid Cylinder = ½mr2 ω = v/r 3. The attempt at a solution First, I step up the energy equation and substitute in my energy expressions. I've used the subscript "1" to represent the 0.20-kg mass, the subscript "2" to represent the 0.25-kg mass, and the subscript "p" to represent the pulley. U1iG + U2iG = U1fG + U2fG + K1f + K2f + Kpf m1gh1i + m2gh2i = m1gh1f + m2gh2f + ½m1v12 + ½m2v22 + ½Iω2 I know what the next step is, but I'm not sure how to get there algebraically. I should end up with: (m1 - m2)gΔh = ½(m1 + m2)vf2 + ¼mpvf2 I understand that v1, v2, and vp are equal, so no problem there. I see that how we substitute ½mr2 for I, and v/r for ω. I don't understand how we end up with (m1 - m2)gΔh. I understand that we pull the common variable g out, and I can kind of understand how the h-values become Δh. However, it looks like this when I set it up: m1gh1i + m2gh2i - m1gh1f - m2gh2f = ½m1v12 + ½m2v22 + ½Iω2 Which becomes: g(m1h1i + m2h2i - m1h1f - m2h2f) = ½m1v12 + ½m2v22 + ½Iω2 Any help is greatly appreciated.