1. The problem statement, all variables and given/known data The coefficient of friction between the block of mass m1 = 3.00 kg and the surface shown is µk = 0.400. The system starts from rest. What is the speed of the ball of mass m2 = 5.00 kg when it has fallen a distance h = 1.50 m? Note that this is just a generic picture of how the problem is set up. M would be m1 and m would be m2 (the ball). 2. Relevant equations W = F ⋅ Δr = F * r * cosΘ Fk = µk * Fn, where Fk is the force due to kinetic friction and Fn is the normal force. ∑W = ΔK, where W is work and K is kinetic energy Ki + Ui + W - Fkd = Kf + Uf, where K is kinetic energy, U is potential energy and Fkd is the work done by friction (non-isolated system) K = (1/2)mv^2 U = mgh (or mgΔy) 3. The attempt at a solution So, I would like to mention that I do have access to the solution to the problem, and I believe I understand the general idea of how it works (treat both masses as a system, and then use the non-isolated system equation listed above with work set equal to zero and so on). Although, before looking at the solution I tried it myself (which turned out to be incorrect). So, I started by stating that ∑W = Wt + Wk (Wt = work done by tension force and Wk = work done by friction), since the normal force and gravity are not doing any work here (m1 isn't moving vertically, so Δr = 0 in the y-direction). I got Wt = (5.00kg)(9.80m/s^2)(1.50m)cos0 = 73.5J. I found friction to be Fk = μk * Fn = (.400)(5.00kg)(9.80m/s^2) = 19.6N, since Fn = mg. I then found Wk = (19.6N)(1.50m)cos180 = -19.6N I found ΣW = Wt + Wk = 73.5J - 19.6J = 44.1J I then used ΣW = ΔK. I know Ki is 0 since it states the system is initially at rest, so ΔK = Kf. ΔK = (1/2)m1vf^2 Setting them equal to each other... vf = √[(2*ΔK)/m1] = √[(2*44.1J)/(3.00kg)] = 5.42m/s The actual answer is vf = 3.74m/s. Why does my attempt not work out to be the correct answer? I know the velocities of both masses are equal due to the tensile force between them and the frictionless (assumed) pulley. I tried thinking of it a different way to try to tell myself that it wasn't correct. I thought of a parallel force pulling m1 with the same force as m2g (49N), and then calculated the total work done and then relating it to the work-kinetic energy theorem to get my answer and it made sense in my head. I'm not understanding how the fact that it's a non-isolated system should get a different velocity than my answer.