1. The problem statement, all variables and given/known data A simple Atwood's machine uses two masses m1 and m2. Starting from rest, the speed of the two masses is 8.7 m/s at the end of 7.4 s. At that time, the kinetic energy of the system is 61 J and each mass has moved a distance of 32.19 m. Find the value of the heavier mass. Find the value of the lighter mass. 2. Relevant equations a = (m2 - m1) / (m1 + m2)g Ke = 1/2mv2 3. The attempt at a solution 61 = 1/2m1v2 + 1/2m2v2 61 = 1/2(8.7)2(m1+m2) 61 = 37.845(m1+m2) m1+m2 = 1.612 a = 8.7 / 7.4 = 1.176 1.176 = (m2 - m1) / (m1 + m2)g 10.976m1 = 8.624m2 m1 = .786m2 .786m2 + m2 = 1.612 2m2 = 2.051 m2 = 1.025 m1 + 1.025 = 1.612 m1 = .587 I was sure that I solved it correctly, perhaps my algebra for solving the acceleration equation is incorrect. Thanks for any help.