1. The problem statement, all variables and given/known data Problem 6.65 Two blocks are connected by a very light string passing over a massless and frictionless pulley (Figure (Figure 1) ). The 20.0-N block moves 75.0 cm to the right and the 12.0-N block moves 75.0 cm downward. Question: Find the total work done on 20.0-N block if there is no friction between the table and the 20.0-N block. 2. Relevant equations ΣF = ma Wtotal = F * s cos(φ), where the force is constant. Wtotal = 1/2(m)vf^2 - 1/2(m)vi^2 vf^2 = vi^2 + 2ad 3. The attempt at a solution So I started with the 12.0N weight. ΣF = m1a = T - 12.0N, where m1 is the mass of the 12.0N weight. I also say acceleration is the same for both weights because they are in the same system. then T = (12/9.8)a+12.0N Then I set up my force equation for the 20.0N weight. ΣF = m2a = T, where m2 = 20/9.8 (20/9.8)a = (12/9.8)a + 12.0N (8/9.8)a = 12.0N a = 14.7 m/s^2 Then I plug this in to my equation for tension. T = (12/9.8)14.7 + 12.0 T=18+12.0 T=30N Then W = T * s cos(0°) W = (30)(.75)(1) = 22.5 J. Which is wrong. 2nd attempt I thought maybe force is not constant, although I'm not sure why it wouldn't be. SO i used this equation. W = 1/2(m)vf^2 - 1/2(m)vi^2, where i assume vi = 0m/s then W = 1/2(m)vf^2 to solve for vf: vf^2 = vi^2 +2ad vf^2 = 2(14.7)(.75) vf^2 = 22.05 Then W = 1/2(20)(22.05) W = 220.5J which is also wrong. I know my displacement vector is correct so it must be my acceleration but I can't see anywhere that i made an error in calculating it.