Two blocks A and B are connected via a pulley. Diagram They are initially at height h = 2 m from the ground. When block A hits the ground, it is moving at 3 m/s. For part a), I already determined the amount of frictional energy dissipated, with E = 45.7 J. This is correct, according to the textbook. Mass of A = 11 kg, Mass of B = 5 kg. Now I'm asked to calculate the tension in the cables. Isolating Block A, we have the TA, mg and the frictional force acting on it. Using the principle of work-energy: 0 + (11)(9.81)(2 m) - (2 m)TA - 45.7 = 1/2(11)(3)^2 Solving, I get TA = 60.3 N For Block B: 0 - (5)(9.81)(2 m) + (2 m)TB - 45.7 = 1/2(5)(3)^2 Solving, I get TB = 83.2 N. However, the back of the textbook has the answers listed as TA = 83.2 N and TB = 60.3 N. I must be on the right track since I have the right numbers, but can anyone point out where I went wrong to get the exact reversal of answers? It makes sense that TA > TB, since A is heavier than B, but I can't find the mistake in my equations. Any help would be appreciated!