I found the answer to this problem using the change in KE, but when I try to relate the work done on the 12.0-N block in terms of potential energy i don't get the same result. Is the change potential energy not equal to the work done? I would also like to know what situations is the change PE equal to work. I know that if an object height isnt changing then i wouldn't need PE, but is this also true in free fall? Thank you! 1. The problem statement, all variables and given/known data 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. Find the total work done on 12.0-N block if μs=0.500 and μk=0.325 between the table and the 20.0-N block. 2. Relevant equations PEi + KEi = PEf + KEf + Elost w=KEf-KEi w=PEf-PEi ? Elost=MGUxD 3. The attempt at a solution I found velocity of for this system which was 1.59 m/s. Found that the work done was simply mv^2/2.