Two 2.4-kg blocks are connected by a string draped over the edge of a slippery table, so that one block is on the table and the other is just hanging off the edge. A restraint holds the block on the table in place, and the string is 0.50 m long. After the restraint is removed, what speed does the pair of blocks have at the instant the upper one is pulled off the table? Ok, so I figured out how to solve this problem, but I just need help understanding the solution. How I solved it was by finding the potential energy of the first block which is: PE=(2.4kg)((9.8m/s)(0.5m)=11.76 Joules Then I took the equation for Kinetic energy and found he velocity like so: 11.76 J= (1/2)(2.4m)*v^2. Except, I had to plug in for both the blocks this time, so: 11.76=(2.4)*v^2 So the answer is that the initial velocity of the blocks is 2.2m/s. So I think my main question is, why do we just find the potential energy of the first block, but then plug in for both blocks when using the Kinetic energy formula? Can anyone shed any light on this? I really don't understand the solution.