I made it into a tension/force problem at first, with the vertical or heavier block being the cause of the tension since the sum of the vertical forces of the tension was 0. So T = mg. So I knew what T was

Then for the block on the incline, I made a free body diagram where mgsin37 and Fk (friction force) pointed towards the left and T (tension) pointed towards the right. For vertical forces on the block, the sum was 0. So I can assume that n=mgcos37.

So my sigma Fx for the block on the incline was

T - mgsin37 + Fk

I calculated it and then plugged it into the work equation.

W= FDcos(theta)

Because the change of kinetic energy is just net work done. I thought W was my final answer.

Can anyone help me? Thanks!