1. The problem statement, all variables and given/known data A conservative force F(x) = bx + a acts on a 2.61 kg particle, where x is in meters, b = 6.3 N/m, and a = 4 N. Calculate the change in potential energy of the particle as it moves along the x-axis from x1 = .749 m to x2 = 4.81 m. 2. Relevant equations work-energy theorem, PE = mgh 3. The attempt at a solution My first thought was simply 0, since it was going along the x-axis and therefore no change in PEg and no springs or such, but that was wrong. I solved that the work done by the force = 87.3556 J. I also solved initial kinetic energy if the final speed was 16.8 m/s, which was 280.9676. KEi + W = KEf. Where does the potential fall in?