1. The problem statement, all variables and given/known data A mass m is resting at equilibrium suspended from a vertical spring of natural length L and spring constant k inside a box. The box begins accelerating upward with acceleration a. How much closer does the equilibrium position of the mass move to the bottom of the box? Diagram: http://www.aapt.org/Programs/contests/upload/olympiad_2008_fnet_ma.pdf Problem 17 (Answer: ma/k) 2. Relevant equations F = ma F = -kx 3. The attempt at a solution Well, the forces acting are: the force of the spring (Fs), the weight of the object(Fg), and the force of the elevator (Fe). Fe + fs - mg = ma Fe - kx = m (a+g) I thought this was correct but am not sure where to progress from here.