1. The problem statement, all variables and given/known data A mass of 1.20 kg. originally at rest, sits on a frictionless surface. It is attached to one end of an unstretched spring (k = 790 N/m), the other end of which is fixed to a wall. The mass is then pulled with a constant force to stretch the spring. As a result, the system comes to a momentary stop after the mass moves 14.0 cm. Find the new equilibrium position of the system. 2. Relevant equations Total Work = Change in kinetic energy = 0 W = F * d Total Work = Work done by force + Work done spring 3. The attempt at a solution My first thought was that since the mass comes to a stop in 14 cm then that must be the equilibrium position since it would only come to a stop if the spring force equaled the applied force. However the answer is actually found by first calculating the work done by the applied force: Total Work = Work by Force + Work by Spring = Change in KE = 0 Work by Force = - Work by Spring Work by Force = 7.4 J Then you determine the magnitude of the applied force: W = Force * Displacement W / Displacement = Force Force = 55 N Then you determine when the Spring Force is equal to 55 N: 55 = (0.5) * k * x^2 55 = (0.5) * 790 * x^2 x = 0.07 m = 7 cm Now, my confusion arises from the fact that in the question it is stated that the spring comes to a MOMENTARY STOP at 14 cm. How is that possible if the applied force is equal to the spring force at 7 cm? How is the mass able to travel an additional 7 cm before coming to a momentary stop? Is it due to the fact that when it reaches 7 cm it has an initial velocity and that kinetic energy is converted to spring energy over a distance of 7 cm? Okay I just checked the math and this works out perfectly. So I guess this answers my question but I would love if someone could confirm this. Thank you for reading.