1. The problem statement, all variables and given/known data A 0.211 kg shoe is dropped onto a vertically oriented spring with a spring constant of 102 N/m. The shoe becomes attached to the spring upon contact, and the spring is compressed 0.117 m before coming momentarily to rest. If the speed of the shoe just before impact is doubled, what is the maximum compression achieved by the spring? 3. The attempt at a solution Before this question, I figured out the speed the shoe was travelling at just before it hit the spring to be 2.08 m/s, which was correct. I also figured out the work done by the spring force to be -0.698 J and the work done by the weight of the shoe to be 0.242 J. I know that when speed doubles, K.E. quadruples, which means the net work done must increase by 4 as well. Wnet= Wgrav+Wspring=0.242 -0.698=-0.456 J. I found the new K.E. to be 1.826 J (with doubled speed). When I equate KE to Wnet, I get a quadratic: m*g*x -0.5*k*x^2= -1.826 J (new K.E.) Is there no other way other than solving this quadratic to get to the new compression? HELP, please!