I have this problem, and I'd just like to make sure if the process I used to achieve an answer is correct. A 0.25-kg block is dropped straight downward onto a vertical spring. The spring constant of the spring is 58 N/m. The block sticks to the spring, and the spring compresses 0.15 m before coming to a momentary halt. What is the speed of the block just before it hits the spring? So I use this equation to solve it: PEg (graviational potential energy) + KE0 (initial kinetic energy) + PEe0 (initial elastic potential energy) = PEg2 + KEf + PEef. And there would be no initial elastic potential energy, no final kinetic energy and no final gravitational potential energy, thus leaving me with: KE0 + PEg = PEef, which expanded would be: 1/2 mv^2 + mgh = 1/2kh (or x)^2 Plug in the numbers accordingly.. 1/2 (.25)(v^2) + (.25)(9.8)(.15) =1/2(58)(.15)^2 which gives me an answer of 1.51 m/s. Is what I'm doing right?