Well.. I've been understanding spring constants up until this point. I just did a lab on them and understood its effects in parallel and series of springs, and I've understood it in regards to a mass and a coil in which the mass is just hanging from a coil.. but now that there is a tabletop and distance the block travels in the air, it's confusing a bit.. so here is the question, equations I think are important, and my work. I was hoping if someone could help me understand this better by telling me where I went wrong and any helpful tips. I really did well on my last test because people (docAl and olderdan, and a few others ^^) were really patient with me and helped me understand it rather than giving me the answer. So I think it's best if I'm not given the answer but only guidelines. Thanks ^^ 1. The problem statement, all variables and given/known data One end of a spring is attached to a solid wall while the other end just reaches to the edge of a horizontal, frictionless tabletop, which is a distance h above the floor. A block of mass M is placed against the end of the spring and pushed toward the wall until the spring has been compressed a distance X. The block is released, and strikes the floor at a horizontal distance D from the edge of the table. Air resistance is negligible. Determine the expresses for the following quantities in terms of M, X, D, h, and g. Note that these symbols do not include the spring constant. (a) The time elapsed from the instant the block leaves the table to the instant it strikes the floor. (b) The horizontal component of the velocity of the block just before it hits the floor. (c) The work done on the block by the spring. (d) The spring constant. It wasn't given, but I'm sure the spring-block is elastic 2. Relevant equations W(elastic) = Fx(cos0) Potential energy = mgh Kinetic energy = .5mv^2 F = kx W = .5kx^2 T = 2pi (sq rt (l/g) 3. The attempt at a solution (a) I think the elapsed time would have something to do with the period. If so, then T = 2pi (sq rt (l/g)) .. and in this case l would be D. So 2pi (sq rt (D/g)) (b) Well, the spring would give M some velocity before it goes off the platform. I think the mgh and 1/2mv2 would have to do something with it, since there has to be kinetic and potential energy. If the potential energy and kinetic energy make up the total energy used, then I can say .5mv^2 = mgh v = sqrt (2gh) (c) The work done is elastic, and should by .5kX^2.. the only problem is I'm not allowed to use k. If k = 4pi^2 x M / T^2, I could just plug in the value I got for T in the T for this equation, then solve for K, and plug that into the work equation.. I think. (d) k = 4 pi^2 x M / T^2 .. so based on my solution for T in the first equation I could then get the answer.