I am investigating how the displacement of an elastic band affects the distance an attached object will travel (the relationship between how far back you pull a slingshot and the distance an object will travel). What I am trying to do is find a formula where distance travelled is a function of elastic band displacement (in other words, a formula which says "x = ... d..."), but I ended up very confused. I tried equation manipulation with the following equations: W = F*d W = ΔE_{Kinetic} E_{Kinetic} = (1/2)*m*v^{2} and E_{Elastic Potential} = (1/2)*k*x^{2} I got close to the final equation, but I couldn't make it. Here are a few things to keep in mind: * In the final equation, where "x = ... d...", I don't want the force, F, to be in the equation. * We are assuming that initial kinetic energy was 0 (meaning that the initial velocity was 0). * We are assuming that the effects of friction and air resistance are negligible. Thank you.
I don't think the F of the elastic band is constant with distance. You could try hooke's law on it and measure the F values for various distances of pullback. Then you can use that to compute the work involved.
You won't need F, but you will need the "sprint constant" k for the sling shot. Note this is an approximation, since the graph of force versus stretch of rubber bands is not a straight line, but a curve. Example graph from a wiki article, which is also somewhat idealized. Usually the slope for the initial stretch is steeper, then decreases to a near straight line, then increases again at the limit of stretch (permanent deformation can occur if stretched close to the limit): Elastic_hysteresis.htm Archived web page showing a stretch versus tension graph for latex rubber used to launch radio control gliders (tension at 300% is about 175 lbs per square inch cross sectional area): rubberdata.htm
Thank you very much, but I have found a solution: First, due to conservation of energy, the elastic potential energy of the elastic band is equal to the kinetic energy of release. So: E_{Elastic Potential} = (1/2)*k*x^{2} E_{Kinetic} = (1/2)*m*v^{2} (1/2)*k*x^{2} = (1/2)*m*v^{2} (1/2)*x^{2} = (m*v^{2})/(2k) x^{2} = (2m*v^{2})/(2k) x = √((m*v^{2})/(k)) By finding v^{2} using the equations of motion, the mass, m, will be the function of spring displacement, x, while v^{2} and k remain constant (v^{2} will change depending on the desired displacement for the object to travel in the x or y axis).