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Kinetic energy/Work

  1. Sep 28, 2005 #1
    I am a full-time student and Physics major. Currently, among my other classes, I am taking Mechanics and Calculus II at once, boy this semester is a bugger. I am studying like mad and not seeing results. I think I understand physics concepts much better than I understand all of the math that is involved! I certainly need to improve my math techniques. Here is one from the homework that has stumped me:

    Imagine three different masses are hung from a spring alongside a ruler. The first mass (F=mg=W=110 N) shows a displacement of 40 mm, the second mass (F=mg=W=240 N) shows a dis. of 60 mm, and the final mass of unknown weight is at 30 mm. Here's where I am having trouble: I need to find out a) which mark the spring will be at if there is no mass on the spring and b) the weight of the final mass.

    I'm confused! If I use Hooke's Law, F= -kx, I can find the spring constant k when I know the force and the position x, but how do we find k if we do not know x? I imagine we use an equation for work by a spring force,

    Ws= (1/2)k*xinitial^2 - (1/2)k*xfinal^2
    which involves finding the initial and final x positions to find work...but I cannot figure out how to set this up. Perhaps I am leaving out another crucial equation or step? Hmmm..... Any help in setting up the problem would be appreciated. Thank ya'll -

    Ever so lost,
    Gin
     
  2. jcsd
  3. Sep 29, 2005 #2

    mukundpa

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    Homework Helper

    Let the spring is at x0 mark with no weight. then
    x1 = 0.04 - x0
    x2 = 0.06 - x0 and
    x3 = x - x0

    Now write the equation

    W1 = k x1 and so on. Solve the three equations for K, x0 and x.
     
  4. Sep 29, 2005 #3
    Thanks for the help. Turns out my main problem was that I was assuming X-Xo was 0, and I kept getting W=k*0, which is just zero, and that certainly didn't make sense for the problem! It was actually irrelevant where we allowed Xo to be, any arbitrary amount works, and of course, that is why you recommended allowing Xo to be the unknown we solve for. Thank you for helping me to see this a different way!
     
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