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Energy Principle with Springs

  1. Oct 29, 2013 #1
    1. The problem statement, all variables and given/known data
    So we have a test this Friday over several topics in modern mechanics. Our professor gave us last year's test to use as a study guide, but hasn't posted the solutions yet. I'm not sure I'm doing this one right so I was hoping someone could check my work! I feel like my number is a bit big.



    A spring whose stiffness is 3500 N/m is used to launch a 4 kg block straight up in the classroom. The
    spring is initially compressed 0.2 m, and the block is initially at rest when it is released. When the block is 1.3
    m above its starting position, what is its speed? Be sure to specify what objects are in your system and what
    objects are in the surroundings. Show all your work, starting from fundamental principles and/or definitions

    system: block, spring, gravity
    surrounding: none

    Us = Potential Energy of Spring
    Ug = Potential Energy of Gravity
    ks = spring stiffness
    si & sf = stretch


    2. Relevant equations

    Ef = Ei + W

    Usf + Ugf + Kf = Usi + Ugi + [STRIKE]Ki[/STRIKE] + [STRIKE]W[/STRIKE]

    Kf = Usi - Usf + Ugi - Ugf

    .5mvf2 = .5ks(si - sf)2 + mg(yi - yf)

    vf2 = (ks(si - sf)2)/m + 2g(yi - yf)

    vf = sqrt((ks(si - sf)2)/m + 2g(yi - yf))




    3. The attempt at a solution

    I might have messed up the stretch, I'm currently looking over it.

    I got vf = 44.038 m/s
     
    Last edited: Oct 29, 2013
  2. jcsd
  3. Oct 29, 2013 #2
    Seems like you need to double check your s's and y's. Are these potential energies(Us and Ug) supposed to add or subtract? Did you use the correct signs for your s's and y's?
     
  4. Oct 29, 2013 #3
    I used the quantity -.2 for both si and yi since the spring is compressed. I'm not sure what you are saying. Usf and Ugf should be subtracting.
     
  5. Oct 29, 2013 #4
    I'm not saying what you did is wrong, but to make sure you are consistent with your signs. What values did you use for delta s and delta y?
     
  6. Oct 29, 2013 #5
    delta s and delta y = -1.5

    The stretch became +2.25 because it was squared

    The thing is though that it is si-sf and yi-yf, so it is different I guess you could say.
     
  7. Oct 29, 2013 #6
    Hmm, I think you can assume all the energy from the spring transfers into the object such that the spring stops decompressing at its equilibrium length. So delta s is just the distance it is compressed, and delta y is how high the block is from where it started.
     
  8. Oct 29, 2013 #7
    Okay I see what you mean. After the spring is released, the block flies off the spring so Usf is actually just 0, right?

    For that, I got Vf = 2.366 m/s which does sound much more realistic haha.
     
  9. Oct 29, 2013 #8
    I think that's around the value I got as well (don't remember, but it's reasonable). Did you use (-)1.3 for delta y? If not, then your velocity is probably a little low.
     
  10. Oct 29, 2013 #9
    No, I used -1.5, because when the block is at 1.3, it is 1.5m above the point it started from.
     
  11. Oct 29, 2013 #10
    The problem says clearly that is 1.3 m above the starting position. Not 1.5.
     
  12. Oct 29, 2013 #11
    Yeah you are right. Sorry I haven't slept in a while, so I'm kind of delusional.
     
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