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Work/Energy with spring

  1. May 9, 2008 #1
    Heres the problem

    http://i14.photobucket.com/albums/a322/guitaristx/adsf.jpg


    Heres my work

    http://i14.photobucket.com/albums/a322/guitaristx/ffff.jpg



    First off I want to make sure im understanding the question. Is it asking to find the speed at B after the collar is shot up by releasing the spring that was depressed 1.5 in.?


    And if that is correct, im trying to figure how to balance the equations incorporating the depressed spring....
     
  2. jcsd
  3. May 9, 2008 #2

    Hootenanny

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    Yes.
    You can find the velocity after the collar has left the spring by using conservation of energy.
     
  4. May 9, 2008 #3
    I understand that I need to use conservation of energy but the spring is not at rest.....so would this work?

    .5kx^2 = .5mv^2 + .5kx^2
    (x,1) ( x,2)




    and would the velocity be v^2 = a*r
     
    Last edited: May 9, 2008
  5. May 9, 2008 #4

    Hootenanny

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    Yes that is correct, but notice that when the collar leaves the spring x=0 (i.e. the spring is completely uncompressed).
     
  6. May 9, 2008 #5
    ok i calculated the numbers and came up with v = 31.12.....however I noticed when i divide 31.12 by the Weight of 6 oz. .... i come up with 5.18 which is the answer in the back of the book......but that doesnt make sense to me....ounces needs to be converted.....
     
  7. May 9, 2008 #6

    Hootenanny

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    Why would you want to divide the speed by the weight?
     
  8. May 9, 2008 #7
    no thats what i meant that it made no sense...im just trying to figure why i came up with 31 and the answer is 5.18 ft/s
     
  9. May 9, 2008 #8

    Hootenanny

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    Thus far you have only calculated the speed at which the collar leaves the spring, not the speed of the collar at point B. Which is what the question asks for.
     
  10. May 9, 2008 #9
    ok, so than I could come up with another equ.

    .5mv^2 = .5mv*2 + mgh

    and plug in the V I found on the left side of the equ.

    but for finding V on the right side of the equation do I need to take into consideration the radius of curvature, such as tangential or normal components..
     
  11. May 9, 2008 #10

    Hootenanny

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    Looks good to me :approve:
    No, since we are using conservation of energy, which is a scalar, direction doesn't matter. We are only interested in the speed.
     
  12. May 9, 2008 #11
    i realized when I did the first calculation with 1.5 inches for x that I didnt convert to feet.........so i converted that as well as everything else....so i get a V of 2.55 ft/s.....thats the collar leaving the spring


    so mass = .012
    10 in. = .833 ft
    1.5 in, = .125ft

    i plugged into the equ.


    .5(.012)(2.55)^2 = .5(.012)(V)^2 + .012(32.2)(.833)

    .039 = .006v^2 + .321

    V is supposed to be 5.18 but its not quite coming out.....
     
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