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Ratios with energy and mass

  1. Jul 27, 2011 #1
    1. The problem statement, all variables and given/known data
    A 2.6-kg block is hanging stationary from the end of a vertical spring that is attached to the ceiling. The elastic potential energy of the spring/mass system is 1.3 J. What is the elastic potential energy of the system when the 2.6-kg block is replaced by a 4.8-kg block?



    3. The attempt at a solution

    [tex]E_i = mgh + \frac{1}{2}kx^2[/tex]

    [tex]E_f = 2mgh' + \frac{1}{2}kx'^2[/tex]

    Now I am stuck....
     
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  3. Jul 27, 2011 #2

    tiny-tim

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    Last edited by a moderator: Apr 26, 2017
  4. Jul 28, 2011 #3
    But there is a mass....
     
  5. Jul 28, 2011 #4

    tiny-tim

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    hi flyingpig! :smile:

    you can only use Ei and Ef (conservation of energy) when there's some process in which energy is conserved

    in this case, the mass is placed carefully in the equilibrium position …

    (or it's allowed to oscillate until it loses enough energy to achieve equilibrium)

    there's no energy conservation process to analyse! :redface:

    use a forces equation (not an energy one) to find a relation between x and m :wink:
     
  6. Jul 28, 2011 #5

    PeterO

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  7. Jul 29, 2011 #6
    How do we know the object was placed at x = 0?
     
  8. Jul 30, 2011 #7

    tiny-tim

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    it wasn't …

    by "equilibrium", i meant the position in which the weight balances the tension :wink:
     
  9. Aug 3, 2011 #8
    I am still stuck...
     
  10. Aug 3, 2011 #9

    PeterO

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    hanging a 2.6 kg mass on the spring will extend the spring.

    the F = kx formula for a spring shows the connection between force applied, extension x and the spring constant k.

    You only know one of these, F , since that is the weight of the block.

    However, you also know how much energy was stored by that extension

    E = 1/2 k x^2 = 1.3 J I think it was? [it was given]

    now 1/2 kx^2 equals x/2 x kx - but you know the value of kx [see above] so you can work out what x is, and thus get out k.

    Once you know that you can find what happens with the new mass.

    Of course if you are good at variation, you can find the answer more directly, but most people aren't good at variation, and if you were you would have the answer by now.
     
  11. Aug 3, 2011 #10
    Are you suggesting [tex]\frac{1}{2}kx^2 = E[/tex], so [tex]F = kx = \frac{2E}{x}[/tex]?
     
  12. Aug 3, 2011 #11

    PeterO

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    Using Variation - not sure you will Follow.

    Energy in a first spring spring is given by E = ½kx²

    Energy in second spring can be shown as E' = ½kx'² =

    ratio of second to first means E'/E = x'²/x² = (x'/x)²

    Now the spring is extended each time according to F = kx, and since it is a weight attached this means mg =kx

    By the same method, the new situation is m'g = kx'

    with ratios we have x'/x = m'/m

    In the energy ratio above that means E'/E = (m'/m)² or E' = E(m'/m)²

    so E' = 1.3 x (4.8/2.6)² which you can now evaluate to 4.4 J
     
  13. Aug 3, 2011 #12

    PeterO

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    Yes.
     
  14. Aug 3, 2011 #13
    Wow so this question combines dyanmics + energy
     
  15. Aug 3, 2011 #14

    But that didn't get me anywhere

    [tex]F = kx = \frac{2E}{x} = mg[/tex]

    [tex]\frac{4E}{x} = 2mg[/tex]
     
  16. Aug 3, 2011 #15

    PeterO

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    it won't get you anywhere is you just multiply each side by 2 ???

    What made you do that???
     
  17. Aug 3, 2011 #16
    Oh right...

    for some reason i thought 2.6 * 2 = 4.8

    So I should multiply by 4.8/2.6
     
  18. Aug 4, 2011 #17

    PeterO

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    I wouldn't. What were you trying to achieve.

    I would use the original data to find x, use that value to then find K.

    then use the new weight to find the new x, then the new x and the k value to find the new energy.

    [actually I would use variation, like the solution I gave, but the sequence outlined above shows the sequence you have to follow otherwise]
     
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