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Potential Energy on an ideal spring

  1. Feb 22, 2008 #1
    [SOLVED] Potential Energy on an ideal spring

    1. The problem statement, all variables and given/known data

    An ideal spring of negligible mass is 12.00cm long when nothing is attached to it. When you hang a 3.15kg weight from it, you measure its length to be 13.30cm.

    If you wanted to store 10.0J of potential energy in this spring, what would be its total length? Assume that it continues to obey Hooke's law. (also says there might be 2 answers)

    2. Relevant equations

    K1 + U1 + Wother
    = K2 + U2

    or

    Wel = 1/2ky1^2 - 1/2ky2^2
    = Uel1 - Uel2

    U = mgy

    3. The attempt at a solution

    12cm = .12m = y1
    13.30cm = .133m = y2
    10J = Wother

    1/2ky1^2 = 10J + mgy1 + 1/2ky2^2

    1/2k(.0144m^2) = 10J + (3.15kg)(9.8m/s^2)(.133m) + 1/2k(.0177m^2)

    -14.11J = 1/2k(.0177m^2-.0144m^2)

    -14.11J = 1/2k(.0033m^2)

    -28.22J = k(.0033m^2)

    -28.22m = (.003m^2)

    is this right? would I just divide -28.22m by .003m^2 and get a #? I'm pretty sure it's not right but I'm not sure of other equations to use, should I use kinematics as well?

    edit: or perhaps I should have another potential energy in there for y1? or use the potential energy for mass in y1 and solve for y2?

    or perhaps use this equation?

    K1 + Ugrav1 + Uel1 + Wother = K2 + Ugrav2 + Uel2

    seems like this would work better than what I was using?
     
    Last edited: Feb 22, 2008
  2. jcsd
  3. Feb 22, 2008 #2

    Kurdt

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    The potential energy of a spring is given by:

    [tex] U=\frac{1}{2}kx^2 [/tex]

    You can work out the spring constant from the initial info they give you.
     
  4. Feb 22, 2008 #3
    you saying I should do:

    10J = 1/2k((.133m)^2-(.12m)^2))

    20J = k(.0033m^2)

    20J/.003m^2 = k

    k = 6060?

    or should I use the gravitational potentiona energy from the 3.15kg mass?

    (3.15)gy = 1/2k((.133m)^2-(.12m)^2))?
     
  5. Feb 22, 2008 #4

    Kurdt

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    You need to work out the spring constant by how far the mass stretches the spring, then use the potential energy formula I gave to solve for x. Since the answer will involve a square root there will be two answers.
     
  6. Feb 22, 2008 #5
    so:

    10J = 1/2k(.133m)^2

    20J = k(.0177m^2)

    20J/.0177m^2 = k

    1129.9 = k

    and mgy = 1/2kx^2

    (3.15(9.8)(.133m)2/k)^1/2 = x?
     
  7. Feb 22, 2008 #6
    x = -8.5cm, 8.5cm ?
     
  8. Feb 22, 2008 #7

    Kurdt

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    The spring constant is a force per unit length. You can work it out by dividing the weight of the mass by how much the spring moves from equilibrium. Then solve for x in the potential energy equation.
     
  9. Feb 22, 2008 #8
    so k = (mg)/y2 = (13.5*9.8)/.133m = 232J?

    and is the potential energy in the equation you gave gravitational or elastic?

    cause I could solve it like this

    mgy = 1/2kx^2

    4.11J = 1/2kx^2

    8.22J = kx^2

    8.22J/232J = x^2

    .035 = x^2

    x = -.188m, .188m = -19cm, 19cm?\

    or is it:

    10J = 1/2kx^2

    20J = kx^2

    20J/232J = x^2

    .086 = x^2

    x = -.29m, .29m = -29cm, 29cm?
     
    Last edited: Feb 22, 2008
  10. Feb 22, 2008 #9

    Kurdt

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    No, the spring stretches 1.3cm from equilibrium. The units of the spring constant are not joules. The only purpose of the information about the mass is so you can determine the spring constant.


    Its the elastic potential energy. That is the energy stored in the spring.

    Now you know how to obtain the spring constant and you know how much energy you want to store you can solve for x in the potential energy equation.
     
  11. Feb 22, 2008 #10
    ooooooooh, it goes from 12cm to 13.3cm which is 1.3cm which .013m, so

    Fx = kx, k = Fx/x = = (3.15kg*9.8m/s^2)/.013m = 30.9N/.013m = 2377N/m

    and 10J = 1/2kx^2

    20J = kx^2

    20J/(2377N/m) = x^2

    x = .092m = -9.2cm, 9.2cm?
     
  12. Feb 22, 2008 #11

    Kurdt

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    Thats good but they want the total length. So that will be 12cm +/- 9.2.
     
  13. Feb 22, 2008 #12
    right the initial length when nothing was attached to it, so it would be:

    2.8, 21.2? thanks alot!
     
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