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Spring constant based on change in potential energy

  1. Mar 7, 2014 #1
    1. The problem statement, all variables and given/known data
    A compressed spring has 138 J of energy stored in it. When it is decompressed by 125 cm, it now stores 82.6 J of energy. What is the spring constant


    2. Relevant equations
    General:
    ΔU = 1/2kxf2 - 1/2kxi2
    U(x) = 1/2kx2
    U1 = 1/2kx12
    U2 = 1/2k(x1 + 1.25 m)2

    3. The attempt at a solution
    U1 = 1/2kx12

    k = 2U1/x12

    U2 = 1/2 (2U1/x12)(x1 + 1.25 m)2

    U2 = U1/x12 (x12 + 2.5m x1 + 1.5625 m2

    U2 = U1 + (2.5 m x1/x1) + (1.5625 m2 U1/x12)

    x12 ΔU = (2.5m U1/x1 + 1.5625m2 U1/x12) x12

    ΔU x12 = 2.5m U1 x1 + 1.5625m2 U1

    Substitute values and get them all to one side:
    55.4J x12 - 345Jm x1 - 215.625Jm2 = 0

    Quadratic formula gives:
    x1 = 6.8 cm and -.57 cm

    138J = 1/2k (.068 m )2

    276J/.00462 m2 = 59740 N/m = k? Obviously that can't be right.
    Any help on this would be great!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 7, 2014 #2

    rude man

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    This equationis wrong; the middle term does not have units of energy.

    I continue to recommend that people leave numbers out until the end, and that they check dimensions term-by-term everywhere.
     
    Last edited: Mar 7, 2014
  4. Mar 7, 2014 #3

    TSny

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    Hello sanctifyd83. Welcome to PF!

    Is ΔU positive or negative?
     
  5. Mar 7, 2014 #4
    Thank you, rude man. Yes I input the middle term incorrectly. It should be 2.5 m/x1. I did it correctly when I worked it out though, I don't believe it affects the rest of my "solution."
     
  6. Mar 7, 2014 #5

    Thank you for that. Yes, the value ought to be negative. This gives the value for x1 slightly different negative values which, when squared, become positive. The result is still an extremely large value for k.

    x1 would equal -5.5 or -.7
    Using -5.5 k would be around 90000N/m and -.7 would give an even larger number.
     
  7. Mar 7, 2014 #6

    TSny

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    I agree with the -5.5 m. (You should see a reason why you have to throw out -0.7 m)

    But I don't see how you're getting such a large value for k. Can you show how the x1 = -5.5 m leads to k = 90000 N/m?
     
  8. Mar 7, 2014 #7
    Forgive me as I'm trying to understand the material but I think that x1[/SUB's units would be in cm? If this is the case -5.5 cm turns into -.055 m which when squared becomes .003 m2. 276J/.003 m2 = 92000N/m.
    How do you know what the variable's initial units are? I assumed we were working with it in cm since the problem initially had the length units in cm. I know that I converted it to meters but it seems that the quadratic function cancelled all units except the ones for x1.
     
  9. Mar 7, 2014 #8

    TSny

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    The energy is given in Joules, an SI unit. Thus, distances should be in meters. So, it's good that you converted the 125 cm to meters. When you solve the equations, the distances will still be in meters.

    EDIT: Take your equation 55.4J x12 + 345Jm x1 + 215.625Jm2 = 0

    The Joules cancel out. Can you then see by inspection of the remaining units that x1 must have units of meters?
     
    Last edited: Mar 7, 2014
  10. Mar 7, 2014 #9
    That absolutely makes sense, I couldn't quite grasp that. It seemed like I was just arbitrarily inputting units. Thanks so much.
     
  11. Mar 7, 2014 #10
    Maybe I'm missing something:
    [tex]138 = \frac{1}{2}kx_0^2[/tex]
    [tex]82.6 = \frac{1}{2}k(x_0-1.25)^2[/tex]
    Divide one equation by the other to get x0 (the k's cencel). Then find k from the first equation.

    Chet
     
  12. Mar 7, 2014 #11
    I'm with you Chestermiller
     
  13. Mar 7, 2014 #12

    TSny

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    Yes, this is much nicer. The OP had already set up and worked the problem correctly except for a sign error, so I didn't want to start over.

    But it's certainly worth while to point out a better method of solution, and I should have done that. :redface:

    Thanks.
     
  14. Mar 8, 2014 #13
    Thanks so much everyone. It helped out a lot.
     
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