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Springs and Hooke's law

  1. Feb 16, 2008 #1
    [SOLVED] Springs and Hooke's law

    1. The problem statement, all variables and given/known data
    An unstretched spring has a force constant of 1200 N/m. How large a force and how much work are required to stretch the spring by 1.00 m from its unstretched length?

    2. Relevant equations

    F= -k*s

    W= F * s

    3. The attempt at a solution

    I used Hooke's law and obtained a force of 1200 N (which was correct). But the displacement is only 1 meter, so work should have also been 1200 (J). But it's wrong anyway.

    The second part of the problem asks "How large a force and how much work are required to stretch the spring by 1.00 m beyond the length reached in part (a)?" so I multiplied the force constant by 2 to get 2400 N, and it was right. However, multiplying 2400 by 2 meters to give W of 4800 J was also incorrect. I don't know what I'm misunderstanding.
    Last edited: Feb 16, 2008
  2. jcsd
  3. Feb 16, 2008 #2

    Doc Al

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    Staff: Mentor

    As you do work on the spring to stretch it, the force is not constant. So you can't just multiply the final force times the displacement. (You can integrate, if you know how.)

    Hint: Have you studied elastic potential energy? How much energy is stored in a stretched spring?
  4. Feb 16, 2008 #3
    The course I'm in is *supposed* to be algebra and trig-based, but I am quickly learning that using calculus would make my life a lot simpler, if I only remembered how to do that stuff. We also haven't studied elastic potential energy yet... I looked on Wikipedia and it said this was found by integrating Hooke's law (which I would love to do if I knew how) and gave a formula. U = 1/2 (kx[tex]^{2}[/tex]) So I plugged in my values for the force constant and displacement, and got elastic potential energy of 600. I used this as my value for the work done in part a), and 600 J was correct.

    However... I tried the same strategy on part b), by using the same force constant and a displacement of 2 meters, but the resulting 2400 J was also wrong. What did you mean by not being able to multiply final force by displacement?
  5. Feb 16, 2008 #4

    Doc Al

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    Staff: Mentor

    That's because the question asked for the additional work needed to stretch the spring from 1 to 2 meters, not from 0 to 2. (Subtract.)
    I was referring to your earlier idea for calculating work. For example, you multiplied 1200 N x 1 m, which is incorrect because the force actually varies from 0 to 1200 N as you stretch the spring.
  6. Feb 16, 2008 #5
    oh! Thank you so much... it definately helps to "talk" to someone about this instead of just giving up. :smile:
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