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Spring constant discrepency (force vs work)

  1. Oct 3, 2008 #1
    y = distance stretched vertically
    k = spring constant
    M = mass of weight on spring

    My Physics book says the work to stretch a spring is .5ky2. If I set that equal to the work done by gravity on the weight (9.8xM) and solve for the constant I get k = (9.8M)/(.5y)

    My Physics book also says that the force from a spring is ky. If I set that equal to the force of gravity on the weight (9.8M) and solve for the constant I get k = (9.8M)/(y)

    Why do I get a different constant??
     
  2. jcsd
  3. Oct 3, 2008 #2

    Doc Al

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    Staff: Mentor

    But the work done by gravity does not equal the work to stretch the spring--it's twice as much as needed to just stretch the spring.

    That's true. It's a restatement of Hooke's law.
     
  4. Oct 3, 2008 #3
    That would explain it. But where does the extra work go? Is that the right question? Why would only half go to the spring?
     
  5. Oct 4, 2008 #4

    Doc Al

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    Staff: Mentor

    The extra work goes into added kinetic energy of the mass.

    Here's what's going on. To just barely stretch the spring, you only need to exert a force equal to the spring force. But that spring force varies from zero (at first) to the full force (mg) at the equilibrium point. So the work needed--Average Force x distance--is only ½(mg)y = ½ky².

    If you gently hang the weight on the spring and bring it down to full extension realize that you are countering the weight of the mass with the upward force of your hand. But if you hang the weight on the spring and just let it drop, then the work done by gravity (when the weight reaches the point y) will be mgy = ky^2, but that extra energy goes into speeding up the weight. It won't just stop, it will keep going until the spring stretches enough to stop the motion. Ignoring friction and other energy leaks, the weight will keep oscillating up and down in simple harmonic motion.
     
  6. Oct 5, 2008 #5
    Thank you.
     
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