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Hooke's law with a spring

  1. Oct 2, 2006 #1
    When a 4.00 kg object is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.5 cm.
    (a) If the object is replaced with one of mass 1.45 kg, how far will the spring stretch?


    -(1.5)(9.8)= -1568x
    I keep being told that this is wrong and yet I have not been able to figure out where I went wrong. Can someone point out what the problem is.

    b) how much work must an external agent do (i.e. a force coming from the 'environment'), to stretch the same spring 4.9 cm from its unstretched position?
    Here I used k=1568 and W=1/2kxi^2- 1/2kxf^2
    W=1.25J or -1.25 J
    Once again I'm being told that this is wrong and would like for someone to point out the problem.

    Thanks in advance:smile:
  2. jcsd
  3. Oct 2, 2006 #2

    Doc Al

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    Staff: Mentor

    I don't see any fundamental problem here (except for your misuse of "F = ma"--see below). But why did you use 1.5 instead of the given 1.45 kg? (You can also solve this using ratios.)

    One thing to point out: You are not using Newton's 2nd law here (F = ma), but you are using Hooke's law (F = kx). In this case, the F is the force that is stretching the spring--which is the weight of the hanging mass: w = mg.

    I only see two problems: (1) Your signs are off. Note that change in spring energy (or anything) is always final - initial. The work done is positive. (2) Check your arithmetic.
    Last edited: Oct 2, 2006
  4. Oct 2, 2006 #3
    technicaly hooke's law is F= -kx and the work done by a spring is infact i-f.
    Last edited: Oct 2, 2006
  5. Oct 2, 2006 #4

    Doc Al

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    Staff: Mentor

    True, but here we are finding the work done by an external agent in stretching the spring--which equals the change in the spring's energy--which is "final energy - initial energy". (The minus sign in Hooke's law tells you the direction of the spring's restoring force, which is opposite to the force applied by the external agent.)
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