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Hooke's law with spring

  1. Sep 28, 2006 #1
    When a 4.00 kg object is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.50 cm. (a)If the object is replaced with one of mass 1.50 kg, how far will the spring stretch?
    b) How much work must an external agent do (i.e. a force coming from the 'environment'), to stretch the same spring 4.00 cm from its unstretched position?

    For a) I used the equation F= -kx to solve for k.
    F=ma so ma=-kx since the object is vertical.
    (4.00)(-9.8)= -k(.025m)
    k= 1568 N/m

    then I plugged it back in to solve for x
    ma= -kx
    (1.5)(-9.8)= -1568x
    x= 0.00938 m

    Is there something I'm doing wrong?
    b) W= 1/2k(xi)^2 - 1/2k(xf)^2
    W= 0-1/2(1568)(.04m)^2
    = -1.25 J

    The homework says this is wrong but I can't figure out what else to do?
    Any help would be appreciated.
  2. jcsd
  3. Sep 28, 2006 #2


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    This is correct. Another way to do this is to notice that Hooke's law means a linear relationship between F and x. So when you F is twice as big (in this case, if you make the mass twice as big), the spring is stretched twice as far. In this case, the mass is 1.5/4.0 times as heavy, so x is altered by the same factor:
    [tex](1.50)/(4.00)*(2.5 cm)=0.938 cm[/tex]
    However, your method gives the value for k, which you need in the next exercise.
    The work done should be positive. You can see this by noting that the force you exert in the same direction as the displacement, or noting that by stretching a spring you increase the potential energy.
  4. Sep 28, 2006 #3
    Thanks for responding. I'm using a computer program for this homework and it's telling me that both answers are wrong. This is after I saw what you wrote. So I'm just double checking to make sure that there is absolutely nothing wrong with this. It could be a glitch in the program.
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