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Hooke's Law

  1. Jan 12, 2005 #1
    A spring with k=45 N/m is .35m when pulled
    down with a 1.0 kg mass, what is the length of the
    spring when the mass is taken off? The answer in the
    back of the book is .13m (13 cm), I can't for the life
    of me figure out how they get that.

    I know that Hooke's Law is: F = -kx

    What I've done so far:

    F = (45)(.35) = 15.75 ---> Force it takes to pull down .35 meters, apparently using 1.0kg. Thus, if I divide 15.75 by 9.81, I should get 1.0kg, yes?

    15.75 / 9.81 = 1.60 ---> :bugeye:

    I must be doing something wrong, lets start backwards.

    F = (45)(.13) = 5.85 --> Force it takes to pull down .13 meters (this x is the answer, but right now i'm trying to find a proper F). 5.85 / 9.81 = .56, again, not 1.0kg. :(

    What am I doing wrong? Can anybody point me in the right direction?
  2. jcsd
  3. Jan 12, 2005 #2


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    Consider the spring without the mass.It has the length [itex] x_{0} [/itex].When acted by a force,its length is [itex] x [/itex].Hooke's law says that the magnitude of the force is proportional to the the length the spring is stretched under the influence of the force:
    [tex] F_{el}=k\Delta x=k(x-x_{0}) [/tex]

    Compute the force and [itex]\Delta x [/itex],then use the expression for the latter to solve your problem.

  4. Jan 12, 2005 #3
    Got it, thanks!

    Man, such a small thing, too.
    *sigh of relief*
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