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Find spring constant

  1. Nov 22, 2015 #1
    1. The problem statement, all variables and given/known data
    It requires 49 J of work to stretch an ideal very light spring from a length of 1.4 m to a length of 2.9 m. What is the value of the spring constant of this spring?

    2. Relevant equations
    u=0.5kΔx2

    3. The attempt at a solution
    [tex]
    u=\frac{1}{2}k\Delta x^2\\
    k=\frac{2u}{\Delta x^2}\\
    k=\frac{2*49}{(2.9-1.4)^2}\\
    k=44 \ \texttt{N}/\texttt{m}
    [/tex]
    Apparently, the answer is 15 N/m.

    EDIT:
    It appears they have done this instead:
    [tex]
    u=\frac{1}{2}kx_f^2 - \frac{1}{2}kx_0^2\\
    u=\frac{1}{2}k(x_f^2-x_0^2)\\
    k=\frac{2u}{(x_f^2-x_0^2)}\\
    k=\frac{2*49}{(2.9^2-1.4^2)}\\
    k=15 \ \texttt{N}/\texttt{m}
    [/tex]
    I thought Δx would be squared after the subtraction?
     
    Last edited: Nov 23, 2015
  2. jcsd
  3. Nov 23, 2015 #2

    SteamKing

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    You aren't given the unstretched length of the spring, only that it is stretched from a length of 1.4 m to 2.9 m.

    The energy it takes to stretch a spring, namely E = (1/2)k x2, is applicable only for a spring which is stretched a distance x from its unstretched length.

    http://hyperphysics.phy-astr.gsu.edu/hbase/pespr.html
     
  4. Nov 23, 2015 #3
    The equation you used is only for when the spring is at rest, Δx is calculated from the equilibrium point. You're work would be correct if they said the spring was stretched 1.5 meters from it's equilibrium point.

    Think about a spring that is as rest compared to a similar one that is stretched quiet far. The force exerted by an ideal spring increases as the spring gets further away from it's equilibrium point. If both are stretched 1.5 meters, the one that's already stretched quiet far would be harder to stretch than the one that was at rest. The equation you used would give the same answer for both springs.

    The equation under "EDIT" is the one you might want to always think of, and just assign zero to the second "x" if the spring is at rest.
     
  5. Nov 23, 2015 #4
    Bob Tran: As SteamKing points out, their answer is not correct either, unless the initial length of the spring is zero (which is silly). I think your original approach is a better interpretation of the problem statement than the given answer (although the problem statement is clearly underspecified).

    Chet
     
  6. Nov 23, 2015 #5

    gneill

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    One can approach the problem in terms of the definition of work done:
    $$ W = \int_{x_o}^{x_f} f(x)~dx $$
    $$ W = \int_{x_o}^{x_f} k~x~dx $$
    and arrive at the formula and result that the book gives.
     
  7. Nov 23, 2015 #6

    SammyS

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    @bob tran ,
    What book is this problem from?
     
  8. Nov 23, 2015 #7
    This is correct only if the unextended length is zero, right?

    Chet
     
  9. Nov 23, 2015 #8

    gneill

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    I believe it should apply over any segment of the stretch.
     
  10. Nov 23, 2015 #9

    SammyS

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    Chet is correct.
     
  11. Nov 23, 2015 #10

    gneill

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    You're right. My bad :oops:
     
  12. Nov 23, 2015 #11
    Not sure. It was from a practice worksheet.
     
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