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Homework Help: Spring constant

  1. Nov 28, 2008 #1
    An elastic cord is 67.7 cm long when a weight of 58.7 N hangs from it and is 85.8 cm long when a weight of 88.1 N hangs froms it. What is te spring constant of this elastic cord?

    how i know which is the long of the spring when it is in equilibrium to then calculate the sprin constant?...

    or anyother help to solve this problem
  2. jcsd
  3. Nov 29, 2008 #2
    Hello once again,

    A review of Hooke's Law will reveal that the k, the spring constant, is a constant of proportionality. This means that F_1/x_1 = F_2/x_2 where x is the extension of the spring. Does this shed any light?

  4. Nov 29, 2008 #3
    ok but if i do that what i can get:


    58.7 N / .677 m = 88.1 N / .858 m

    and that doesn't help me too much... because i dont know the distance of the spring when it is at equilibrium
  5. Nov 29, 2008 #4
    how i know what is the extension of the spring?...
  6. Nov 29, 2008 #5
    That's not exactly true. Remember, x is the extension of the spring, not the length of the spring. So, if we let the equilibrium value be l , the extension in one case is (0.677 - l). Does that clear things up?
  7. Nov 29, 2008 #6
    ok i know that x is the value of the extension of the spring but how i know whta is the extension of the spirng when is at equilibrium to then calculate the extension from the diferencies of my value sof x with the equilibrium distance?....
  8. Nov 29, 2008 #7
    Since the extension x = (length of spring with load - length of spring at equil.):

    The eqn shld be:

    58.7 / (0.677 - l) = 88.1/ (0.858 - l)

    where l is the equil. length, solve for l, with is simple algebraic manipulation...
    Last edited: Nov 29, 2008
  9. Nov 29, 2008 #8
    edited the eqn, sorry, made a typo
  10. Nov 29, 2008 #9
    ok let me try and tell u....
  11. Nov 29, 2008 #10
    solving for l y get 0.316.... then i use F=kx solve for k and get

    k= F/x = 58.7/(0.677-0.316)=162.60 is this right?
  12. Nov 29, 2008 #11
    Yes, your answer agrees with my calculated answer.
  13. Nov 29, 2008 #12
    ok many thanks....
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