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Hooke's Law - Springs

  1. Sep 26, 2012 #1

    MG5

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    Hooke's law describes a certain light spring of unstretched length 37.0 cm. When one end is attached to the top of a door frame and a 8.00-kg object is hung from the other end, the length of the spring is 40.5 cm.
    (a) Find its spring constant.

    (b) The load and the spring are taken down. Two people pull in opposite directions on the ends of the spring, each with a force of 150 N. Find the length of the spring in this situation.


    I found out (a) is 2.24 kN/m but I don't know how to get to the answer.

    So far I've just converted all the units, so 37 cm to .37 m and 40.5 cm to .405 m.

    And I guess y would be .405-.37= .035 m
     
  2. jcsd
  3. Sep 26, 2012 #2
    do you have the definition of Hooke's Law?

    You should be able to use that to find k.
     
  4. Sep 26, 2012 #3

    MG5

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    Fs=-kx

    But I don't have the force. And I wouldnt use F=ma would I? doesnt seem right since its a spring.
     
  5. Sep 26, 2012 #4

    PhanthomJay

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    You do have the force. The weight of the object is the force. The object hangs at rest when the spring displaces 0.035 m.
     
  6. Sep 26, 2012 #5

    MG5

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    8/.035 does not give me 2.24. And how can that be the force its just the mass of the object.
     
  7. Sep 26, 2012 #6
    The weight of the object is the force. So far you've only used the mass. Also, be sure to check your units when you get the answer.
     
  8. Sep 26, 2012 #7

    PeterO

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    A mass of 8kg does not have a weight of 8N on the earth !!!! It is is even more than 8N on the moon!!
     
  9. Sep 26, 2012 #8

    MG5

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    Yeah I know I didnt see that he said the weight of the object and not the mass. I dont even know how to find the weight though
     
  10. Sep 26, 2012 #9

    PhanthomJay

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    Weight is mass times the acceleration of gravity, W = Mg, where g is 9.81 m/s^2 on Planet Earth. When the Mass is in kg and the acceleration is in m/s^2, the weight has units of kg*m/s^2, which is called Newtons for short. So an 8 kg mass has a weight on Earth of ___??____ Newtons. Then solve for the spring constant, k, per Hooke's law. What is the value and units of the spring constant?
     
  11. Sep 26, 2012 #10

    MG5

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    So the weight is 78.4 N. Just 9.8m/s^2 * 8 kg. And I don't know the value of the spring constant yet but I know its going to be N/m.

    And I know PEs=1/2kx^2

    Would x be the difference between the distances when the 8 kg weight hangs from the spring? So .035 m? .405-.37=.035

    So would it be PEs=1/2(.035m^2)k and then if thats right im not sure what to set the equal to or what else to solve.
     
    Last edited: Sep 26, 2012
  12. Sep 27, 2012 #11

    NascentOxygen

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    Yes.

    You earlier said Hooke's Law is F=-kx
    and you have determined F for a particular x,
    so it seems you are very close to determining k. :smile:
     
  13. Sep 27, 2012 #12

    MG5

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    I still don't know what x is though.
     
  14. Sep 27, 2012 #13

    PhanthomJay

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    You have already stated the value of x more than once. See your original post, where you called it y.
     
  15. Sep 27, 2012 #14

    MG5

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    What, .035m? F=kx, 78.4=k(.035) then solving for k doesnt get me 2.24
     
  16. Sep 27, 2012 #15

    PeterO

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    snip from initial post:

    I found out (a) is 2.24 kN/m but I don't know how to get to the answer.

    have you noticed that little k in the unit kN/m ?
     
  17. Sep 27, 2012 #16

    MG5

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    Yeah I was wondering what that was. Do I have to convert stuff.

    Edit: Oh I divide the answer I get from 78.4/.035 by 1000 and get 2.24. Would someone like to explain why 1000? My professor never went over this. Only N/m
     
  18. Sep 27, 2012 #17

    PeterO

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    What is the significance of the little k in

    Mass kg
    Distance km
     
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