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Stretching a spring

  1. Apr 11, 2008 #1
    A force of 52N stretches a spring 0.73m from equalibrium. What is the value of the spring constant?

    F=-k * delta s
    52N= -k * 0.73m
    52N/0.73N = -k
    k= -71.2 N/m or -71 N/m with sig figs.
    The negative sign means that the force is opposite of the stretch.

    Is this right?
     
  2. jcsd
  3. Apr 11, 2008 #2

    Dick

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    Of course it's right.
     
  4. Apr 12, 2008 #3

    alphysicist

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    The spring constant k will be a positive number.

    The force F in Hooke's law is the force of the spring, not the external force that pulls on the spring. Also, F and (delta s) are vectors so we need to keep track of their directions.

    So, just to pick a specific direction, let the spring be pulled to the right, and also choose rightwards to be the positive direction. Then (delta s)=+0.73m. But to get the sign of F we need to find which direction the spring is pulling.
     
  5. Apr 12, 2008 #4
    So:
    both the force and the displacement are in the same direction since the sping is being stretched, but the k constant is pulling in the opposite direction so k = 71 N/m pulling the spring in the opposite direction of the stretch.
     
    Last edited: Apr 12, 2008
  6. Apr 12, 2008 #5

    alphysicist

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    If you pull a spring to the right, the spring pulls back against you to the left. So if delta s is positive, then F is negative.
     
  7. Apr 12, 2008 #6
    ok..

    -52N=-k*0.73m
    -52N/0.73m = -k
    -71 N/m = -k
    71 N/m = k

    Which is what I got before.
     
  8. Apr 12, 2008 #7

    alphysicist

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    If my posts made it seem like I was saying the magnitude of your answer was wrong, I definitely did not intend that. I was just pointing out that the sign of your original answer was wrong, and that it might be coming about because maybe you were thinking that F was the external, applied force when it is really the force of the spring.
     
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