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Potential energy of 2 sprinds on a single point

  1. Nov 9, 2008 #1
    1. The problem statement, all variables and given/known data
    Identical, massless springs, with spring constants k = 20 N/m, are connected in series, as shown. The springs are attached to rigid supports at x = ±L, where L = 59 cm. The equilibrium length of the springs is much smaller than the stretched length and can therefore be neglected in this problem.

    There is also a diagram where the springs are on either side of a point and the positive x direction is to the right and the positive y direction is up



    The connection point between the springs, initially at x=0 cm, y=0 cm, is pulled to x=17 cm, y=27 cm, and held there.
    2. Relevant equations

    What is the potential energy of the system now?



    3. The attempt at a solution

    It said hint:For this problem it is a good idea to use vectors in component form

    so I said Usp(y)=.5(20)(.27^2) and then multiplied that by 2 for the y component

    for the x component I said Usp(x)= -.5(20)(.59+.17)^2 + .5(20)(.59-.17)^2 for the x component.

    I then tried to use the pythagorean theorum to find the missing side as the question needs only one answer ie. it can't be in vector form.
     
  2. jcsd
  3. Nov 10, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi dal11! Welcome to PF! :smile:

    Yes, PE = .5k(r12 + r22) = .5k(x12 + x22 + y12 + y22).

    (but why the minus sign?)

    I don't understand what you mean by the missing side :confused:
     
  4. Nov 10, 2008 #3
    Oh thank you, um when I said the missing side I meant that I made a triangle out of the x PE and the Y PE because I thought that the X component of PE on the left was bigger so it would cancel the right out. I will try your method thanks.
     
  5. Nov 10, 2008 #4
    I have tried the question using your method but I am still not getting the right answer:(
     
  6. Nov 10, 2008 #5
    The potential energy is not a vector so it does not have components.
    The potential energy of the spring is 1/2 k L^2 where L is the length of the spring.
    You have to add the energy of the first spring and the energy of the second spring.
    The lengths of the springs are given by
    L1^2=(59+17)^2+27^2 (in cm)
    L2^2=(59-17)^2 + 27^2

    By using these lengths you'll find the energy in the new position.

    (If you want to find the change in energy then you subtract the initial energy when both spring had L1=L2=59 cm)
     
  7. Nov 10, 2008 #6

    tiny-tim

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    Show us your calculations. :smile:
     
  8. Nov 10, 2008 #7
    Ok but what do I do when I have the potential energy in each spring, add them together or subtract them?
     
  9. Nov 10, 2008 #8
    I got the answer, thank you so much
     
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