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Potential energy for a spring law

  1. Sep 19, 2014 #1
    Hi guys! Thanks for this great forum. I have a question I need explanation about please...
    I just started 11th grade and it's my first time taking springs. In the book, they said :
    Ep= 1/2 k x^2
    I already understand all the previous laws like [w= -1/2 k (x2^2 _ x1^2)] and know where thay came from
    But THIS law is not explained in the book and I didn't get where it came from...
    What's x^2 ?? Why 1/2 and not -1/2??
    Can u please help me with this??
    Thank you :)
     
  2. jcsd
  3. Sep 19, 2014 #2

    f95toli

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    The force from a spring is given by k*x
    Energy is the integral of the force with respect to distance (x in this case)

    So we have ∫kx dx=1/2 kx^2
     
  4. Sep 19, 2014 #3
    Potential energy is an amount of energy available to do work.

    You said you understand this formula
    which describes the change in potential energy as the spring elongation (or compression) changes, or W = Ep(2) -Ep(1).

    A description not involving calculus
    --------------------------------------
    You should know that when you push an object by a force (F) a distance (x), the work you have done W = Fx. You could say that some amount of potential energy that you once had, has now been used to do work. This previous amount of potential energy is Ep = W.

    For a spring, the force is not constant but varies from a value of 0 at free extension to a value of F at compression.

    So, draw yourself a graph with force as the ordinate and distance the spring is compressed as the abscissa.
    The graph of the spring compression should be a straight line starting at (0,0) with slope k, and at the final compression of the spring the coordinates of force and compression are (x,F).

    Constant force:
    If the force to compress the spring had been constant the energy put into the spring would have been the work done by you or,
    Ep = Fx
    and obviously that is not the energy input into a spring during compression.

    Non-constant force:
    With a non-constant force during compression, but one that is proportional to distance of compression, you can see that the energy input is Fx/2, or
    Ep = Fx/2

    Knowing that F = kx, a substitution evaluates to
    Ep = 1/2 kx^2
     
  5. Sep 19, 2014 #4
    Why is it 1/2 and not -1/2.
    Potential energy of the spring has what value when it is not compressed?
    What is the potential energy with compression?
    What is the change in potential energy?
     
  6. Sep 19, 2014 #5

    russ_watters

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    Non-calculus explanation. W=f*d, but f varies linearly with d. So average force is half of peak force (force at full d).
     
  7. Sep 19, 2014 #6



    Well THANK YOU SOOOOOOOOOOOOOOOOOOOOOOOOOOO MUCH! :!!) :thumbs: I get it now!! I really thank you :-D
    And about -1/2 I tried getting the formula of the potential energy from the ( wfs = -1/2 k (x2^2 - x1^2) )
    But now I get it. It's my first time taking the lesson so excuse my stupidity :shy: :blushing: thanks again ^.^
     
  8. Sep 19, 2014 #7
    Thank you guys! I get it now! Thanks to all of you ;) <3
     
  9. Sep 19, 2014 #8

    jtbell

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    Note the "area" of a region on this graph has units of force times distance = work. The work done in compressing the spring is the "area" of the triangular region below the graph that 256bits decribed. What's the area of a triangle with width x and height F?

    Now substitute F = kx into that...
     
  10. Sep 19, 2014 #9

    sophiecentaur

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    This same factor of a half comes into the calculation for distance travelled in time t, under acceleration a :
    s = at2/2
    and in many other instances, like the energy stored in a Capacitor:
    E = CV2/2
    It's the same basic bit of Calculus that produces the same factor, all over Physics.
     
  11. Sep 20, 2014 #10

    sophiecentaur

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    I obviously need help with LaTex. The following looks OK on my preview but comes out in its original form when I post it.
    Thus
    $$ E = \frac{C \ V^2}{2} $$

    Wassup?
     
  12. Sep 20, 2014 #11
     
  13. Sep 20, 2014 #12

    sophiecentaur

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    That problem was crazy. I can now see the formula just as i want it. I may have been impatient at first. It seems to take a while to cope with the tex rendering.

    But, back to the question and my comment - there are so many times where a formula fits into different contexts with different variables. Just cos there's a smartypants answer for why, doesn't make it any less remarkable. :)
     
  14. Sep 20, 2014 #13
    well i have a serious problem with physics... i study all formulas and understand them and get examples and all... then i put a problem and i just like o_O ... like PLZ OMG I UNDERSTAND EVERYTHING! :( if I don't get full a full mark in physics next year (12th grade) and this current year (11th grade) all my life will fall apart...
     
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