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Homework Help: Potential Elastic Energy & Kinetic Energy

  1. Jan 14, 2009 #1
    [RESOLVED] Potential Elastic Energy & Kinetic Energy

    1. The problem statement, all variables and given/known data
    A mass of 8.0 kg arrives at a spring, with the K coefficent (constante de rappel in french) of 250 N/m with a speed of 5.0 m/s It hits the spring and returns in the other way.

    2. Relevant equations
    [tex]E= 1/2mv^{2}[/tex]
    [tex]E= 1/2Kx[/tex]

    3. The attempt at a solution
    [tex]1/2mv^{2} = 1/2Kx[/tex]
    [tex](1/2)(8)(5)^{2} = (1/2)(250)x[/tex]
    [tex](1/2)(8)(5)^{2} = (1/2)(250)x[/tex]
    [tex]100 = 125x[/tex]
    [tex]100/125 = x[/tex]
    [tex]0.8 = x[/tex]

    This gives me a distance of 0.8 meters, I have indicated that the answer was 0.89 meters, now I am not sure if I mistakingly indicated 0.89meters or I made a mistake at some point.

    I appreciate your help in advanced.
    Last edited: Jan 14, 2009
  2. jcsd
  3. Jan 14, 2009 #2


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    If this is vertically dropped then there is the additional PE from gravity that needs to be accounted for in the potential energy transferred to the spring.

    Namely you have

    ½mv² + mgx = ½kx²
  4. Jan 14, 2009 #3
    Thank you for your help so far.

    I believe the formula for elastic energy that we learned is ½kx and not ½kx² but anyhow.

    I have tried with both formulas and did not get an answer closer to the 0.89 written answer, it also did not indicate at any point that it was horizontally dropped and there was nothing about gravity or any friction therefore there is none.

    I apologize in advance for any french-ized terms but I study physics in french so I try my best to translate this to english.
  5. Jan 14, 2009 #4


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  6. Jan 14, 2009 #5


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    If it is horizontal and frictionless then there is a straight conversion of Kinetic to Potential energy at maximum detent.

    If it is vertical then there is the additional change in gravitational potential over the distance of the detent.
  7. Jan 14, 2009 #6


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    If you know the answer is .89 then it is horizontal and frictionless.

    ½*8*52 = 100 = ½*250*x2

    x2 = 100/125 = .8

    x = .89
  8. Jan 14, 2009 #7
    Heh, I was writing the way on how I figured it out and you wrote it.

    THANK YOU VERY MUCH! I can't say how much I appreciate your help.

    Here's how I did it step by step
    [tex]1/2mv^{2} = 1/2Kx^{2}[/tex]
    [tex](1/2)(8)(5)^{2} = (1/2)(250)x^{2}[/tex]
    [tex]100 = 125x^{2}[/tex]
    [tex]100/125 = x^{2}[/tex]
    [tex]0.8 = x^{2}[/tex]
    [tex]\sqrt{0.8} = x[/tex]
    [tex]0.89 = x[/tex]
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