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Energy conservation, and mgh

  1. Feb 28, 2008 #1
    my question is, when using energy conservation when do i put the potential energy of gravity negative? if its under where it set my potential energy to 0? or am i wrong, is gravity's potential energy always positive (except at my set u=0 point) ?

    can someone explain this to me, having trouble getting ANYthing done because i do not understand how to set up these homework questions

    for example, this is a ball pushed into a barrel, compressing the spring, ball has initial velocity 0
    here are the values,
    k = 667 N/m
    mass of ball = 1.5kg
    ball is pushed into tube .25 meters
    if u=0 is set at the mouth of the barrel, then the initial potential energy is mgh+.5kx^2?
    or is it -mgh+.5kx^2? -mgh-.5kx^2?
    Last edited: Feb 28, 2008
  2. jcsd
  3. Feb 28, 2008 #2
    You get to define where the potential energy is 0. Everything above that is positive, below that negative.
  4. Feb 28, 2008 #3
    what about the springs energy is that negative as well?
    meaning if i set the muzzle of the gun to be potential energy 0,, is the total potential energy at the compressed initial part
  5. Feb 28, 2008 #4
    Yup, because you also defined the springs 0 potential energy mark at y=0
  6. Feb 28, 2008 #5
    im trying to find the velocity once it is at y=0 (once it leaves the gun)
    i am setting

    [tex] -mgh - \frac{1}{2}kx^{2} = \frac{1}{2}mv^{2}[/tex]

    i solve this for v and i receive


    i do not receive the correct answer, i get 5.72

    here are the values,
    k = 667 N/m
    mass of ball = 1.5kg
    ball is pushed into tube .25 meters

    i am placing -.25 into the formula since it is below the muzzle, anyone see where i went wrong?
    Last edited: Feb 28, 2008
  7. Feb 28, 2008 #6
    U = mgh+.5kx^2

    You get the negatives from the the springs compression and the height, you didn't use
    -mgh-.5kx^2 THEN also set x=-.25 did you?
  8. Feb 28, 2008 #7
    yes i did, i just tried again with correct negatives i think (i see where i went wrong last time)
    and i got answer: 3.38m/s, wrong still though

    [tex] mgh + \frac{1}{2}kx^{2} = \frac{1}{2}mv^{2}[/tex]
    Last edited: Feb 28, 2008
  9. Feb 28, 2008 #8
    That would be the problem:


    Where x=-.25
  10. Feb 28, 2008 #9
    The potential energy of the ball is +mgh. If the ball is below th etop of the barrel, h is negative, so mgh will then be negative. The elastic energy of the spring is always equal to +kx^2/2 if its unstretched position is at the top of the barrel.
  11. Feb 28, 2008 #10
    jesus wrong again, i got 22.89166 m/s this time, lol

    i have used

    o nvm i forgot the sqrt, i got correct answer now! 4.784

    thank you feldoh, kamerling
    Last edited: Feb 28, 2008
  12. Feb 28, 2008 #11
    4.784 m/s once it leaves muzzle
    solved now though, i didnt know how to properly set any of the initial and final energies, so i couldn't solve it initially, i got it now though.
    thanks again
  13. Feb 28, 2008 #12
    Ah, that's good. I was beginning to think I just made up everything I told you since it's been a while since I've done that lol XD
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