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Gravitational potential energy near the earth

  1. Oct 7, 2012 #1
    I am a bit confused about gravitational potential energy near the earth, namely the formula given by mgh.

    I know that potential energy is defined as U(x) = -W(x0 to x) where x0 is our chosen reference point. Let's take the earth's surface as the zero point and let's travel upwards to a point x. Since mg is pointing down and my displacement is pointing up, the dot product is negative so W(x0 to x) = -mgx. Therefore, U(x) = -(-mgh) = mgh.

    however, let's say i want to travel downwards instead. then since mg is pointing down and my displacement is also pointing down, the dot product is positive this time so W(x0 to x) = mgx (x is negative). but then U(x) = -(mgx) = -mgx. If i say travel downwards from x = -2 to x = -4, i get U(-4) - U(-2) = mg(4) - mg(2) > 0, which doesn't make sense since if i am going downwards, shouldn't my change in potential energy be negative?
     
  2. jcsd
  3. Oct 7, 2012 #2

    Simon Bridge

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    Do you have a higher PE on the ground or higher up?
     
  4. Oct 7, 2012 #3
    Using the formula U(x) = -mgx derived from taking the work going from the zero point (ground) to a point below the surface, it seems that points above the surface have negative potential.

    what i don't understand is why both approaches don't give me the same U(x). for U(x) = mgx, i traveled upwards so i can only use positive values of x (above the surface). If i want to use negative values of x, I travel downwards but i come up with U(x) = -mgx which seems to contradict with the function i got from traveling upwards since taking points above the surface, one gives me a positive PE and the other gives me a negative PE.

    is there a way to reconcile them?
     
  5. Oct 7, 2012 #4
    You lost a minus sign here.
     
  6. Oct 7, 2012 #5

    Simon Bridge

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    U=-W, W=F.d

    F=-(mg)j
    d=(y-y0)j if we go "up" (final minus initial)
    ... since y0<y, d>0 - for later reference, put h=|d|
    U= -(-mg)[+(y-y0)] = mgh

    if you go "down" the d=(y0-y)j = - (y-y0)j
    U=-(-mg)[-(y-y0)] = -mgh

    Keeping track of the minus signs is, indeed, the trick to keeping things consistent.
     
  7. Oct 7, 2012 #6

    Doc Al

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    As others have already explained, you are essentially double counting the minus sign.

    ΔU = -(F)*(Δx) = -(-mg)*(Δx) = (mg)*(Δx)

    When Δx is positive, ΔU is positive; when Δx is negative, ΔU is negative.
     
  8. Oct 7, 2012 #7

    Simon Bridge

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    @demonelite: this sound good to you?
     
  9. Oct 9, 2012 #8
    sorry about my late response. i'm still a bit confused. for the case "going up" i agree that F = (-mg)j and d = (y - y0)j since as you said it was final point (y) - initial point (y0).

    but for the "going down" case, F = (-mg)j but how come d = (y0 - y)j? isn't final point - initial point still y - y0?

    thanks for all the replies so far everyone.
     
  10. Oct 10, 2012 #9

    jtbell

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    In the "going down" case the object is going "from" y "to" y0, so final point - initial point is y0 - y which is negative.
     
  11. Oct 10, 2012 #10

    Simon Bridge

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    Thank you jtbell :)
    @deminelite123 - since y > y0, in order to go down you have to start at y.
     
  12. Oct 10, 2012 #11
    ah ok. what you said earlier makes sense now. thanks!
     
  13. Oct 10, 2012 #12
    Like everyone said, you missed a minus sign. :)
     
  14. Oct 10, 2012 #13

    Simon Bridge

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    No worries, happens to the best of us.

    In general, it is a good discipline to have the same labels refer to the same things. In this case: locations - when you were going down, you forgot to account for your y0>y so that y-y0 < 0. Try it with a diagram.
     
  15. Oct 10, 2012 #14
    Wait a second... The professor at MIT said that when I do positive work, gravity does negative work and when I do negative work, gravity does positive work. How does this make sense. When I do negative work from A to B then gravity is also doing negative work right? since it is -mg?
     
  16. Oct 10, 2012 #15

    Simon Bridge

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    Keep track of the directions ...

    If you lower a mass at constant speed, then you are exerting a force in the opposite direction to the displacement, and gravity is acting in the same direction as displacement.

    In your example, A must be higher than B so the displacement from A to B is negative.
     
  17. Oct 11, 2012 #16
    Thanks a lot
     
  18. Oct 11, 2012 #17
  19. Oct 11, 2012 #18

    Simon Bridge

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    No worries.
    Tracking the minus signs can be tricky - especially if things are slowing down as well as changing direction.
     
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