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Work and Energy

  1. Dec 6, 2012 #1
    Can someone explain to me when you use 1/2mv^2-mgh as opposed to 1/2mv^2 or mgh
     
  2. jcsd
  3. Dec 6, 2012 #2

    K^2

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    In what context?
     
  4. Dec 6, 2012 #3
    So we can know were you are coming from, are you able to answer the following -

    What is 1/2mv^2 ?

    What is mgh ?
     
  5. Dec 6, 2012 #4
    I know that 1/2mv^2 is Kinetic energy and mgh is potential energy. My question is in some example problems they use 1/2mv^2 and in others they use 1/2mv^2-mgh.
     
  6. Dec 6, 2012 #5

    CWatters

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    You will have to tell us what problems.

    In some case the total energy of a system is KE + PE. It's possible that the direction of "h" in your question has been defined such that PE = mg(-h) making your equation really...

    1/2mv^2 + mg(-h)

    or it might be nothing of the sort. Tell us about the problem where you saw this equation.
     
  7. Dec 6, 2012 #6
    Good,
    So if a problem asks for the kinetic energy of an object, which formula would I use?

    If a problem asks for the potential energy of an object, which formula would I use?

    Do you know about the concept of conservation of energy? When a ball is falling what is happening to its kinetic energy? Its potential energy?

    See if you can figure out from these clues where 1/2mv^2 - mgh would be useful.

    EDIT to fix sign: should be 1/2mv^2 + mgh
     
    Last edited: Dec 6, 2012
  8. Dec 6, 2012 #7
    While the kinetic energy increases, the potential energy will decrease. Is that correct? I've thought about it, but i still do not understand why I would use 1/2mv^2-mgh
     
  9. Dec 6, 2012 #8
    Good, yes that is correct.

    Conservation of energy means that the *total* energy remains the same (this is what conservation laws are, some quantity remains unchanged even though its constituent parts are changing).

    So if the kinetic energy increase by some amount (say x), now much will the potential energy have to decrease?

    What will happen to the quantity 1/2mv^2 - mgh is this case?

    EDIT: should be 1/2mv^2 + mgh
     
    Last edited: Dec 6, 2012
  10. Dec 6, 2012 #9
    So 1/2mv^2-mgh would also increase by x?
     
  11. Dec 6, 2012 #10
    No, think about it carefully (don't get frustrated, you are very warm).

    Start by answering: If the kinetic energy increase by some amount (say x), now much will the potential energy have to decrease?
     
  12. Dec 6, 2012 #11
    the potential energy will decrease by x. does that mean that 1/2mv^2-mgh also decreases by x?
     
  13. Dec 6, 2012 #12
    No,

    Here is what you have so far:

    Kinetic energy is 1/2mv^2

    Potential energy is mgh

    Kinetic energy increases by x

    Potential energy decreases by x

    What happens to 1/2mv^2 - mgh?

    EDIT: should be 1/2mv^2 + mgh
     
    Last edited: Dec 6, 2012
  14. Dec 6, 2012 #13
    would 1/2mv^2-mgh remain the same?
     
  15. Dec 6, 2012 #14
    Bingo!

    1/2mv^2 - mgh is the total energy and it is conserved. In other words when a ball is falling, the quantity (1/2mv^2 - mgh) will remain constant no matter where that ball is in its flight.

    (of course we are ignoring air resistance and other complicating stuff).

    *So*, consider this problem, I a drop a ball from 10meters and want to know how fast it was going when it hit the ground.

    Calculate (1/2mv^2 - mgh) before the drop (1/2mv^2 = 0).

    Calculate (1/2mv^2 - mgh) at the ground (mgh = 0)

    And trust that the difference will not change.
     
  16. Dec 6, 2012 #15
    I understand that mgh would = 0 when it hits the ground, but why is it 0 before the drop?
     
  17. Dec 6, 2012 #16
    Look at my previous post carefully, what did I say was = 0 before the drop?
     
  18. Dec 6, 2012 #17
    Oh so the kinetic energy is 0 because it is not in motion?
     
  19. Dec 6, 2012 #18
    Exactly!

    I am just realizing that we have a sign error here, perhaps this was the source of your confusion.

    It should be 1/2mv^2 + mgh = constant right? Sorry I did not notice this before.
     
  20. Dec 6, 2012 #19
    thanks so much for your all your help and time!
     
  21. Dec 6, 2012 #20

    CWatters

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    ONLY for certain problems.

    Consider a pendulum swinging back and forth with no air resistance or friction.

    At the top of the swing:

    PE = Maximum = mghmax where h is measured from the bottom of the swing
    KE = Zero because it stops momentarily

    At the bottom of the swing:

    PE = Zero
    KE = Maximium (= 0.5mV^2)

    At any point in between:

    PE + KE = Constant = mghmax
     
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