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Help with Kinetic/Potential Energy

  1. Jan 26, 2006 #1
    Hi, I am trying to find the percent PE to KE converted using PE/KE times 100. However, when I do this I get extremely high percentages (in the thousands or so). I know this is not right, what am I doing wrong?

    Ex) PE=.039
    KE=.0028

    When I do it, it comes out to 1392.9% converted. I know this cannot be right. Please help!! Thanks!!
     
  2. jcsd
  3. Jan 26, 2006 #2

    berkeman

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    Do it with respect to the total energy. The PE and KE will each start out as some percentage of the total energy, and after some of the energy trades off, there will be two new percentages with respect to the total (the total should stay unchaged). That change in the percentages is the percent change....

    Oh, and be sure to include units in your calculations. PE = 0.039 whats?
     
    Last edited: Jan 26, 2006
  4. Jan 26, 2006 #3
    Wait....So I calculate the PE + KE and then multiply that by 100 to get a percent and subtract the percent I got from 100%? I think I am kind of lost....

    and my values are Joules!!! Thanks for the catch tho!!!
     
  5. Jan 26, 2006 #4

    berkeman

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    Well, in your example, PE = 39mJ and KE = 2.8mJ. The total energy is the sum, which is 41.8mJ. The PE is 93.3% of the total, and the KE is 6.7% of the total energy.

    Are those the starting or ending numbers for the PE and KE? What mechanism is causing a change from PE<-->KE? Like, is it a ball falling off a platform or something?
     
  6. Jan 26, 2006 #5
    It is a penny going down a curved chute and out the end. So that is the starting PE. Then we measured the distance the penny traveled from the chute, and calculated the velocity. We used the velocity to find the KE.
     
  7. Jan 26, 2006 #6
    So, then the total energy to begin with if the penny is at rest is just the PE, and at the end it has only the KE. So the percent is KE/PE * 100 I think.
     
  8. Jan 26, 2006 #7
    The question is: State the percent PE to KE. I thought PE would get divided by KE? Am I wrong here?
     
  9. Jan 26, 2006 #8

    Gokul43201

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    Chrisdapos : PLEASE write down the original question EXACTLY as it appears in your text/homework/lab notes/class notes/etc. What you just wrote down is very ambiguous and bordering on meaningless.
     
  10. Jan 26, 2006 #9
    Think about it. The percent of the initial energy, PE converted to KE is KE/PE * 100
     
  11. Jan 26, 2006 #10

    Gokul43201

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    No, the percentage conversion would more likely be something like [itex]\frac {\Delta PE}{PE_{initial}} * 100 [/itex]
     
  12. Jan 26, 2006 #11

    berkeman

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    As a side note, you can't use just the exit velocity of the penny to calculate its KE. What's special about the penny that gives you an extra energy term in this setup?.....
     
  13. Jan 26, 2006 #12
    delta PE = zero, there is no potential energy left.
     
  14. Jan 26, 2006 #13

    berkeman

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    Right, which is why you can't just divide them. In a regular falling object problem, 100% of the PE gets converted into KE, because the datum of PE=0 is where the falling object passes (or hits). But again, in the case of this penny, some of the PE is going somewhere else.....
     
  15. Jan 26, 2006 #14
    So you are saying that 0.0028 is not 7.2 % of 0.039?
     
  16. Jan 26, 2006 #15

    berkeman

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    Dunno about the numbers, but what I'm saying is that the delta PE did not all go into the 1/2mv^2 linear velocity (hint) KE of the penny.....
     
  17. Jan 26, 2006 #16
    The percent errors work out if I do KE/PE times 100, and they make sense. I dont know why my teacher had as calculate KE from exit velocity either, because that is a new system.....it doesnt really matter tho, thanks for all of your help!! The initial question was What is the percent PE to KE, if anyone has the method of solving this, I would be glad to know it for future uses. Thank you!!!
     
  18. Jan 27, 2006 #17

    Gokul43201

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    There is no method of solving this because the question, as stated above, is meaningless...heck, it's even ungrammatical.

    The reason this thread got nowhere is that no one knew what the question really was. Berkeman's concern about the rotational KE is valid, but it's not clear whether that needs to be calculated at all.
     
  19. Jan 27, 2006 #18

    berkeman

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    I agree with Gokul. The question as stated is meaningless, because PE has to be defined with respect to some datum. Usually in KE<-->PE tradeoff questions, there is an obvious datum, and the object of the question is to understand how the total energy is conserved, and use that fact to help calculate the overall changes in KE and PE.

    But my guess in this problem with the penny (even though it is not stated by the OP) is that the professor wanted the sharper students to realize that the rotational energy of the penny takes some of the available energy. So I think the correct way to solve this problem is to define the bottom plane of the ramp as the zero PE datum, calculate the initial PE from the starting height and mass of the penny, measure the final linear velocity of the penny, and calculate the moment of inertia for the penny to get the linear KE and angular KE for the penny. Then show the percentage of the initial PE that went into the final linear and rotational energy terms. The initial PE is 100% of the available energy, and the final rotational plus linear KE also add up to that 100% total. The interesting thing in this problem is how the rotational energy factor slows down the linear velocity at the bottom of the ramp. The lower the moment of inertia of the rolling object, the higher the exit velocity at the bottom of the ramp.

    That's my guess as to what the prof was trying to get at. Too bad the problem was stated so badly.
     
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