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Momentum Formula Disagreement

  1. Oct 13, 2016 #1
    • Member advised to use the homework template for posts in the homework sections of PF.
    Hi, everyone, this is my first time posting.

    I have a problem set that states "A machine at a packaging facility places stationary packages of mass m onto a horizontal conveyor belt that is moving packages steadily and horizontally at speed v. Once placed on the belt, the packages start moving at speed v; that is, they are rapidly accelerated."

    It asks that I calculate the average mechanical power required by the belt by dividing the kinetic energy per package by the time interval between packages.

    I have P = 1/2 m v^2 / T.

    It then asks to compute the average force the belt is applying to packages.

    I have F = mv/T

    Then it says "power is force times velocity so the force answer can be turned into a power answer and compared to the power you computed above."

    I get P = mv^2/T

    Finally, the set asks "Do you have a discrepancy? If so, why? Which answer is more correct?"

    I have wracked my brain for hours on this and cannot figure out why the two power formulae do not match. I've had one idea, but it's not really leading me to an answer; using the total time interval might be messing things up because power is only concerned with the time interval over which work is done. I thought if I knew the value for that time interval and plugged in the numbers the results of the formulae would match. However, I tried assigning values to the variable and plugging then in and the formula did not match. Can someone help please? Thank you!
     
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  3. Oct 13, 2016 #2

    haruspex

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    You have effectively used two conservation laws. Each has its limitations. Which might be in question here, and why?
     
  4. Oct 13, 2016 #3
    Hi, haruspex, thanks for the response. I think it's conservation of energy versus conservation of momentum. Perhaps it's accurate to say that power based on force and velocity is more accurate than that based on kinetic energy because there are likely factors that complicate the conservation of energy such as friction in the conveyor belt system. Does that sound right?
     
  5. Oct 13, 2016 #4

    haruspex

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    Yes.
    That is still assuming conservation of work, effectively. What would render conservation of momentum invalid? Is it valid here?
    Not necessarily friction. If the package drops onto a rubber belt it might be that there is no slippage, but losses occur as oscillations in the rubber, which dissipate. Any time there is a sudden acceleration you should suspect work is not conserved.
     
  6. Oct 13, 2016 #5
    Now that I think about it, an outside force can complicate both conservation of energy AND conservation of momentum, so neither power formula is totally valid here. I need to think about which one is more "messed up" if outside forces come into play.
     
  7. Oct 13, 2016 #6

    haruspex

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    What outside force? The force of the belt, right? But that is the force we are interested in.
    What else do you need in order to find the momentum change produced by a force?
     
  8. Oct 13, 2016 #7
    I was thinking the outside force would be something like the force of a box being dropped, causing oscillations, like you said, or friction.

    To figure out momentum change, I need to know the time during which the force is applied. Ft = mΔv
    That leads me back to my original thinking; we're not using the actual time during which the force is being applied; we're using a much longer time period finding the average.
     
  9. Oct 13, 2016 #8
    A classmate and I have come up with somewhat of an answer:

    Power based on ΔKE is half the power based on force. This is because when based on force, the power formula uses the average velocity, while the ΔKE-based power uses only the final velocity. Given that the actual power required is likely higher than using conservation laws would suggest, because of forces such as friction opposing forward motion, it is probably more accurate to use force-based power given that its magnitude is higher.

    I am not confident in this answer, but it's the best I can come up with. If anyone can let me know what they think of this, I'd appreciate it.
     
  10. Oct 13, 2016 #9

    haruspex

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    You know the momentum change on each package from mΔv, and you want the average force over all time, not just the time while one package is changing speed.
    That is right, but a bit hard to put up as a rigorous argument. Better to use the momentum approach.
     
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