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Is reference frame important when looking at work done?

  1. Jul 31, 2012 #1
    It must be, right? Obviously, if you're pushing a block by exerting a force F on it over a distance D on the ground, if you are in the frame of you or the block, your distance is 0 so it appears you're doing no work.

    I ask this question because I was doing a practice problem that should be really simple, but is bugging me. In it, there's a guy (mass m) in an elevator, which is accelerating upwards at constant acceleration a. At the given moment, its speed is V. It's easy to show that the normal force on him is F = mg(1 + a/g), which makes sense.

    But then they say that he now has a ladder in there, and is climbing up at a constant velocity v, and ask what his power output is. Naively, I say it's just P = Fv. However, the answer says P = F(v + V). This seems wrong for several reasons to me: a) If we look at the limit v = 0 (he's just sitting there, not climbing), it says he's still burning energy, and b) In the limit of a = 0 (the elevator is going up at a constant speed), he shouldn't be able to tell he's even in an elevator as opposed to a stationary room, so it should definitely degenerate into mgv, not mg(v + V).

    It seems like extra energy he'd have to expend climbing the ladder when a =/= 0 is included in F. Can anyone illuminate this for me?
     
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  3. Aug 1, 2012 #2
    Energy and power are indeed, as you thought, frame dependent quantities; something can be moving relative to me at one velocity in one frame and I say it has a different kinetic energy than it does when I am in another frame where it has a different relative velocity. In your problem, depending on what frame I am in, the velocity for the man will be different so certainly the power [itex]P=\vec{F}\cdot\vec{v}[/itex] will be different as well. The reason [itex]P=F(V+v)[/itex] is not making sense in both of your limiting cases is because you are interpreting it in the frame of the elevator. The man is moving at [itex]v+V[/itex] in the frame of someone at rest back on Earth's surface, and in the case of [itex]a=0[/itex] the force on him does reduce to [itex]F=mg[/itex] but he still is moving at a velocity of [itex]v+V[/itex] in the other person's frame. Most of the time you'll want to avoid calculating quantities in non-inertial (accelerating) frames anyway, because usually this requires special treatment.
     
  4. Aug 1, 2012 #3

    Dale

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    Yes, except that not only does it appear you are doing no work, you are in fact doing no work in such a reference frame. Note that humans can be quite inefficient machines in some circumstances, burning lots of energy but doing no external work.

    The answer is correct.

    You are neglecting the power input by the elevator to him.

    Consider a cable attached to a tractor towing a trailer. The power output of the cable is P, but the power input of the cable is also P, so the cable does not burn energy. This is true in all reference frames, despite the fact that P varies greatly between reference frames.
     
  5. Aug 1, 2012 #4

    cjl

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    Well, you are doing no work on the block, assuming that the frame is not accelerating. You are, however, doing work on the ground, and the block is doing equal and opposite work on the ground in that frame.
     
  6. Aug 1, 2012 #5
    Ok, so let me see if I'm getting this straight: His TOTAL power output is the answer, F(v + V), but that's only because the elevator is putting power FV into him. His NET power output is just what he's doing himself, Fv. Eh?
     
  7. Aug 1, 2012 #6

    Dale

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    Yes, I should have been more clear, thanks.
     
  8. Aug 1, 2012 #7

    haruspex

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    How is that part of his power output?
    Quite so, and this is still true when a > 0. It should be no different from an increase in gravity by that amount. His power output is m(g+a)v = Fv. The book(?) is wrong.
     
  9. Aug 2, 2012 #8

    Dale

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    If power is input to a system and the systems internal energy is unchanged then that same amount of power must be output or energy would not be conserved.
     
  10. Aug 2, 2012 #9

    jbriggs444

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    One way of looking at it is to step back to a free body diagram. What forces is the climbing man exerting? And what are the velocities of the objects on which these forces are exerted?

    The soles of his feet are exerting a downward force of F = m*(g+a) downward on a ladder that is rising with velocity V. He is exerting an upward force of f = m*g on a motionless earth due to gravity. The latter does does no work and accordingly does not contribute to power.

    By this analysis, the power that he is applying to his surroundings is negative and is given by F*V.

    That's not a very satisfying answer. I maintain that it is technically correct, but it certainly does not match any intuitively appealing notion of "power output".


    Another notion is that the "power output" of a man climbing a ladder measures the work performed by the man's legs on his body. Rather than being the downward force of the man on the ladder multiplied by the distance moved by the ladder, this is equal to the upward force of the ladder on the man multiplied by the distance moved by the man's body.

    By this measure, the power that is applied to the man by his legs is given by F*(V+v)

    EDIT:

    This result neglects the power output by the man's legs on the ladder which is given by -F*V. The correction yields the sought-after result: power output = F*v = m(g+a)v

    [resume original text]

    This answer is not very satisfying either. It does not match the principle that the man's effort should be independent of the velocity of the elevator.


    Now, of course this F*(V+v) power is identically equal to the power imparted to the man by the elevator/ladder. If the man were simply standing still on the ladder, his legs would have still been performing work on his body a rate of of F*V. It seems like cheating to count the elevator's motion as a contribution to the "power output" of the man's legs. So we are driven to look for another measure.

    Define the "power output" of the man's legs as the force they apply to the man's body multiplied by the velocity of the man's body relative to the platform on which he stands

    Now we get a figure of F*v = m(g+a)v that fits with our intuition of what the question should be asking.
     
    Last edited: Aug 2, 2012
  11. Aug 2, 2012 #10
    Assuming what I asked previously is right, I think I may have found a good way of thinking about it.

    I think what bothered me at first were the limiting cases I mentioned, and the fact that it seems pretty crazy to talk about total power output, for some reason. But let's say V is incredibly high, and there's now a window is the elevator. If he reaches his arm out and punches someone (like...vertically, I guess, in the direction the elevator is going) as he goes past, that person is going to feel not only the power of the punch from his muscles, but mostly the power from the elevator. But still, it definitely went through him, so he really did kind of output that power (even if most of it came from an outside source).

    Does that make sense?
     
  12. Aug 2, 2012 #11

    haruspex

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    Exactly. The power that comes from the elevator's acceleration is transmitted through his legs, but that does not make it part of their power output.
     
  13. Aug 2, 2012 #12

    Dale

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    Why not? It is power and it is delivered by a force he exerted. All power output eventually came from something else and is only being transmitted through any force. If that disqualifies something from being power output then there is no such thing as power output (except maybe the big bang).
     
  14. Aug 2, 2012 #13

    haruspex

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    According to that view, we can discuss the power output of the ladder, or of my running shoes... It is a bizarre use of the term 'output'. Throughput I would accept.
     
  15. Aug 2, 2012 #14

    Dale

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    "Throughput" works for me. I just don't know where you draw the line on "output", after all, an engine clearly has "power output" but all of that power came from the fuel so it really is "throughput" rather than "output" too. In any case, I don't think that "power output" is a well-defined term anyway, so it probably doesn't really matter.
     
  16. Aug 3, 2012 #15

    jbriggs444

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    How about "net power output".

    Consider that the man's legs have an interface toward the floor and an interface toward the rest of the man's body. Their "net power output" turns out to be invariant with respect to choice of inertial reference frame.

    If the elevator is moving upward rapidly, the legs are harvesting power from the ladder while delivering extra to the man's body.

    If the elevator is moving downward rapidly, the legs are harvesting power from the man's body while delivering power to the ladder.
     
  17. Aug 3, 2012 #16

    Dale

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    Sure, that is fine by me also. I don't think that "net power output" is a generally defined term elsewhere, so you are free to define it as you wish, and this seems a reasonable way to do so. You don't want to redefine existing terms, but making new ones is not a problem at all.
     
  18. Aug 3, 2012 #17

    haruspex

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    That doesn't resolve the issue of how to interpret the original question. Absent a formal definition, it strikes me as unreasonable to read it as anything other than net power output, and highly unlikely that it meant otherwise. The book answer is flat wrong.
     
  19. Aug 3, 2012 #18

    Dale

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    Obviously the authors of the book felt differently. I would have answered with "total power", not "net power", so I agree with the book and its answer. Depending on the rest of the text (i.e. whether or not they defined their terms earlier in the chapter) it may have been unclear, but it is certainly not "flat wrong".
     
    Last edited: Aug 3, 2012
  20. Aug 3, 2012 #19

    haruspex

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    Or made a mistake.
     
  21. Aug 3, 2012 #20

    Dale

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    It is certainly not "flat wrong". At worst, it was "unclear".
     
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