## Is reference frame important when looking at work done?

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?
 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 $P=\vec{F}\cdot\vec{v}$ will be different as well. The reason $P=F(V+v)$ 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 $v+V$ in the frame of someone at rest back on Earth's surface, and in the case of $a=0$ the force on him does reduce to $F=mg$ but he still is moving at a velocity of $v+V$ 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.

Mentor
 Quote by VortexLattice 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.
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.

 Quote by VortexLattice 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).

 Quote by VortexLattice 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).
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.

## Is reference frame important when looking at work done?

 Quote by DaleSpam 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.
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.
 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?

Mentor
 Quote by cjl Well, you are doing no work on the block, assuming that the frame is not accelerating. You are, however, doing work on the ground
Yes, I should have been more clear, thanks.

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 Quote by DaleSpam You are neglecting the power input by the elevator to him.
How is that part of his power output?
 he shouldn't be able to tell he's even in an elevator as opposed to a stationary room,
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.

Mentor
 Quote by haruspex How is that part of his power output?
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.
 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?

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 Quote by jbriggs444 It seems like cheating to count the elevator's motion as a contribution to the "power output" of the man's legs.
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.
 Mentor 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).

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