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Suppose you are in a hot air balloon of mass $M$ which is floating in the air, and you get out and start climbing down a rope at constant velocity $v$. By conservation of momentum we can show that the balloon moves up with velocity $mv/(m+M)$ where your mass is $m$.

You can also check that that energy is not conserved when climbing down the rope, but when you stop climbing the the energy is "lost".

So what is the mechanism for the creation and loss of the energy?

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  • $\begingroup$ If you just proved that energy is actually not conserved in such a situation, well congratulations for your breakthrough! The physics community will be glad to you if you show us your calculation so that we can check if you are right or (somehow more likely) you missed something :) $\endgroup$
    – glS
    Commented Jan 23, 2015 at 18:51
  • $\begingroup$ Yes I obviously did something wrong. What I mean is initially everything is stationary with respect to the ground, then you start climbing, so both the balloon and the climber have kinetic energy. So how is that possible? $\endgroup$
    – math_lover
    Commented Jan 23, 2015 at 18:53

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The situation you describe is more complicated than you think. If you just start climbing down the air baloon with a rope, you can't naively apply momentum conservation because there are a lot of other interaction you are missing.

The balloon is held in position by helium (or some other gas lighter than air, it depends on the kind of balloon I guess) particles "pushing" up from inside the balloon more than the air molecules outsides are pushing it down (i.e. Archimedes' principle).

When you start climbing down, assuming you are very heavy, you can think that the balloon will receive a slight push upward the moment your weight is no longer inside it, but this would not be due to conservation of momentum, but due to the gas particles in the balloon now for a moment "winning" against the air particles outside.

The energy that allows you to climb down the rope comes from potential (mainly chemical) energy stored in your muscles that allows you to move and of course the gravitational potential energy due to your elevated position. The energy that pushes the air balloon up comes mainly from the air surrounding it due to the Archimedes' principle.

A situation in which momentum conservation would be more significantly applied is if you somehow are on the inside roof of the balloon and you push yourself down very strongly. Then the balloon will gain a vertical momentum equal to the vertical (downward) momentum you managed to achieve pushing yourself down (and again, the kinetic energy "created" in such a process comes from the energy stored in your muscles, so as always Nothing is created, nothing is destroyed, everything changes).

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  • $\begingroup$ I don't think this answers the question: initially energy is 0, but then you have some kinetic energy created. You have to have some negative energy so that energy is conserved, the question is where is the negative energy? $\endgroup$
    – math_lover
    Commented Jan 23, 2015 at 19:19
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    $\begingroup$ @Joshua Benabou - That question was already answered by the part saying "The energy that allows you to climb down the rope comes from potential (mainly chemical) energy stored in your muscles". In particular your body uses the potential energy in ATP molecules to store energy, when the bonds of this molecule are broken energy is released which can be used to drive other reactions including ones that cause muscular contractions (see here for details). $\endgroup$
    – Hypnosifl
    Commented Jan 23, 2015 at 19:24
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    $\begingroup$ @JoshuaBenabou when we say that "energy is conserved" we mean that the sum of potential and kinetic energy is conserved. The kinetic energy comes from the potential one, as I explained $\endgroup$
    – glS
    Commented Jan 23, 2015 at 19:24
  • $\begingroup$ Sorry about that - read too quickly. Now I understand. $\endgroup$
    – math_lover
    Commented Jan 23, 2015 at 19:35
  • $\begingroup$ @JoshuaBenabou no problem. By the way was the downvote yours? You can revert it if you want now :) otherwise please explain how do you think the answer should be improved. $\endgroup$
    – glS
    Commented Jan 23, 2015 at 19:40

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