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Gravitational potential energy question -- Ojbect sitting on the Earth
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[QUOTE="kuruman, post: 6489417, member: 192687"] If you could say that, then you would be equally justified to say that the Earth (when considered as your system) will have the potential energy and not the object. So which is it that "has" the potential energy, the object or the Earth? The answer is neither, the potential energy belongs to the [B]two-component[/B] Earth-object system. When the common potential energy changes, so does the kinetic energy of the system's components. The correct mechanical energy conservation to write in this case is (subscripts o = object and E = Earth) $$\Delta K_{\text{o}}+\Delta K_{\text{E}}+\Delta U_{\text{o+E}}=0.$$It's only because the Earth's vertical speed does not change noticeably over the time that the object falls, that we ignore the Earth's change in kinetic energy as commented by [USER=85613]@A.T.[/USER] You can separate the two and consider the object as a single-component system, but you have to be careful of what you say and how. If you drop the object when it alone is your system, it does not lose potential energy and gain kinetic energy as sometimes is the claim. The dropped object gains kinetic energy because the Earth does positive work on it through the force of gravity. This positive work is equal to the loss of potential energy of the two-component system. To summarize, you can think of the transfer of energy to the object as taking place internally in the two-component system and externally in the single-component system. [/QUOTE]
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Gravitational potential energy question -- Ojbect sitting on the Earth
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