# Conservation of energy.

Asked by JMCSD (243) March 19th, 2008 from iPhone

Visualize this, a long pole sticking out of the ground on a rotating base, at an angle of let’s say 20–30 degrees. Now imagine a heavy weight hanging off the side at a reasonable good distance attached by a stiff connection. Ok, so if you were to take the entire body of this shaft and layer it with generators that converts the rotation into power, and let the weight hanging off be affected by gravity, why would there be no multiplication in energy output because of the multiple generators? Ps. Sorry if once all said and done this makes no sense.

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## 9 Answers

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kevbo (25634)

something simliar: producing energy while walking:O
http://www.sciam.com/article.cfm?id=powering-cell-phone-battery

Mtl_zack (6759)

are you kidding, I was just curious and the simple definition for me just doesn’t do it. If one generator equals x power, why don’t 3 equal 3x? On top of that I’m not in school so don’t think you’re doing me any favors. Jeez.

JMCSD (243)

My apologies. We get a lot of hw questions and this sounded like a textbook example.

Off the top of my head, I’d say that the added friction of each generator “dilutes” the kinetic energy of the weighted pole. the

kevbo (25634)

I understand. See what I don’t get is while it makes sense that the friction would require more energy input, why wouldn’t gravities pull on the offset weight take care of all the additional friction?

JMCSD (243)

The weight/gravity is constant. You’d have to add more weight to compensate for the added friction.

kevbo (25634)

Interesting. See I find the idea of free energy very interesting and this question kind of spawned from a theory of mine. I’ll have to draw something up. Thanks.

JMCSD (243)

No problem. Again, sorry for the misunderstanding.

kevbo (25634)

oh no problem, appreciate the input.

JMCSD (243)

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