Understanding MCAT Question Confusion: Energy and Force on Steep Hills

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SUMMARY

The discussion centers on the MCAT question regarding energy conservation while biking up a steep hill. The correct answer indicates that the energy expended remains constant regardless of the path taken, as gravitational potential energy is independent of the route. The participants clarify that while the distance increases when zigzagging, the force required decreases proportionally, resulting in no net change in energy. The conversation emphasizes the importance of understanding conservative systems in physics, particularly when non-conservative forces like friction are ignored.

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mrlucky0
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A girl riding her bicycle up a steep hill decides to save energy by zigzagging rather than riding straight up. Ignoring friction, her strategy will:

A. require the same amount of energy but less force on the pedals
B.
C.
D.

I missed this question and when I saw the correct answer, I couldn't entirely understand why the the energy remains the same. Since dE = F*D, and she has more distance to travel (to the top of the hill) shouldn't energy increase?

Edit: I understand the force is decreased by the same factor distance increases, making energy the same. But what about the increased distance to bike up the hill? Though I think I may be reading too deeply into the question making it more difficult, I'm not satisfied with the MCAT answer. Perhaps someone could correct my reasoning.
 
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Gravitational Potential Energy is independent of the path.
 
mrlucky0 said:
Edit: I understand the force is decreased by the same factor distance increases, making energy the same. But what about the increased distance to bike up the hill? Though I think I may be reading too deeply into the question making it more difficult, I'm not satisfied with the MCAT answer. Perhaps someone could correct my reasoning.

Didn't you just answer your own question?

On a side note, since you're ignoring friction, a non-conservative force, your system is a conservative system. Thus, the path one takes has no bearing on how much energy is expended to get up there. You're simply dealing with an initial energy and a final energy; in this problems case, potential energies.
 
Pengwuino said:
Didn't you just answer your own question?

On a side note, since you're ignoring friction, a non-conservative force, your system is a conservative system. Thus, the path one takes has no bearing on how much energy is expended to get up there. You're simply dealing with an initial energy and a final energy; in this problems case, potential energies.

Thanks, I've got it now. Somehow I completely overlooked the fact that, friction neglected, the system would be conservative. I mean, from a "practical" standpoint from which I based my thought process on when I encountered the question, it's just hard to believe any sane person would want to bike up a hill like that, believing they could conserve energy.
 

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