Understanding MCAT Question Confusion: Energy and Force on Steep Hills

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A girl riding her bicycle up a steep hill zigzags to save energy, leading to confusion about energy conservation. The discussion clarifies that gravitational potential energy is independent of the path taken, meaning the total energy expended remains the same regardless of distance traveled. Ignoring friction, the system is considered conservative, so the energy required does not change with the zigzagging route. The initial and final energies are what matter, not the path. Ultimately, the realization is that the method of ascent does not affect the total energy needed to reach the top.
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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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