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When my book explains the work by gravity it doesn't explain why it passes from the shift (edit: the blue line in the picture) to the ray in this integral enter image description here

Of course, I understand that the work depends just from the height but why the integral from A to B of ds is not the length of the shift itself?

I also add the example picture: enter image description here

I apologize in advance if the question is too stupid but I just don't figure it out.

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  • $\begingroup$ Because the integral on the left is more general: it would be true, for instance, if $\vec{F}$ was not constant, or did not even represent a conservative field. $\endgroup$ – tfb Nov 25 '17 at 10:51
  • $\begingroup$ I think a key point is that, if you can take the force out of the integral, then it must be constant, and that means the field must be conservative and the work can't be path-dependant. But of course there are conservative fields (such as Newtonian gravity, and in fact almost anything) where the force is not constant: for those you have to work a bit harder. So I think what they're probably trying to do is to help develop intuition about this stuff. $\endgroup$ – tfb Nov 25 '17 at 11:47
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The workdone in this problem by gravity should be $mg (z_A - z_B)$. This is assuming that the gravity force is constant throughout and pointing towards $-z$ direction.

When you said "shift" did you mean the length of the blue line? If so, then workdone cannot be equal to that. This is to do with the fact that gravity of a conservative force, which means you can write it in terms of potentials. It also means that the workdone only depends on the starting and ending points and is independent of the path taken.

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  • $\begingroup$ Yes, it's what it says after but it doesn't exaplain the r(B)-r(A). Do you think that is it wrong? $\endgroup$ – Peto Nov 25 '17 at 10:32
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    $\begingroup$ NO. It's not wrong. They treat g as a vector and they take it's dot product with $r_{AB}$. That gives you the zA-zB part. $\endgroup$ – Ari Nov 25 '17 at 10:41
  • $\begingroup$ Maybe I understood it wrong. When you said "shift" did you mean the length of the blue line? If so, then workdone cannot be equal to that. This is to do with the fact that gravity of a conservative force, which means you can write it in terms of potentials. It also means that the workdone only depends on the starting and ending points and is independent of the path taken. $\endgroup$ – Ari Nov 25 '17 at 10:44
  • $\begingroup$ Yes sorry for misunderstanding (i'm not so good in english :/). Oh right, finally, it's what I thought from the start: the trouble was that the book will explain the conservative force 3 paragraphs ahead... $\endgroup$ – Peto Nov 25 '17 at 10:50

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