- #1
pfr
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I've been struggling with this for the past hours and I can't find a good answer.
Using the integral for work [itex] W = \int_a^b \vec{F}\cdot d\vec{s} [/itex], when [itex] a > b [/itex], and the force is directed from a to b, i keep getting a negative result. I am considering [itex] d\vec{s} [/itex] as the infinitesimal difference of the position vector along the integral route, so it has the same direction of the force. Is this wrong? Where am I messing up, and what [itex] d\vec{s} [/itex] should truly mean here?
E.g The work performed by a gravitational field upon a particle while bringing it from infinity
[tex] W = \int_\infty^R \vec{F}\cdot d\vec{s} = \int_\infty^R (\frac{-GMm}{r^2})\hat{r}\cdot (-dr)\hat{r} = \int_\infty^R \frac{GMm}{r^2}dr = \left.\frac{-GMm}{r}\right|_\infty^R = \frac{-GMm}{R} [/tex]
Which is false, since the work is obviously positive.
Using the integral for work [itex] W = \int_a^b \vec{F}\cdot d\vec{s} [/itex], when [itex] a > b [/itex], and the force is directed from a to b, i keep getting a negative result. I am considering [itex] d\vec{s} [/itex] as the infinitesimal difference of the position vector along the integral route, so it has the same direction of the force. Is this wrong? Where am I messing up, and what [itex] d\vec{s} [/itex] should truly mean here?
E.g The work performed by a gravitational field upon a particle while bringing it from infinity
[tex] W = \int_\infty^R \vec{F}\cdot d\vec{s} = \int_\infty^R (\frac{-GMm}{r^2})\hat{r}\cdot (-dr)\hat{r} = \int_\infty^R \frac{GMm}{r^2}dr = \left.\frac{-GMm}{r}\right|_\infty^R = \frac{-GMm}{R} [/tex]
Which is false, since the work is obviously positive.