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Astrophysics
 one year ago
@empty hey hey
Astrophysics
 one year ago
@empty hey hey

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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2dw:1436077425511:dw leggo

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2I didn't have the chance to till now, so now I'll dooooo it! \[W = \int\limits \vec F \cdot dr\]

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Good luck I'll be watching haha.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Oh that dr should be a vector but w/e ok lol

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{ \vec dr }{ dt } = \vec r'(t) \implies \vec r'(t) dt\] \[\vec r'(t) = <1,2(t2)>\] \[W = \int\limits <0,g> \cdot <1,2(t2)> dt\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2dw:1436077905456:dw just seeing the drawing again XD \[W = \int\limits_{0}^{4} (0) \vec i + (2g(t2)) \vec j dt \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2is: \[\Large {\mathbf{F}} = \left( {0,  mg} \right)\]?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Got 0? It's a made up question haha

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2But yeah it would be mg otherwise

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2since, I see: \[\Large {\mathbf{F}} = \left\langle {0,  g} \right\rangle \]

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Yeah I just made this question up on the fly last night really late, so yeah throw an m in there haha.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2dw:1436078560687:dw hey empty quick question, I can't really remember, but would I have just a constant in the i direction as I'm integrating respect to 0

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Mhm sec, maybe I made a mistake

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2I messed up at distributing XD

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2please note this: \[\Large W = \int_0^4 {dt\left( {2gt  4g} \right)} \]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2ok let me redo it dw:1436078733805:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Thanks for checking @Michele_Laino !!

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so: \[\Large W = \int_0^4 {dt\left( {2gt  4g} \right)} = 16g  16g = 0\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2:) @Astrophysics

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Ok so you have the calculation done correctly which is excellent, but what is the significance of what you've done? @Astrophysics That's what I want you to explain to us.:D

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2I think that a possible answer can be this: "the work done by the weight force, along that trajectory is 0"

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2The work done by the field is pulling us down via trajectory giving us 8g, but if we had a mass it would be W = mg.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Well Michele put it much nicer than me

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2we gave the same answer! :)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Haha, yeah well work problems are pretty fun

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2yes! they are related to vector calculus, and I like very much the vector calculus! as you know @Astrophysics eh eh :)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2But if we had a circle lets say, that's the field and the circle is going counter clockwise, how would we define the work exactly? We'll just assume x = cost, y = sint, w/e dw:1436079570813:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2dw:1436079731168:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2I think that we have to use this parametrization: \[\Large \begin{gathered} x = R\cos t \hfill \\ \hfill \\ y = R\sin t \hfill \\ \end{gathered} \] dw:1436079812068:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Haha yeah, you're probably right, all of these questions in this post are just made up out of our heads and I guess that's not very useful...xD but to say the least it's fun, I just want to know what happens with the field and the path taken

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2If we give it an interval [0,2pi], we would get 2 pi, so let me just ask you a question to clarify, aside from the math what exactly does this mean, it's opposing the field, or just saying it's going in the opposite direction?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2what is the vector field?

Empty
 one year ago
Best ResponseYou've already chosen the best response.0This is because you have a nonconservative field in this case. In the previous case, we could have written that particular vector field as the gradient of a scalar field while this one has no corresponding scalar field. So think, what is the corresonding scalar field to the Force field from the first question? Just let yourself be stumped on this for a while so you will be satisfied when learning the answer since it is important and should be remembered, as it has a fairly well known associated equation. :)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2I got this: \[\Large W = \int_0^{2\pi } {\left( {R\sin t,R\cos t} \right) \cdot \left( {  R\sin t,R\cos t} \right)} dt\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Ah, I see thanks empty, I did not think of it as a nonconservative field, and no worries Michele, the math is fine, but I was just trying to figure out the theoretical part of the problem!

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2oops.. I have made an error of sigh, here is the right integral: \[\Large W = \int_0^{2\pi } {\left( {R\sin t,  R\cos t} \right) \cdot \left( {  R\sin t,R\cos t} \right)} dt\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Thanks @Michele_Laino and @Empty !! :)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2we can note that your field can be rewritten as follows: \[\Large {\mathbf{F}} = \left( {y,  x,0} \right)\] now according to the observation of @Empty , we have: \[\Large \nabla \times {\mathbf{F}} = \left( {0,0,  2} \right)\] namely our vector field is an irrotational field
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