Question

explain conservative forcefield

Answer 1

\[W \int\limits_A^B = K.ds\]\[K=\int\limits_A^B(ds)_K=K(s_B-s_A)\]

Answer 2

this is what it says in my textbook

Answer 3

i find it very confusing

Answer 4

this is about work apparently..

Answer 5

I can't help you explain that math, sorry.

Conservative fields, more specifically conservative vector fields, are a mathematical thing. Maybe the math section will help with understanding the math.

What I remember is independence of path.

Answer 6

true

Answer 7

it's the math i don't get though

Answer 8

Something about the path of a line, when you do something with a line integral in a vector field...

I hope someone else in can explain this soon!

About work, I might be able to explain that somewhat....

Answer 9

If you have a conservative field to describe the work, then the work done is the sum of the work from one point to another, and it does not matter what path connects the points. For example, we look at work by one charge as it moves in a field of another charge. Two positive charges, say. The charge moves away and positive work is done on it. You move the charge towards the other, and negative work was done. You move it right back to where it started, and no work was done.

Answer 10

A block on a table, though, you push it, and friction always opposes movement. You move it back and forth, and you did work on it. The path does matter.

Charges - move the charge all over. Work is sort of done and undone.

Friction - move the block, and you're increasing the total work done as you push.

Answer 11

Sorry I'm not fully aware of this stuff! Good luck finding out!

Answer 12

alright, thanks anyway

Answer 13

Umm although I haven't even studied forcefields but the word "conservative" rings a bell. I can say that the work done by this forcefield doesn't change no matter the path taken by the particle. Hence mechanical energy is conserved. @theEric what do you think?

Answer 14

A conservative field (like gravitational and electric fields) are conservative when the work done to go from point A to point B is independent of the path. This implies that if you leave point A and go somewhere, any where in whichever direction and then return to the same spot, in a conservative field, you have done no work.
When you have a coil in a circuit, the path you take is important. You get an induced EMF of one value if you take a certain Path A to B and if you take a different path from these same two points, you may have a different value for the work.

Most textbooks utilize Kirchoff's Rule to analyze circuits, but this can only be used if you have no magnetic coils in the circuit. Faraday's law should be used to analyze circuits in the presence of nonconserviative fields, because the close-loop integral will NOT be zero.

Answer 15

alright, so what about the math? is it correct?

Answer 16

You wrote\[W \int\limits_A^B = K.ds\]That does not make sense, since something has to be in the integral.
Maybe that should be\[W =\int\limits_A^B K\dot\ ds\].

I'm not sure if the physical interpretation is coreect though, sorry...

Answer 17

correct*

Answer 18

\[\overrightarrow W =\int\limits_A^B \overrightarrow K\dot\ d\overrightarrow s\]I guess that is more correct.

Answer 19

and yet my professor wrote it like that..

Answer 20

As for your equation, "The gradient theorem, also known as the fundamental theorem of calculus for line integrals, says that a line integral through a gradient field can be evaluated by evaluating the original scalar field at the endpoints of the curve.

http://upload.wikimedia.org/math/c/a/8/ca8b2c89c9693e534bf73ab7c65d411f.png

The gradient theorem implies that line integrals through irrotational vector fields are path independent. In physics this theorem is one of the ways of defining a "conservative" force. By placing φ as potential, ∇φ is a conservative field. Work done by conservative forces does not depend on the path followed by the object, but only the end points, as the above equation shows."

Source: http://en.wikipedia.org/wiki/Gradient_theorem

Answer 21

So work, regardless of whether or not the function in the integrand is the gradient of some function (e.g. \( \phi \) above) , can be written generally as:

\(W=\int\limits_{A}^{B}f(r)dr\) Sum of all force times distances between A and B

This does not imply path independence.

However, if f is the gradient of some function, say \( \phi \) (i.e. f= \( \bigtriangledown \phi(r)\) ), then by the fundamental theorem of calculus for line integrals, we write work as:

\( W = \int\limits_{A}^{B}\bigtriangledown \phi(r)dr=\phi(B) -\phi(A) \)

The Wiki article above goes further:

"The gradient theorem also has an interesting converse: any conservative vector field can be expressed as the gradient of a scalar field. Just like the gradient theorem itself, this converse has many striking consequences and applications in both pure and applied mathematics."

More info: http://en.wikipedia.org/wiki/Gradient (see the section on conservative fields)

Answer 22

This is not easy to digest. I suggest taking some time to think about it.