Question

Can someone help me with the following problem? Energy and force.

URL: http://i.imgur.com/pyiR4tB.png

Answer 1

I need in-depth explanation of the entire procedure, I just don't understand this problem.

Answer 2

yes. where have you got to?

Answer 3

To be honest I was only able to get the answer A), but that one is pretty easy. I was told something about derivatives or some sort, not quite sure tho.

Answer 4

OK

potential energy function. is that per unit mass? what is your answer?

we can do the rest.

Answer 5

sorry just noticed, answers are given.

ok leave that one and onto part (b).

do you know how work and force inter relate. put simple W = F.x or W = ∫ F dx

i do not know what level you are working at but this is the start of the idea of a potential energy function....

Answer 6

so within a force field such as the earths gravity field or the electric field in a capacitor, you can generate a potential function from the field equation that allows you to describe the energy content of a particle or a charge anywhere in that field.

am i waffling or is this helping you?

we can cut to the math if you like.

Answer 7

No, I'm actually understanding, I'm aware that W is the integral of Fdx, yeah :D

Answer 8

great. if:

\[U(r) = kr = k \sqrt{x^2 + y^2 + x^2 }\]


what is
\[\frac{dU }{ dy} ??\]

hopefully, the actual derivative is easy enough, but what does it means in terms of these ideas of integrating a force field to get an "energy equation"??

Answer 9

is that too cryptic, or waffly?

Answer 10

if you get the derivative, you get the Force, I believe, right?

Answer 11

make that ∂U/∂y....

Answer 12

yep, that's the idea. we are just doing it in 3 dimensions so we use partial derivatives, ∂U/∂x, not dU/ dx.

Answer 13

Oh, I see so I need yo make the partial derivative of my energy equation, but the deal is that I don't know the math either, can you help me with it? with the drawing tool or sumthing? I know I'm asking for a lot, but I have a test tomorrow and that's the only question I'm having issues with.

Answer 14

no problem. you want to do the derivative, ∂U/∂y? right?

Answer 15

Yeah, so I can get both answers, B and C

Answer 16

treat x and z as constants and do it just like you would dU/dy ......

Answer 17

btw I want to thank you in advance for all the help, and let me see...

Answer 18

Apparently yeah, I used wolfram and it gives the answer on B, but anyways I'm not really familiarized with deriving, could you give me a hand?

Answer 19

to make it totally clear that x and z are constants in this, i am going to replace x by pi, and z by e, the natural number. these are constants that we cannot add up so we must preserve then through the process. with that artifical device in place, we are asked to calculate:

\[\frac {∂U}{∂y} = \frac {d}{dy} (k \sqrt{\pi ^2 + y^2 + e^2}) = k \frac {d}{dy} ( \sqrt{\pi ^2 + y^2 + e^2})\]

\[= k \frac {d}{dy} ( {\pi ^2 + y^2 + e^2)}^{1/2} = \frac {1}{2}k \frac {d}{dy} ( {\pi ^2 + y^2 + e^2)}^{-1/2} (2y)\]

\[= \frac {ky}{\sqrt{\pi ^ 2 + y^ 2 + e ^ 2}}\]

that os clearly NOT the final answer, but is starting to look close.

NB we are still missing a minus sign and that is something we can address in the final answer which is very close.

make sense?

Answer 20

all we need to do now is swap x and z for pi and e, and explain the minus sign. we also need to introduce the "funny d" notation to make it clear that we are doing partial derivatives, ie we are freezing x and z and looking only at changes in the y direction....

Answer 21

I think that we have to use this formula:
\[\Large {\mathbf{F = }} - \nabla {\mathbf{U}} = \left( { - \frac{{\partial U}}{{\partial x}},\; - \frac{{\partial U}}{{\partial y}},\; - \frac{{\partial U}}{{\partial z}}} \right)\]

Answer 22

Yeah, that formula sounds familiar, and oh, yeah, that derivative makes sense!!

Answer 23

good, we have just calculated Fy, which is the force along the y axis, or ∂U / ∂y.

the tweak we need to make conerning the minus sign is definitonal. because we are actually only interested in differences between potential energies, and not the actual values themselves, the sign is a bit tricky if you are new to this.

broadly speaking, in hydrodynamics we are finished. in electricity ad mechanics, in the **definition** of the potential energy a minus sign is added, the reasons for this belong in the physical world and it looks to me like you are just going at the maths so we can introduce that into our solution.

the final solution would look something like this:

\[F_y = - \frac {∂U}{∂y} = -k \frac {∂}{∂y}(\sqrt{x^2 + y^2 + z^2})\]

\[ = -k (\frac{1}{2} )(x^2 + y^2 + z^2 )^{-1/2}(2y) = \frac{-ky}{\sqrt{x^2 + y^2 + z^2}}\]

i think there was a type somewhere above but this is where we now are...

ok?

Answer 24

yeah, so far so good

Answer 25

the final part is some simple math.

first calculate k .... U(r) = kr

then plug the coordinates (x,y,z) into your equation for Fy as derived above.

ok?

Answer 26

plus, i hope/assume you see that this is completely symmetrical about the origin.

from the U(r) equation, you will know that, the further away from the origin you move, the more energy you have, and the relationship is linear.

that is also why the force Fy has a negative sign. it is attractive. ie the forces pulls you back toward the origin just as gravity pulls you in the direction of earth.

Answer 27

Thaks a lot! and sorry for replying just now, I went to college and I just got back, so sorry, thanks a lot!