Work done by force on a non-linear track.

[Anza Power]

We have a force which changes according to what point you're at, e.g: f=(4x²,3z-2y,2z), we need to calculate the work done by this force from point (0,0,0) to (2,1,3) on different curves

I don't want an answer I just want an explanation for how this is done, this is what I understand so far:

If we want to calculate the curve (a , a²/4 , 3a³/8), aren't we just supposed to put in the equations of the track as x y z in the force equation and do an integral for a from 0 to 2? or do we have to multiply it by the differential of the curve equation? that's the part I'm not getting...

f=(4x²,3z-2y,2z)
f=(4a² , 9a³/8-a²/2, 3a³/4)

I integrated that from a=0 to a=2 but it turned out wrong...

Also, if a force does the same amount of work on two different tracks, does that mean it's conservative for all tracks?

 

[tiny-tim]

Hi Anza Power!

We have a force which changes according to what point you're at, e.g: f=(4x²,3z-2y,2z), we need to calculate the work done by this force from point (0,0,0) to (2,1,3) on different curves

I don't want an answer I just want an explanation for how this is done, this is what I understand so far:

If we want to calculate the curve (a , a²/4 , 3a³/8), aren't we just supposed to put in the equations of the track as x y z in the force equation and do an integral for a from 0 to 2? or do we have to multiply it by the differential of the curve equation?

The https://www.physicsforums.com/library.php?do=view_item&itemid=75" is ∫ F.ds

Here they've given you a instead of s, so you need ∫ F(a) ds/da da

(ie, ds/da = 1/(da/ds) is the difficult part: you need to find that first … is this a real exam question?)

Also, if a force does the same amount of work on two different tracks, does that mean it's conservative for all tracks?

No, it could just be a coincidence.

However, in this case you could shorten the calculations by seeing that (4x²,-2y,2z) is conservative … you can probably see the potential, so just subtract potentials …

and that only leaves you to calculate the work done by (0,3z,0), which is a lot easier!

 

[Anza Power]

It's a homework question, thanks for the shortening tip, but why would I need s instead of a? aren't they the same?

I've already read that library post but it didn't help much, do you know maybe where I might find some questions with detailed answers of the same type?

 

[tiny-tim]

… why would I need s instead of a? aren't they the same?

nooo … s is arc-lenth, but a can be any old parameter …

for work done, it's essential to use arc-length

see http://en.wikipedia.org/wiki/Arc_length" for some details

 

[Anza Power]

^ Ahaaaaa, now everything finally makes sense, thanks a bunch dude...