Is it accurate to say work is motion against an opposing force?

[A.T.]

2. Can the familiar notion of "work" apply to electromagnetism, gravity and other fictitious forces at the same time?

Going back to the OP and the statement in question:

"work is done to achieve motion against an opposing force"

I really don't see how the above would apply to two charges repelling/attracting each other. There is only one force on either of the charges that accelerates it, and does work on it. Nothing is "opposing" that force. So even for interaction forces, that do have a Newton 3rd reaction (the force on the other charge), the statement makes no sense.

 

[sophiecentaur]

A lump of iron is affected. A lump of plastic is not.

I chose magnetic force as an example because it would be larger and work over a distance. I could have suggested a Coulomb force .

You can define "work" so that both fictitious forces and interaction forces are considered to do work. For @A.T. and myself, this is the default posture.

I would need a lot of effort to construct a world where there would be a difference (two different classes of Work) I think I am convinced.

There is only one force on either of the charges that accelerates it, and does work on it. Nothing is "opposing" that force.

Unfortunately, that would have to apply to all the electrical forces even between two 'solid' objects. At a microscopic level, molecules are held in place by non-contact, electric forces; one object would just be pushed by a field and the other would be pushed back by a field there would be no N3 pairs at all. Take away on object and the other object would have no force. This, of course is where I have trouble with gravitational interaction. An object in a gravitational field is creating a force on the object creating the field.

Only when you bring in the business of frames does the Maths say something different.

 

[jbriggs444]

Unfortunately, that would have to apply to all the electrical forces even between two 'solid' objects. At a microscopic level, molecules are held in place by non-contact, electric forces; one object would just be pushed by a field and the other would be pushed back by a field there would be no N3 pairs at all. Take away on object and the other object would have no force.

Whoah there. In the simplistic picture of Newtonian gravity and the Coulomb force, there is no field as a tangible entity. There are just forces-at-a-distance. Real honest N3 pairs.

Untold numbers ($\frac{n(n-1)}{2}$ for a system with $n$ charged or massive particles) of force pairs, certainly. But that is no problem. We have the continuum approximation and Newton's shell theorem to dial things back to a managable number of interacting bodies.

To be sure, if one looks closely at the details, there is a time lag from action to effect and the notion of a field with properties (e.g. momentum, energy, electromagnetic waves and gravitational waves) becomes necessary. But from a first year physics point of view, action at a distance and N3 are adequate.

 

[A.T.]

one object would just be pushed by a field and the other would be pushed back by a field there would be no N3 pairs at all.

Whatever pushes the objects back, it does work on them even if there is no "opposing force", so they can separate freely. Thus the statement in question doesn't apply in general.

 

[sophiecentaur]

even if there is no "opposing force",

With Coulomb force repulsion there are two forces; one on each charge. When there is 'clear air' between them it's just the same as when two solid objects push on each other. If this is not true then where would you draw the line?
I now appreciate that "the statement" doesn't apply universally but you can't chuck out all instances of 'non-contact' interaction.

BTW I was just searching my memory for frame dependent forces where there is no possibility of finding a reaction force somewhere. Coriolis Force comes to mind. My experiences on fairground rides suggest that, to overcome the Coriolis force, you still have to apply a real force (to move your arm you have to restrain your body). That's a N3 pair but, as has been pointed out, that pair doesn't actually involve the Coriolis force.

 

[A.T.]

I now appreciate that "the statement" doesn't apply universally ...

Which was my sole point. One counter example should be enough, and I have given one for interaction forces and one for inertial forces.