Zero acceleration = zero net force?

[A.T.]

It's the same with Work By and Work Done. It's just Work with a sign which depends on your measurement frame.

The sign of work doesn't just depend on the reference frame, but also on whether the value represents work done by A on B, or by B on A. These are two completely different issues, that can affect the sign of work.

 

[sophiecentaur]

These are two completely different issues

Maybe but they are both viewed in an intuitive way.

but also on whether the value represents work done by A on B, or by B on A.

It's only by knowing the displacement and the force that you can decide what's on what. In a scenario of say two jet engines pushing against each other you can pass through the condition where the net FTD changes sign. Or are you saying that the frame of reference is involved? Perhaps if you made your statement in a different way, I'd understand what your are saying.

 

[A.T.]

Maybe but they are both viewed in an intuitive way.

That is no reason to conflate them.

It's only by knowing the displacement and the force that you can decide what's on what.

Yes, from these two you can compute work.

Perhaps if you made your statement in a different way, I'd understand what your are saying.

This has been explained to you so many times in previous threads. Just reread them instead of hijacking another one.

 

[berkeman]

This thread has run its course, and after some editing it has been closed. Thanks to all who tried to offer clear explanations.