I'm trying to find a good way to think about work.(adsbygoogle = window.adsbygoogle || []).push({});

From what I can tell, it's what the object experiences, not really what the force does, correct?

If a train is heading west and superman grabs it, stops it, and starts pushing it in the opposite direction until it's at the same speed as it was before, he's done zero work right?

And this is because before it stops, we have W = FDcos(180) = -FD, while after it stops we had W = FDcos(0) = FD? For a net work of FD - FD = 0.

Which coincides with W = K(final) - K(initial)

This makes perfect sense to me (assuming I got that right) but it just feels much weirder when we're talking about changing it's direction in a different way (not just the reverse.)

If the train were going along +x and superman was drunk and decided to push it until it were going along +y at the same speed (and no longer x displacement at all), he's still done 0 work, but how do I show it as above? It's not as simple as two things canceling out.

My thinking is that this would be more complex to "write out" because the angle relating the force to displacement would constantly change while this diagonal force were applied, not just once as above. If that is the case... would there need to be some function for a relationship between that angle and the duration of the force's application to integrate, pretending we didn't know that the work would come out to zero?

I realize all of this is unnecessary to actually solve problems, but I'm not one of those "just take it as axiom" kind of people.

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# Work - Kinetic energy theorem Need more fundamental understanding

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