[beachstartup]

> That is quantitatively zero work, buddy

this statement is both technically wrong, and stupid because of your inane try-hard elitism. with google you can calculate down to several sig figs how many joules of work was done by picking up a weight, and putting it down.

https://en.wikipedia.org/wiki/Work_(physics)

maybe you meant 'qualitatively'? even then, still wrong. moving heavy shit around is work, even if it ends up in the same spot.

[cyphar]

I agree that the statement is elitist, but since gravity is a conservative force I think that the strict definition of "work" from physics (W = \int_a^b{\vec{F} \dot d\vec{s}}) would mean you get as 0 a result.

[beachstartup]

if we're going to get pedantic, what is the chance that the weight ended up in exactly the same spot that it started in, when moved by a person?

[cyphar]

The position doesn't matter, as gravity doesn't act along the surface of the Earth -- it only acts radially. So the work is only affected by the height above the ground (or radial distance from centre of the Earth if you prefer).

So if the person puts the weight on the floor where they picked it up from they will have done zero work on it. Your attempt at pedantry was misdirected -- I would've said that because of the change in density of the Earth's crust that technically the force of gravity does change and thus is not an entirely conservative force.

Also, I wasn't getting pedantic. I was quoting the definition of work, which I believe that GGP misunderstood.

[beachstartup]

i don't buy it. you're telling me a draw bridge, or a lifting door, does no work? get out of here.

[cyphar]

The overall work done by a system is the integral of the force vector dot product the direction of movement. If you have a conservative force setup (no friction) then any motion that takes you back to where you started does no work. We did this in first year physics at university (we also did it in high school, but other school systems probably don't work the same way).

You're confusing work and energy. Energy is expended, but work is not. Work relates to the change in potential energy of a system -- which will not change in a conservative system (assuming you return to the original point). https://en.wikipedia.org/wiki/Work_(physics)

[beachstartup]

yeah, i completely agree.

however, you will never, ever, achieve a zero work vector in a weight room, whether you're off by a micron or a mile. give it a shot, tell me how it goes.

you're arguing for an impossible result using textbook definitions, pretty much what i would call 'useless pedantry'.