Solving Work Done Problems in Physics

[Friache]

I'm having some trouble wrapping my head around a couple work concepts. I've been reading the following thread: https://www.physicsforums.com/threads/displacement-and-work.322461/
and I don't think I've fully grasped how work is accomplished. My first question is if I lifted a (w=3.0N) water bottle 3.0m vertically and the returned it to its original position, has work been done? I realize that work is done from lifting and lowering it down, but once it returns to its origin does it matter?
The second is if two people press against a wall and person A gives up twice as person B then have persons A and B accomplished any work? And if so then has person B accomplished more for pushing longer? I realize that straining your muscles in any way requires energy, especially for long periods of time, but in a physics sense (We're talking college physics I) does it matter?

 

[Delta2]

A necessary (but not sufficient) condition is that work is done when the point of application of the force is moved(not necessarily in the same direction as that of the force). So

A) Work is done by the force of your hand and by the weight of the bottle during the ascend (because the point of application of these forces which we consider to be the center of mass of the bottle, is moving) . But, assuming that the bottle is moving with constant velocity, this means that the force of your hand cancels the force of weight, therefore it is somewhat easy to prove that the work by force of hand is equal and opposite of the work of the weight. So the total work done for the ascend is zero during the ascend, though the weight and the hand force have done equal but opposite work (lets say that the work of the hand is positive and the work of the weight is negative during the ascend). So [itex]W_{hand}+W_{weight}=0 [/itex](1).Similarly during the descent and assuming that the bottles also moves with constant velocity we conclude that the total work done is also zero, but this time the work of weight will be positive and the work of the hand will be negative. So [itex]W'_{hand}+W'_{weight}=0 [/itex] (2).Now what also comes into play is that the force of weight is a conservative force which means that the work done by the weight in any closed trajectory(closed means that the start and end points coincide) is zero, no matter what the trajectory is. So it will be [itex]W_{weight}+W'_{weight}=0[/itex] and from this in conjuction with (1) and (2)we can conclude that [itex]W_{hand}+W'_{hand}=0[/itex]. So we see that the total work of weight, and the total work of the hand are each zero, though the forces have done work during the ascend , or during the descent, but it is equal and opposite.

B) Since the point of application is not moved (we consider the wall to be immobile) the work done by A is zero, and also the work done by B is zero. Person B might be more tired but that's because when we apply a force (regardless if the point of application is moved or not) we strain our muscles therefore we consume energy but this energy doesn't necessarily go as kinetic energy of the object we aplly the force to.

 

[Friache]

A necessary (but not sufficient) condition is that work is done when the point of application of the force is moved(not necessarily in the same direction as that of the force). So

A) Work is done by the force of your hand and by the weight of the bottle during the ascend (because the point of application of these forces which we consider to be the center of mass of the bottle, is moving) . But, assuming that the bottle is moving with constant velocity, this means that the force of your hand cancels the force of weight, therefore it is somewhat easy to prove that the work by force of hand is equal and opposite of the work of the weight. So the total work done for the ascend is zero during the ascend, though the weight and the hand force have done equal but opposite work (lets say that the work of the hand is positive and the work of the weight is negative during the ascend). So [itex]W_{hand}+W_{weight}=0 [/itex](1).Similarly during the descent and assuming that the bottles also moves with constant velocity we conclude that the total work done is also zero, but this time the work of weight will be positive and the work of the hand will be negative. So [itex]W'_{hand}+W'_{weight}=0 [/itex] (2).Now what also comes into play is that the force of weight is a conservative force which means that the work done by the weight in any closed trajectory(closed means that the start and end points coincide) is zero, no matter what the trajectory is. So it will be [itex]W_{weight}+W'_{weight}=0[/itex] and from this in conjuction with (1) and (2)we can conclude that [itex]W_{hand}+W'_{hand}=0[/itex]. So we see that the total work of weight, and the total work of the hand are each zero, though the forces have done work during the ascend , or during the descent, but it is equal and opposite.

B) Since the point of application is not moved (we consider the wall to be immobile) the work done by A is zero, and also the work done by B is zero. Person B might be more tired but that's because when we apply a force (regardless if the point of application is moved or not) we strain our muscles therefore we consume energy but this energy doesn't necessarily go as kinetic energy of the object we aplly the force to.

Thanks for taking the time to go in detail about this. Its been driving me insane.