Len: My instinct says that is the same work since the same weight moved the same distance against the same gravity. However, a variable has changed: the muscles are putting more energy into the system, so something else has to give. I would say that time has to be what gives. Same work; faster rate. Watts are the rate of work (joules are the actual measure of work itself, right?) So we've increased the wattage.
Coming back to MrAl's comment: the thing that ties all of this together is heat; the mechanical equivalent of heat. That I'll have to investigate further; but for now I can accept that I can measure all of these with the same unit because they convert into heat and back.
Ratch: I'm already reading an introductory textbook (electronics) more or less for fun. I'll put thermodynamics next on my list, thank you, but for now there seem to be people here who are willing and able to explain some of my questions. If I stop and go to a more fundamental text every time I have a question, I'll wind up in pure mathematics, from which there is no escape. <grin>
Hi again,
Yes heat is used as the equivalence most basically, friction heat vs power heating.
There's a more direct route however. Way back in the 18th century James Watt equated the power of his horse to what it takes to lift a heavy weight from the bottom of a deep well. He found that it took a certain amount of force over a given time to lift the weight, and this gave something like 22000 ft lb/min to the horse, but that didnt include the effects of friction which brought it closer to 33000 ft lb/min. A slightly more accurate number might be 32572 ft lb/min so lets use that.
work=force*distance
power=work/time=force*distance/time=32572 ft lb/min=543 ft lb/sec=746 watts.
work=power*time=543 ft lb/sec*1 sec=543 ft lb=746 watts*1sec=1HP*1sec
Thus, a loss free motor of 746 watts running for 1 second can provide 543 ft lb of work.
So the power itself is not the work, it's the power used over time that provides the actual work.
Note that if the motor has losses of say 50 watts then the motor consumes 796 watts but only 746 watt seconds of work is done over 1 second. Thus, when we calculate the mechanical equivalent we dont include losses yet because that would mess up the calculation. We include that later to see what it is going to take to actually run the motor and get a certain amount of work accomplished.
The above of course assumes that the motion is in the same direction of the force for all time, and that the motion is constant. A more general definition is like this:
Work=LineIntegral F.dr
where the dot between F and dr indicates the dot product, and this would be the work done on a particle moving from r(a) to r(b) caused by a force F for example.
(I can just barely see that dot on my monitor so if we replace the dot with an asterisk it would look like this: F*dr)