In all my years, I’ve seen physics problems requiring me to figure out how much force, or work, or power, etc. is used/required/etc. to lift an object. Essentially, the force required to lift a 10 kg object is 10kg. But how much energy is required to put something down? Or to go downstairs? The only way I can think about such a problem is to think about freefall, due to gravity, and to think about counteracting that force by moving something downward at a rate slower than gravity would accelerate it. How do you think about these types of questions? I’ll pose one for us to consider. I want to move a 10 kg mass one meter. If it takes me one second to move it, how much energy was required? How much force was used? How much work did I do? How much power did it take?
Of, course, in my question posed above, I want to move the mass DOWN one meter. That was the point.
I can’t help you with the math, but couldn’t you approach the problem like this?
To figure out how much energy, force, etc. is required to lower a 10 kg mass one meter in one second, just do the calculations to raise the mass one meter but reverse the sign on the force of gravity.
Does that make sense?
Well the energy is easy:
E=force x distance moved
So for a 10kg mass, on the Earth (assume gravity has a value of 10m/s^2, the force is 100Newtons, and the energy required is simply
E=100*1 = 100J
Regardless of how long it takes you to move it…
For the force required, you moved 10kg with gravity at a constant velocity. So, you had to balance the force exerted on the object by gravity. Hence the force required is simply 100N.
You did the same amount of work as the energy it required, hence you did 100J worth of work.
If you moved the object in one second, then using the definition of Power=Work done/time taken, then, Power = 100W.
Feel free to correct me if you think I’m wrong - its more than likely… 
Actually, the “force” is a combination of the mass of the object and the acceleration applied to it. Holding a 10kg object doesn’t take 10kg of “force”, but moving it one meter in one second requires 10 Newtons.
Dropping an object doesn’t really require force; most of the work is done by simple gravity, after all. Putting it down under control is another matter. If you want to put an object down so that it has a gentle landing, you will have to exert some force on it to resist the downward acceleration gravity applies. Exactly how much force is involved depends entirely on how slowly you want to lower the object.
Gravitional acceleration is 9.8m/s², which I’m going to round up to 10 for simplicity. Therefore, a 10kg object exerts a downward force of (10x10) 100N just by being in Earth’s gravitational field. When this object is resting on a table, the table must resist this force with tensile strength. Now, if the table is a metre tall and you decide to put the 10kg object on the floor without damaging it, you may decide to lower it gently, say taking 2 seconds to do so. Now YOU have to absorb the 100N of downward force, but since you are lowering the object, it is the equivalent of letting it fall under control. Going from motionless to covering 1 metre in 2 seconds means an average acceleration of 0.5 m/s². Multiplied by 10kg means the object moves with a force of 5 Newtons. The object wants to push down at 100N, but you are resisting all but 5N, meaning you have to exert 95 Newtons for those two seconds.
So, holding the object motionless requires a steady 100 Newtons. Lowering the object slowly requires (100-5) 95 Newtons for 2 seconds. Lowering the object more quickly means you exert less resistance to its natural acceleration. Letting the object fall out of control requires 0 Newtons, at least from you. Trying to throw the object down so it falls faster than gravity will require adding force again.
Boiled down, Newton’s Second Law says F=MA, or force (in Newtons) equals Mass (in kg) times Acceleration (in m/s²). Unless you include acceleration, the whole question is meaningless. Just make sure you include 9.8 m/s² (gravity) as a factor in all your equations, if it applies.
I don’t think that you’re going to like the answer to this question. As I learned it in physics, the differential definition of work is dW = F • ds, where F is the force exerted on an object and ds in the direction of motion. This means that if you are exerting an upward force on a downward-moving object, you are doing negative work. In other words, you are not putting energy into the system; you are getting energy out of the system.
Consider the example of water falling over a dam. If the water turns a turbine, the turbine exerts an upward force that slows the fall of the water. However, the turbine does not need to expend energy to do this; it generates energy. In practice, only a small fraction of this energy can be converted into usable forms due to the second law of thermodynamics and all that.
Why, then, does a human being require energy to lower a heavy object? For the same reason that it requires a human to carry a heavy object sideways, or simply to hold it in one place, neither task requiring any work at all. Unfortunately, I’m not quite sure what this reason is. It has something to do with the way human muscles work; I think it may be that they require a constant input of energy to remain tense.
Anyway, that’s why you never see this coming up in a physics problem. Our real-world experience seems contrary to the actual result, and they don’t want to confuse people with this complexity.
Yeah, the definitions are different.
Work is Force times Distance, where Distance is the distance an object is moved by the force. So, if you push very hard on an immoveable object, you will do no Work.
But you’ll get very tired, right? And use a lot of energy somehow. So the problem we have to analyze involves the internal structure of the human body–and the textbook intro physics problems don’t go into that amount of detail.
Same thing with lowering a weight.
>> Of, course, in my question posed above, I want to move the mass DOWN one meter.
Just let it go and it will go by itself. Where’s the problem?
It’s kind of like the legal definition of “insanity.” That’s why I never got physics, I tried to really understand it instead of just accepting its definitions.
In the case of freefall you can say Energy = mass X g X height it falls from
So, you do E = (10kg)(10 m/s^2)(1 m) = 100 Joules
If it was going down a slanted slope, like a flight of stairs, you first figure out the angle of incline between the stairs and level ground. Lets say its a 30 degree incline. Then you do: E = F d where F is the force moving the object parallel to the surface, and d is the distance it moves. Using trigonometry we know that the hypotenuse of the slope (the path it will actually take) is 2 meters. Here’s where this gets tricky. We assume that the hypotenuse of the slope is the X axis to make things simpler. So now we need to know what component of the force of gravity is in the x axis. To get this we multiply its total weight (100N) by the sine of 30 degrees. (100)(sin 30) = 50 N. Now we multiply this times the distance, 2 meters, and we get: (50 N)(2 m) = 100 J
I hope this makes sense, its hard to explain over the net.
I can’t answer your question, but I can ask some more which might lead to an answer.
What is the obect? More spcifically, what impact force will cause it damage? If that force is greater than than freefall over your specified heights, then the only force necessary is the force to “unlock” it and let it drop.
What is the “holding up” mechanism? A human hand? A crane? The design of this mechanism will dictate a necessary force, which may vary wildly from one to the next.
Or perhaps you’re asking, “How much energy is expended in moving a given object of a given mass a given distance, given both a gravity force and direction (is vector the correct term for this?-- I think a vector must include a velocity, but I’m not sure)”?
I dunno, haven’t got a clue, just making it up as i go along.
But I do know this. Energy is conserved, by conversion from one form to another. So as far as I can tell, it takes NO energy to do ANYTHING. 
So why does my back hurt after moving a box of books? It is beyond knowing, grasshopper. ;j
I think the question is, CC knows they are doing work in lowering a weight–because it makes their muscles tired–but as you point out, the simple application of physics would show that you shouldn’t do any work.
So far, it’s not clearer for me. As to the work: yes, I know work is calculated by FxD. But, the fact is that I am, indeed, exerting force, when I keep it from falling at 9.8 m/sec/sec. But I am also moving it some distance. So, it SEEMS as if I’m doing work. At the same time, I know that it seems like I’m doing work when I exert a force and nothing moves, and that’s NOT work. Sheesh. Are there any conceptual physics folks out there who can help this pinhead?
That was my point. The force has to be exerted in the direction of the distance displacement in order to be considered positive work. If you push sideways on something that is moving perpendicular to your effort, the object does move but the “work” done is still none.
Like I said, an analysis of the energy expenditure in such a case would need an analysis of internal human body structure–there’s work being done at the cellular level, surely.
No, Luke, I can insult Darth Vader even without The Force.
Okay, show of hands. Who thought this was going to be a Star Wars question coming in?
And if this WERE a Star Wars question, the answer would be “sizr matters not.”
Force and Work and Human Muscles in the Real World.
OK, let’s first start with the idealized physics world: Force is one thing, and Work (or Energy) is another. Force is how hard you’re pushing on something. Work is how much energy is expended moving something. Key point : A stationary force doesn’t do any work.
Example: A barbell sitting on a table. Gravity is pulling down on a barbell (exerting a force on it, in other words). If nothing exerts an upwards force, it will fall. However, the table is exerting an upwards force on the barbell exactly equal to gravity pulling down. So it’s not moving, and no work is done.
If, on the other hand, the upwards force is greater than the downwards force (we just replaced the table with a very comressed spring), then the barbell starts moving up, and whatever is pushing up is expending energy (in this case, it is energy stored in the spring). Coming down, you’re getting energy back (Think about using a cranked up weight to power some kind of machine). In the ideal frictionless physics world, moving something sideways takes no work because there’s no opposite force.
OK, now on to the real world: two things happen. First is friction, which is why moving things sideways takes force, and since it’s moving, it takes energy/work. As we all know, reducing friction (greasing the skids, using wheels or rollers, etc.) will reduce the force you need to push against, and the amount of energy you need. Second real world thing is that human muscles, because of the biology of how they’re built, use up energy any time they exert a force, whether they’re moving or not. The table holding up a barbell is doing no work, but if you’re holding a barbell, you are expending energy just keeping your muscles happy. That’s why lowering a barbell takes energy for humans, even though in physics terms you should be getting energy back.
Try thinking of this as a tug-of-war, which you are slowly losing. Earth is your opponent, pulling with a steady amount of force. If you put in a lesser amount of counterforce, the object will slowly be lowered to the floor. If you put in an equal amount of force, the object will not move. In neither of these cases are you doing “work” in the sense that your efforts are accomplishing anything positive. Work would only count if you put in more force than your opponent, which means lifting the object (or winning the tug-of-war).
You can exert force without accomplishing work, if your force is countered by something equal or greater.