Calculation of Work Typically I see problems laid out along these lines. If you are lifting a 100 kg weight 1 meter how much work do you do?
Work = Force * Distance
Work = (100 kg) (9.8 m/s^2) * 1 m = 980 J
But it seems to me that if I'm moving it I'm not just overcoming the downward acceleration of gravity. That would just hold it in place. I'm making headway (eventually it gets to 1 m of displacement). I could be accelerating it very quickly against gravity. I know that would be taken into account in Power, but it seems like it should apply to Work as well.
Can someone explain how I should be thinking about this?
Edit I'm guessing this is all about looking at potiential change of the box, but I'm not sure that really accounts for my total work, given the various accelerations I could have exerted.
 A: The idea of net work encompasses the fact that while you may have to accelerate the mass initially to get it to move, it will also decelerate later (in other words, you can stop applying force a little bit before you reach the end, and let its momentum carry it the rest of the way).
Let's show this with a simple example. Assume I am trying to lift a mass M over a distance of h.  Now we take into account that I am initially accelerating the mass. Let's say I accelerate it over a distance d to get enough velocity that it can "coast" to the top. This means that the extra work I have to do over the first $d$ meters is just the work done to get the object to a total height $h$ - the velocity it needs to reach is such that $\frac12 m v^2 = mg(h-d)$. I will leave it up to you to prove that the extra force needed over the distance $d$ is indeed such that while I do more work in the first part of the lift, the total work done is still the same - it is exactly the work needed to overcome the potential energy
All this ignores certain physiological factors - like the fact that "perceived work" on your body is not linear, and that even holding a weight still appears to take work (in the physiological sense that is true - but not in the physics sense).
A: You can consider resolving the work needed to lift mass m as just throwing it up by initial speed 
$V _ 0 =$ and the we know the height to which the mass is reached is .$  \  \ H=V_0^2/2g$
And $V_0^2=2gH$ and enrgy you gave to it is $E =1/2 mV^2=mgh$
which is the work needed to lift mass M to height H!
A: Although Floris and kamran have answered your question, I want to add something pertaining to your edit. Even if you exerted different accelerations, going along different paths, as far as you moved the object from a lower point A to a higher point B in a gravitational field, the work done by you would always be calculated as Force$\times$displacement.
This path independence is a special property of fields where there are no losses, and such fields are called conservative fields. If the gravitation field were not conservative, or lets say you were working against a non-conservative field (e.g. some frictional force), all the different routes you took would give you different amounts of work done.
The key really is whether there are losses (dissipation) or not. In fact, the absence of dissipation in a gravitational field is what allows us to write it purely in terms of a potential. You can't define such potentials in the presence of dissipation, or for non-conservative fields.
To imagine your situation, first remove the human and consider a machine doing the lifting. If the machine is holding the weight in place at a certain height, well lubricated wheels (negligible friction) will allow you to move the machine along with the lifted weight by just pushing it once -- your applied force is converted to kinetic energy without losses and you don't have to do further work to maintain the speed. Now imagine doing this in a flowing stream; you will have to continuously push and do work against the dissipative force of the stream even for horizontal movement, i.e. even if you are not doing work against gravity.
