In the formula for work, $W=\vec{F}\cdot \vec{d}$, is $\vec{F}$ the resultant force? Is the force $F$ the resultant force on the object, or exactly the force needed to move the object?  
e.g.: I might 'need' only 1 N to move an object over 1 metre, but if I apply a 10 N force to that object over 1 metre, am I still doing the same work as if I applied a 1 N force over the same 1 metre distance?
Or, if I use a resultant force of, say, 100 000 N upwards, to lift an object of 1 kg, 1 metre above the ground, what is the work being done?
 A: Work is done by a thing applying a force.  Whatever force you name defines the work done.
If you apply 10N to a an object over the distance of 1 metre, you did 10N*1m=10J of work.  If you wish to phase it as there being a "needed" portion of the force and an "unneeded" portion, you could say that the "needed" portion of the force did 1N*1m=1J of work.
The latter is not typically useful, but I use it to point out how you can divide up the forces in any way that you find meaningful, and we can talk about the work done by each and every one of those forces.
You can even play games such as saying that that 10N force was actually 100N in the direction of motion and 90N against it.  There's nothing that prevents it, though it might be awkward.  In such a case, you would say the thing applying the 100N did 100N*1m= 100J of work on the object, and the thing applying the 90N in the other direction did -90*1m=-90J of work.  As you can see, if you add up the 100J of work done by one force, and the -90J done by the other, you still get the same 10J of work that was done in total.  Any way you split it up is fine.
A: If you use the net force, then you are calculating the net work
$$ W_{net} = \int\limits_\text{path} \vec{F}_{net} \cdot \mathrm{d}\vec{s} \;.$$
This is useful because the net work appears in the work-energy theorem.
If you use some other force, say the normal force or the weight (gravitational force), 
\begin{align}
W_N &= \int\limits_\text{path} \vec{F}_N \cdot \mathrm{d}\vec{s} \tag{normal}\\
W_g &= \int\limits_\text{path} (-mg\vec{y}) \cdot \mathrm{d}\vec{s} \tag{gravity} \;.
\end{align}
then you are calculating the energy transferred by the force you considered only. This is useful when bootstrapping the existence of potential energies, and leads to the whole infrastructure of energy conservation computed as a set of terms associated with individual interactions.
