Does a different opposing force affect work? Suppose a man exerts $10~N$ as he lifts a $1~kg$ box a distance of $2~m$ against Earth's gravity.
To determine work we can use the following equation:
$$
W = F \cdot d \\
W = (10~N) \cdot (2~m) = 20~J
$$
The work in this case is $20~J$.
Would work be the same if the man performed this task on the moon rather than the Earth?
Mathematically, the equation shows that if the man exerts the same force ($10~N$) over the same distance ($2~m$), then the work will remain the same ($20~J$) -- but I'm having trouble conceptualizing this.
$$
MoonGravity < EarthGravity
$$
The force opposing our movement when lifting the box on the moon would be less than that on Earth since the gravity on the moon is far smaller. Conceptually, it seems that it'll be "easier" to lift the box on the moon, and thus should take less work.
Where is the fault in my logic?
 A: The fault is assuming that the same average force will be able to lift your object to the same height on earth and in moon. Suppose, we consider lifting an object with zero speed to a height of $x$ meter on earth. Then the force required would be the gravitational force on the object ($mg_{earth}$). But on the moon, same lifting (without any acceleration) will require lesser force ($mg_{moon}$). Therefore, if we talk about the same average forces, lifting the same object, the distance to which it will be able to lift will be different for different gravity (or opposing force in general).  
But in all cases the work done by the object (always given by $\vec F.\vec{ds}$) will be equal to the rise in the gravitational potential energy of the object ($mgh$ for small displacements) which will be much less on moon as compared to earth for the same height due to lower gravity.
A: the fault is in the assumption that work done on the box only goes towards gravitational potential energy. take note that if the applied force exceeds the weight of the box at any time, the box will accelerate and also attain kinetic energy.
assuming $g_{earth}$ = $10N/kg$, then the box never accelerates, and the full $ 20J$ work done goes into increases the box's potential energy.
assuming $g_{moon}$ = $1.6N/kg$, then by $W_{PE}\approx mgh$, $W_{PE}= 3.2J$. the "missing" $16.8J$ goes towards kinetic energy (this box is moving very fast by the time it reaches the top).
I'll leave it to you to confirm that the KE is indeed 16.8J. (hint $v^2=2as$, where $a=\frac{(10N-1.6N)}{1kg} $and $s=2m$
