How does potential energy increase with no work? If you're dragging an object up a hill at a constant velocity, work is technically 0 (as acceleration is 0), but potential energy constantly increases. How would you represent this situation mathematically, and how does the potential energy increase despite a lack of work?
 A: If the speed stays constant, the net work is zero, but the work done by the individual forces may not be. In your case,
\begin{align}
W_{\text{net}} = W_g + W_{\text{drag}} = 0
\end{align}
So both you and gravity are doing work, it's just that whatever work you do by dragging, gravity does minus that: $W_g = - W_{\text{drag}}$. By doing this work, you're storing energy as potential energy in the Earth-sled system, since
\begin{align}
\Delta U_g = -W_g = W_{\text{drag}}.
\end{align}
A: The work-energy theorem should always be your starting point:
$$\boxed{K_1 + W = K_2} \quad \text{or} \quad \boxed{\Delta K = W}$$
You should read this as:

Change in object's kinetic energy equals total work done on the object.

Total work means work done by all forces! If change in kinetic energy is zero, that means the total work done on the object is zero. But if gravitational force has done some work, which we know it did because object changed its altitude, then there must have been some other force (or forces) which did exactly the opposite work:
$$W_F + W_G = 0$$

Now say you push the object down the hill, i.e. there is no external force except for the initial push which we will neglect at the moment. The object starts at $K_1 = 0$ and along the way the gravitational force did some work. What is the final kinetic energy? Assuming there are no other forces such as friction, the whole gravitational potential energy is converted to the kinetic energy. In general, work done by gravitational force is defined by
$$\boxed{W_G = - \Delta U_G = -(U_{G,2} - U_{G,1})}$$
where $U_G$ is the gravitational potential energy.
A: The net force ends up being 0, but you are still applying a force because gravity is pulling down.
Gravity exerts a force down the hill, so to keep the block at a constant velocity, you must exert a force opposite that. This means that work will be applied.
For example, suppose a block is falling down a vertical shaft. At the top end is a pulley (this is equivalent to pulling a block up a 90 degree incline). If the block has a mass of 10 kg, and we'll approximate Earth's gravitational acceleration with $10 \frac{m}{s^2}$. By Newton's second law, 100 N of force are being applied down the shaft on the box. To counteract that you must pull up with 100 N of force (via the pulley). Suppose you pull it up 10 meters.. This means the the work done is 100 N * 10 m, or 1000 J.
A: Work is displacement times force - it is by no means zero when drugging an object uphill, even at constant velocity.
The work that enters the potential energy is the work done by the field (gravity in this case), and it is finite. There is also work done by the person drugging the object, as well as, possibly, work done by friction - but these do not affect gravitational potential energy.
