Say I'm riding a bicycle at speed $v$. There's air that causes drag force, let's suppose it's constant, equal $F$, and it doesn't change with speed (we know it does increase quadratically with speed, but it doesn't matter here).
Now, is there any work done by air drag force? The formula says force times distance. We have distance here, because the bike is moving, and force as well. I need to do more work to maintain my speed because of the air drag force, so I'm certainly adding some energy to the bicycle, and the air drag is removing that energy at the same time (thus the speed is constant).
The net work is zero, but when looking from the perspective of individual forces, there IS identical non-zero work on both sides, but with opposite signs.
Now, let's consider a different situation: me and my friend are pushing a box, standing on opposite ends of the box with force $F$. The box isn't moving, but is there any work done by me or my friend? You could say – there's no work, because all the energy your muscles generate is transfered into heat.
But what if the box WAS moving, say, at constant speed $1$ kph? Would there be work then, even though the net force applied to the box by me and my friend was $0$?
There are many places on the Internet where you can find people arguing the can be non-zero work (looking from the perspective of a single force) even if the net force is $0$ (and net work is $0$ too), like these:
https://www.physicsforums.com/threads/work-done-on-object-when-net-force-is-zero.619110/
Best Answer
Let's say that you are standing on the ground and you observe two horizontal forces of equal magnitude, but acting in opposite directions on the block, because of which the net vector sum of all horizontal forces is zero, i.e. $$\sum F_x=0 \implies \left(a_{x}\right)_{net}=0 $$ Thus the block moves with constant velocity. Now, you wan't to calculate the net work done by the forces. The equation for work done by force is given by, \begin{align} W =&\int\limits_{x=x_1}^{x=x_2}\vec{F(x)}\cdot \vec{dx}\\ &=\int\limits_{x=x_1}^{x=x_2}|\vec{F(x)}|\times|\vec{dx}|\times \cos\theta \end{align} where $\theta$ is the smaller angle between force vector $\left(\vec{F(x)}\right)$ and displacement vector $\left(\vec{dx}\right)$, when joined Head-to-Head Or Tail-to-Tail.
Now, since in your case the force vector remains constant throughout the motion, our work equation becomes, \begin{align} W&=\int\limits_{x=x_1}^{x=x_2}|\vec{F(x)}|\times|\vec{dx}|\times \cos\theta \\ & = |F|\times\cos\theta \int\limits_{x=x_1}^{x=x_2}|\vec{dx}|\\ & = |F|\times\cos\theta\times\left(|\vec{x_2}|-|\vec{x_1}|\right)\\ \end{align} One the forces, would be acting along the direction of displacement, the angle made by this force with the displacement vector would be $0^°$, let us call this force say $F_1$. The other force would make an angle of $180^°$ with displacement vector, let us call this force say $F_2$. Over here, $|F_1|=|F_2|$( magnitude of $F_1$ and $F_2$ are equal).
Other than these two forces there can be two additional forces, if we assume that the experiment is being conducted in space where there is gravity, thus a gravitational force; and a normal reaction force if the block is moving on a surface, thereby compressing it due to its weight(because of gravity).The gravitational force always act in vertically downward direction, and the normal reaction force always act normal to the surface. Over here we assume that the block is moving on a horizontal surface, thus the gravitational force and the normal reaction would be anti- parallel to each other and both would make an angle of $90^°$ with the horizontal displacement.
Since there is no acceleration in vertical direction, i.e. $$\left(\vec{a_{vertical}}\right)_{net} = 0$$ $$|\text{gravitational force}| = |\text{normal reaction force}|=|mg|$$
And since there was no initial velocity in vertical direction and the acceleration is also taken $0$, there will be no displacement in vertical direction.
Thus, work done by $F_1$ would be, \begin{align} W_1& = |F_1|\times\cos 0 \times\left(|\vec{x_2}|-|\vec{x_1}|\right)\\ & = |F_1|\times 1 \times\left(|\vec{x_2}|-|\vec{x_1}|\right)\\ & = |F_1|\times\left(|\vec{x_2}|-|\vec{x_1}|\right)\\ \end{align}
Similarly work done by force $F_2$ would be, \begin{align} W_2& = |F_2|\times\cos 180^° \times\left(|\vec{x_2}|-|\vec{x_1}|\right)\\ & = |F_2|\times -1 \times\left(|\vec{x_2}|-|\vec{x_1}|\right)\\ & = -|F_2|\times\left(|\vec{x_2}|-|\vec{x_1}|\right)\\ \end{align}
For Gravitational force, work done would be, \begin{align} W_G& = |mg|\times\cos 90^° \times\left(|\vec{x_2}|-|\vec{x_1}|\right)\\ & = |mg|\times 0 \times\left(|\vec{x_2}|-|\vec{x_1}|\right)\\ & = 0\\ \end{align}
For Normal reaction force, work done would be, \begin{align} W_N& = |\text{normal reaction}|\times\cos 90^° \times\left(|\vec{x_2}|-|\vec{x_1}|\right)\\ & = |mg|\times 0 \times\left(|\vec{x_2}|-|\vec{x_1}|\right)\\ & = 0\\ \end{align}
Therefore, now the net work done by all the forces would be sum of work done by all individual forces.
That is, \begin{align} W_{net}&= W_1 + W_2 + W_G + W_N\\ &=|F_1|\times\left(|\vec{x_2}|-|\vec{x_1}|\right) + \left(-|F_2|\times\left(|\vec{x_2}|-|\vec{x_1}|\right)\right) + 0 + 0\\ &=|F_1|\times\left(|\vec{x_2}|-|\vec{x_1}|\right) - |F_1|\times\left(|\vec{x_2}|-|\vec{x_1}|\right)\\ &= 0\\ \end{align}
Thus, $$W_{net}= 0$$
The net work done on the block by all the forces is $0$, work done by individual forces may or may not be $0$.
This is also evident from the fact that the Kinetic Energy, (given by $\frac{1}{2} m v^2$) remains constant through out the motion as the velocity of the block remains constant, because $\left(a_{x}\right)_{net}=0$ because two equal anti-parallel forces are applied on the block.
And kinetic energy being constant $\Delta K= 0$, and we know that $W_{net} = \Delta K$, thus $W_{net} = 0$
We had earlier assumed that the observer was standing stationary on the ground, now assume there is another observer who is moving with same velocity as the block. For him, in his inertial frame of reference the block hasn't moved, because the displacement of the frame and the block would be equal as they move with the same velocity.
Thus this observer will say that the work done by individual forces is $0$, thus $W_{net}= 0$.
As I had earlier said that the work done by "individual forces may or may not be $0$" it depends on the frame of reference. We will always calculate the work of any individual force in any reference frame from our basic work equation that is,
$$W =\int\limits_{x=x_1}^{x=x_2}\vec{F(x)}\cdot \vec{dx}$$
But all the observers in their respective inertial frame of reference would always agree on $\bf{W_{net}}$. That is why nature as a whole is a conserved system.
In your experiment where you are riding the cycle, you said that the cycle was moving with constant velocity $\implies \Delta K= 0$, which means that $W_{net}$ must equal to $0$. Now, if we take the rider and the cycle as our system, we observe that there are two external forces acting on the system: Air Drag and Muscle Force (excluding the gravitational force and the normal reaction from ground), we also exclude the internal forces as they work in pair according to Newton's Third Law of Motion, thus canceling out the work of each other as a whole (as we observed in the above example how $F_1$ and $F_2$ cancel out each other's work as a whole).
We can write $W_{net}$ as $W_{external} + W_{internal}$, in this case $W_{internal}=0$, as stated above. Now external forces are : Air Drag and Muscle Force.
Caution: Don't confuse Muscle force as an Internal force, we regard it as an internal force, because any organism (like human being) can generate energy whenever it wants from the fat and other nutrients inside body, thus it is regarded as an external force. You can look at it as $\text{Sun's light energy }\underrightarrow{photosynthesis }\text{ food }\underrightarrow{ digestion }\text{ ATP = Muscle Energy}$ thus in fact Muscle energy is Sun's Energy, an external source.
now, \begin{align} & W_{net} = W_{external} = W_{Air Drag} + W_{Muscle} = \Delta K = 0\\ & \implies W_{Air Drag} + W_{Muscle} = 0 \\ & \implies W_{Air Drag} = - W_{Muscle}\\ \end{align}
Thus the rider is trying to keep the energy of the system constant(for a constant velocity) by continuously compensating the energy lost due to Air Drag by putting in his muscle energy.