From what I have learned work is only done when a force displaced something against another force. (For example if you lift something you lift it opposite to gravity)

From what you are showing, Wikipedia is simply using the example of someone doing work to lift an object against gravity, and gravity doing work on the freely falling object converting the potential energy into kinetic energy according to the work- energy theorem. I don't see it saying that all work is done "against another forc" (maybe you left something out?)

If you have a mass at rest on a frictionless horizontal surface and you apply a constant horizontal force $F$ through a distance $d$ in the direction of the force, the work done will equal $Fd$ and there is no opposing force (assume no air drag). This work was not done "against another force". In addition, if the mass started at rest by the time it has gone a distance $d$ it will acquire kinetic energy according to the work-energy theorem which states that the net work done on an object equals its change in kinetic energy, or

$$\frac{mV^2}{2}=Fd cos θ$$

Hope this helps.

From what I have learned work is only done when a force displaced something against another force. (For example if you lift something you lift it opposite to gravity)

From what you are showing, Wikipedia is simply using the example of someone doing work to lift an object against gravity, and gravity doing work on the freely falling object converting the potential energy into kinetic energy according to the work- energy theorem. I don't see it saying that all work is done "against another force" (maybe you left something out?)

If you have a mass at rest on a frictionless horizontal surface and you apply a constant horizontal force $F$ through a distance $d$ in the direction of the force, the work done will equal $Fd$ and there is no opposing force (assume no air drag). This work was not done "against another force". In addition, if the mass started at rest by the time it has gone a distance $d$ it will acquire kinetic energy according to the work-energy theorem which states that the net work done on an object equals its change in kinetic energy, or

$$\frac{mV^2}{2}=Fd cos θ$$

Hope this helps.