Why do some forces' work depends on path and others not.

Nonconservative forces cause *energy loss* during displacement. For example, friction when an objects moves over a surface converts the stored energy to heat that disappears and is wasted. Therefore the final state depends on how long the path was, because that determines how much energy is lost along the way.

Conservative forces cause no loss in energy. Therefore, energy associated with such forces can only be converted into other stored forms in the object (kinetic energy) or system (potential energy). In fact the work done by a conservative force is what we describe as *potential energy*. The word "potential" gives the feeling that it is stored; it is merely a name for the work that the conservative force will do when released. And when released, that potential energy will be work done on the object and it turns into kinetic energy, which is still stored in the body. If you are told what the start and end speeds are, you therefore know that the difference in kinetic energy must be stored. Regardless of the path.

We can consider the conservation in terms of energy like here or entropy and maybe others as well. I personally find the energy approach the most intuitive.

When I hold a thing in my hand and make it follow a short and long random path in different cases and come to the some position 'x' , I feel I have done different amount of work in both cases. But gravitation being a conservative force says I have done equal amount of work in both cases. Where am I mistaken?

Gravity might be a conservative force, but the force you exert on the object is not.

Also why do 2 forces exist? Forces are forces they must be same nature.

Which two are you thinking of?

In any case, yes, forces are "the same thing" so to speak. It doesn't matter what "kind" of force or what created the force - forces are forces and they can be added, for example in Newton's laws, where we don't care about the "type" of force.

Lastly, -can I state all unidirectional forces are conservative?

What do you mean by unidirectional force?

If gravity pulls downwards, so a box slides down an incline, there can still be a friction in only one direction on the incline. The directionality is not a measure of if a force is conservative or not.

Instead think about what kind of energy that force causes. Is it potential or kinetic, then the force is conservative. Is it heat or alike, then not.

-Are there other classifications of forces?

There are many "types" of forces: Electric, magnetic, chemical, gravitational, elastic etc. Those are just names that tell us the origin of them. As stated above, the "type" or origin is of no importance; all forces can cause acceleration in the same manner.

## Best Answer

Your conclusions are not correct. Here is a simple counter-example. Consider this force $$\vec{F}=k(x\hat{y}-y\hat{x})$$ where $\hat{x}$ and $\hat{y}$ are the unit-vectors in $x$ and $y$-direction, and $k$ is some constant.

From this definition we see, the magnitude of the force is $F=k\sqrt{x^2+y^2}=kr$, and its direction is at right angle to $\vec{r}=x\hat{x}+y\hat{y}$. So we can visualize this force field like this:

The force circulates the origin in a counter-clockwise sense.

This force clearly satifies your first condition

But it is not of the form $\vec{F}=f(r)\hat{r}$.

And this force violates your second condition

To prove this consider the following two paths:

Then the work for path A is (because here $\vec{F}$ is always parallel to $d\vec{r}$) $$W_A=\int \vec{F}(\vec{r}) d\vec{r}=kR\cdot\pi R=\pi k R^2.$$

Then the work for path B is (because here $\vec{F}$ is always antiparallel to $d\vec{r}$) $$W_B=\int \vec{F}(\vec{r}) d\vec{r}=-kR\cdot\pi R=-\pi k R^2.$$

You see, the work is different for the two paths, although the start and end point of the paths are the same.

This is a simple example of a non-conservative force. The non-conservativeness can easily be checked by calculating its curl and finding it is non-zero.

$$\vec{\nabla}\times\vec{F} =\vec{\nabla}\times k(x\hat{y}-y\hat{x}) =2k\hat{z} \ne \vec{0}$$