Don't be surprised that physics has a lot of definitions that are circular. Ultimately, we are just describing the universe.
Work and energy have been defined in a certain way in newtonian physics to explain a kinematic model of reality. This is a model, not reality - you will find no such thing in reality. However, in many scenarios, it is close enough to reality to be useful.
For example, let's say that a human has a 10% efficiency at converting food to mechanical work. So if you spend 1000 kJ of food energy to press against a wall, are you doing 1000 kJ of work, or 100 kJ of work, or 0 kJ of work?
In strict mechanical sense, you did no work whatsoever, and all of the energy you used was wasted as heat. If you instead used this energy to push a locomotive, you would have wasted "only" 900 kJ of the energy as heat, with 100 kJ being work. But the locomotive has its own friction, and it wil stop eventually, wasting all the energy as heat again. And overall, you did expend all those 1000 kJ of food energy that is never coming back.
All of those are simplifications. Kinematics is concerned with things that move. Using models is all about understanding the limits of such models. You're trying to explain thermodynamics using kinematics - this is actually quite possible (e.g. the kinematic theory of heat), but not quite as simple as you make it. Let's look at the fire example. You say there is no displacement, and therefore no work. Now, within the usual context kinematics is used, you are entirely correct - all of that energy is wasted, and you should have used it to drive a piston or something to change it to useful work.
Make a clear note here: what is useful work is entirely a human concept - it's all 100% relevant only within the context of your goals; if you used that "waste" to heat your house, it would have been useful work. It so happens that if you look closer, you'll see that the heat from the fire does produce movement. Individual molecules forming the wood wiggle more and more, some of them breaking free and reforming, and rising with the hot air away from the fire, while also drawing in colder air from the surroundings to feed the fire further. There's a lot of displacement - individual molecules accelerate and slow down, move and bounce around... But make no mistake, the fact that kinematics can satisfactorily explain a huge part of thermodynamics is just a bonus - nobody claimed that kinematics explains 100% of the universe. It was a model to explain how macroscopic objects move in everyday scenarios. It didn't try to explain fire.
For your specific questions, you really shouldn't ask multiple questions in one question. It gets very messy. But to address them quickly:
- There is no potential energy in the kinematic model. The concept is defined for bound states, which do not really exist as a concept in kinematics. In other models, you might see that there's a difference between, say, potential energy and kinetic energy - no such thing really exists in reality. You need to understand the context of the model.
- In a perfectly kinematic world, this is 100% correct. However, as noted before, kinematics isn't a 100% accurate description of reality, and there are other considerations that apply, such as the fact that humans have limited work rate, limited ability to apply force, and the materials we are built of aren't infinitely tough, perfectly inflexible and don't exist in perfect isolation from all the outside (and inside) effects. In real world applications of models, these differences are usually eliminated through understanding the limits of given models, and using various "fixup" constants - and if that isn't good enough, picking (or making) a better model.
- You're mixing up many different models at different levels of abstraction and of different scope so confusion is inevitable. Within the simplified context of kinetics, there is no concept of "potential energy". You simply have energy that can be used to do work, and that's it; it doesn't care about how that energy is used to do work, about the efficiency of doing so etc. In another context, it might be very useful to think of energy and mass as being the same thing - and in yet another, they might be considered interchangeable at a certain ratio, or perhaps in a certain direction, or at a certain rate. It's all about what you're trying to do.
- How is that equation useful? That's the only thing that matters about both definitions and equations. I can define a million things that are completely useless if I wanted to - but what's the point?
- Within the original context, those aren't considered at all. Within a more realistic context, both heat and sound are also kinematic.
The reason you have so much trouble finding the answer to your questions on physics sites and forums is that the question doesn't make much sense in physics. It's more about the philosophy of science, and the idea of building models of the world that try to describe reality to an approximation that happens to be useful to us. You think that those words have an inherent meaning that is applicable in any possible context - this simply isn't true. From the very inception of the idea of physics, people have known that it isn't (and never will be) an accurate representation of reality; and we've known for a very long time that, for example, different observers may disagree on the energy of one object. You just need to understand where a given model is useful, and pick the right model for the job. Don't try to drive a screw with a garden rake.
The potential is defined as a function $U$ such that the conservative force $\vec F$ that we are studying is given by the gradient $\vec F = -\nabla U.$ Since you probably have not seen vector calculus yet, let me be very careful to write this out as the components, $$F_x = -\frac{\partial U}{\partial x},\\
F_y = -\frac{\partial U}{\partial y},\\
F_z = -\frac{\partial U}{\partial z}.$$These "partial derivatives" are evaluated as normal derivatives treating the other variables as constant, so for example the potential $U=k~x^2~y + p~z^2$ would generate $F_x = -2k~x~y, F_y = -k~x^2, F_z = -2p~z.$
Partial derivatives are the natural way to understand calculus on a function of many variables. In single-variable calculus, you were trying to approximate a curve with a tangent line; in this multi-variable calculus we are trying to approximate surfaces with planes. In particular, if you can make this approximation then it means that a function can be expanded around a point, $$f(x + \delta x, y+\delta y, z+\delta z)\approx f(x, y, z) + \frac{\partial f}{\partial x}~\delta x+ \frac{\partial f}{\partial y}~\delta y+ \frac{\partial f}{\partial z}~\delta z.$$Here by $\delta q$ I just mean "a little change in $q$", whatever $q$ is. One could also move this term $f(x,y,z)$ to the left-hand side and refer to that difference as $\delta f$, if you'd like. This understanding of expanding out a multivariable function will be important.
The power exerted by a force on a particle is the dot product of that particle's velocity with the force, and the work done over the path is the time integral of power exerted by that force. It is common to denote the position of the particle as a vector $\vec r(t)$ with components $r_{x,y,z}(t)$ and then this is:
$$ P(t) = F_x~\frac{dr_x}{dt} + F_y~\frac{dr_y}{dt} + F_z~\frac{dr_z}{dt}\\
W = \int_{t_0}^{t_1}dt~P(t).$$
The combination of these two definitions is what you seem to be asking about: but it is not very complicated at all. Combine the two and then stare for a second at the following: $$P(t)~\delta t = -\frac{\partial U}{\partial x} ~\frac{dr_x}{dt}~\delta t -\frac{\partial U}{\partial y} ~\frac{dr_y}{dt}~\delta t -\frac{\partial U}{\partial z} ~\frac{dr_z}{dt}~\delta t.$$What should now stand out to you is that this is very much like the above expression for $\delta f$ above, if we defined $\delta x = \frac{dr_x}{dt}~\delta t$ and so on for $\delta y, \delta z.$ And those are very natural definitions, as $r_x$ represents an $x$-component of position and if we take this time-derivative we get a component of velocity, and multiplying against a short time $\delta t$ we get a small change in this $x$-component due to the particle's current motion.
There is a more formal way to do this and it is to invoke the chain rule, which says that when we apply some function $U(x, y, z)$ to these time-varying components $x = r_x(t)$ and so forth, we find that:$$\frac{d}{dt} \Big(U\big(r_x(t),~r_y(t),~r_z(t)\big)\Big) = \frac{\partial U}{\partial x}~\frac{dr_x}{dt} + \frac{\partial U}{\partial y}~\frac{dr_y}{dt} + \frac{\partial U}{\partial z}~\frac{dr_z}{dt}.$$
Therefore what we have found above is simply, $$P(t) = - \frac{d}{dt} \Big(U\big(r_x(t),~r_y(t),~r_z(t)\big)\Big).$$The work is the time integral of power, but integrals perfectly undo derivatives, and therefore when we do this definite integral we get from the fundamental theorem of calculus, $$\begin{align}
W &= -\int_{t_0}^{t_1} dt~\frac{d}{dt} \Big(U\big(r_x(t),~r_y(t),~r_z(t)\big)\Big) \\
&= -\Big(U\big(r_x(t_1),~r_y(t_1),~r_z(t_1)\big) - U\big(r_x(t_0),~r_y(t_0),~r_z(t_0)\big)\Big)\\
&=-\Delta U.\end{align}$$
That's really all there is to it: for conservative forces $\vec F = -\nabla U$ the power $\vec F \cdot \frac{d\vec r}{dt}$ is immediately seen to be a chain rule expression $\frac{d}{dt} U(\vec r) = \nabla U\cdot \frac{d\vec r}{dt}$, which identifies it as a total time derivative, and therefore the work, which is just the time integral of the power, must be the overall change in the quantity: in this case the quantity is $-U$ and so $W = \Delta(-U) = -\Delta U.$
Best Answer
Positive work indicates positive energy transfer from the e.g field to a mass. via the increase of kinetic energy
$\int F \cdot dr$
$\int F \cdot vdt$
$\int m (dv/dt) \cdot vdt$
$m \int d/dt(1/2 v^2)dt$
$1/2 mv^2$
Plugging in bounds from v0 to v1 where shows this is the difference of kinetic energy
so net positive work indicates a positive kinetic energy change, and net negative work equals negative kinetic energy change
The definition
W= -change in U
Means that if there is a positive change in U, then negative work is done
and if there's a negative change in U, positive work is done
There is no contradiction, as U is defined as the amount of work done against the field from A to B, if positive work is done against the field moving something from A to B then the field does negative work on the object moving from A to B meaning a decrease in Work done on the object
As change in U is defined as $\int_{a}^{b} -F \cdot dr $
This represents the work I would have to do against the field moving something from a to b. Taking the minus out,
$\int_{a}^{b} -F \cdot dr =- \int_{a}^{b} F \cdot dr $
The term on the right is - work done by the field so if Potential difference from A to B is positive then F.dr has to be negative, aka negative work done
EDIT:
To conceptualise this,
consider the equation.
$ke + \int_{a}^{x} F \cdot dr = 0$
Where ke is the amount of energy that the object has initially, e.g in the form of kinetic for this example.
This equation states that, given it has some kinetic energy, in moving from a distance (a to x) the field F does work on the object, such that the total kinetic energy when the object has reached a distance x is zero.
Rearranging for ke,
$Ke = - \int_{a}^{x} F \cdot dr $
This is the kinetic energy an object should have, if I want to reach a position x in the presence of a force field such that the object stops at x
This is what potential is defined as ,but instead of saying a particle has some initial ke, the energy could be given over a distance. same thing.
This is WHY this expression represents work done by an external force(me) to move something from a to b against a force field F , clearly if positive work is required to move something from a to b in the presence of forcefield F, then the forcefield F must do negative work on the object moving through that distance ( as I have to give it energy to move it there, because the field does negative work along the way)