Converting potential energy to kinetic energy

Question

I know how potential and kinetic energy interrelate when you drop anything.

But is it a general understanding in the community that a change in potential energy is always kinetic energy? I am supposing heat and vibrations in general as kinetic energy, as they are so, in tiny scales.

Answer 1

But is it a general understanding in the community that a change in potential energy is always kinetic energy?

Not always. For example a change in gravitational potential energy could lead to a change in elastic potential energy, or chemical potential energy, or electrical potential energy, instead of a change in kinetic energy.

Also, I would not generally consider thermal energy to be a form of kinetic energy. While it can be true for an ideal gas, the behavior of other systems microscopically may be closer to elastic potential energy than to kinetic energy. Also, on the scales where you consider thermal energy you are deliberately ignoring the microscopic scales where thermal energy is kinetic energy for an ideal gas.

Answer 2

A change in potential energy of an object does not have to be accompanied by a change in kinetic energy of that object.

To see this, we know that, according to energy conservation $$K_1+\sum_iU_{1,i}+W_{ext}=K_2+\sum_iU_{2,i}$$ where $1$ and $2$ relate to two points in time, $K$ is kinetic energy, $U_i$ is any potential energy, and $W_{ext}$ is any work done by external/nonconservative forces. Or we can also write this as $$\Delta K+\Delta U=W_{ext}$$ where I have condensed $\sum_i\Delta U_i$ into just $\Delta U$

So let's consider a situation where I am lifting a block at a constant velocity. Since the block is moving at a constant velocity, $\Delta K=0$. But, we also have $\Delta U=W_{ext}>0$. Therefore, I am doing work and increasing the potential energy, but there is no change in kinetic energy.

Another scenario not involving external forces, imagine I place a box on a spring and then the spring compresses. Well then comparing the initial and final states of the box (both at rest), we have $\Delta K=0$ and $\Delta U=\Delta U_{grav}+\Delta U_{spring}=0$. Therefore, we have converted all of the gravitational potential energy into elastic potential energy with no net change in kinetic energy (of course while the box moves from the initial to the final state it will have some kinetic energy from the initial potential energy, but not all of the potential energy goes to this kinetic energy, which it then eventually loses anyway).

So in general, we only for sure get a complete conversion of potential to kinetic energy when the following criteria are met:

1. There is no external/nonconservative work being done
2. There is only one conservative force acting on the object

Then we have $\Delta K=-\Delta U$ where $\Delta U$ only involves one type of energy.

Answer 3

Well, there are only two "types of energy": potential and kinetic.

And there is an energy conservation law.

Logical consequence of these two statements is that total change in potential energy always results in change of kinetic energy.

But I do not thinks there is some deep meaning in this result.