What do you mean by "the amount of energy that is available at any moment"? What kind of energy are you talking about? In what forms do we find this energy, for a point particle of mass m and no internal structure? How can this energy be extracted?
On that note, I am wondering whether yaldon theory uses calculus at all. Does it? Or can the originators of yaldon theory only cope with algebra?
Not knowing calculus would help to explain why you got the work-kinetic energy theorem wrong.
Obviously you have not read our work, due to the fact that we use Calculus throughout
The Yaldon Particle Theory.
When a mass (
m) is under a net force that is equal to zero, then the energy will be potential energy.
For example: When a mass is elevated and kept at a height of
y meter(s), it will have potential energy $$(E_p=mgy)$$. In the same way, when a mass travels in empty space with a constant velocity, it will have a net force that is equal to zero (Newton's First Law). Then this mass will have potential energy that is equal to $$E_p=mv^2_c$$; where $$v_c$$ is the constant velocity. When this mass (
m) has a variable velocity, then it will have kinetic energy.
We will translate the above statement to Math using Newton's Laws:
For Kinetic Energy:
$$F(v)=\frac{m}{t}v$$
$$[F(v)=\frac{m}{t}v]\cdot dv$$
$$\int\limits_0^{v_{max}}F(v)\cdot dv=\frac{m}{t}\int\limits_0^{v_{max}}v\cdot dv$$
$$Fv_{max}=\frac{1}{2}\frac{m}{t}v^2_{max}$$
$$F[v_{max}\cdot t]=\frac{1}{2}mv^2_{max}$$
Where $$F\cdot s_{max}=E_k$$
$$E_k=\frac{1}{2}mv^2_{max}$$
For Potential Energy:
$$F(s)=m\frac{v}{t}$$
Where $$m\frac{v}{t}$$ is a constant value; then the velocity will also be a constant $$(v_c)$$.
After multiplying both sides of the above equation by
ds and taking the limit from zero to $$s_{max}$$:
$$F(s)\int\limits_0^{s_{max}}ds=m\frac{v_c}{t}\int\limits_0^{s_{max}}ds$$
$$F\cdot s_{max}=m\frac{v_c}{t}\cdot s_{max}$$
Where $$F\cdot s_{max}=E_p$$
$$E_p=mv_c\frac{s_{max}}{t}$$
Where $$\frac{s_{max}}{t}=v_c$$, since both are properties of the same mass (
m)
$$E_p=mv^2_c$$
When $$E_k=E_p$$ for the same mass (
m)
$$\frac{1}{2}mv^2_{max}=mv^2_c$$
Then:
$$\frac{1}{2}v^2_{max}=v^2_c$$
Or:
$$v^2_{max}=2v^2_c$$
The above mathematical proof is not needed to explain
The Yaldon Particle Theory, we just want to show the relation between the Laws of Newton and energy. Since the formulas $$E_k=\frac{1}{2}mv^2_{max}$$ and $$E_p=mgy$$ are accepted, then one must accept the Laws of Newton; due to the fact that these formulas have been derived from Newton's Laws. As one can see, we don't need a "rest mass" versus "relative mass" philosophy.
As one can see, when a mass moves with a constant velocity; it will have potential energy.
The Yaldon Particle Theory has arrived at this conclusion; due to the correct interpretation of Newton's Laws.
