yaldonTheory
Registered Member
The value of energy is a quantity that is dispersed or gained, regardless of time. Then the value of energy is independent from time.
For example: A candy bar will have a certain amount of energy within it. Say that, we also have a stick of dynamite with the same amount of energy as the candy bar. The energy contained within the candy bar will be dispersed slowly, with a long period of time, after consumption. But, the energy contained within the stick of dynamite will be dispersed rapidly, over a shorter period of time, after it is ignited. As one can clearly see from this example, time is not a factor for the quantity of energy contained within a system.
Now once more, we will derive the kinetic energy formula, hopefully from an approach that you will prefer.
Starting from this equation:
$$v=at$$
The equation above can be plotted on a Cartesian graph, as seen in the following image: https://drive.google.com/file/d/0B-vl7kjDEZ8HXzByUWtrSUd5TDA/view
After multiplying the above equation by dt:
$$[v=at]\cdot dt$$
Performing integration:
$$v\int\limits_0^{t_c}dt=a\int\limits_0^{t_c}t~dt$$
$$v_{max}t_c=\frac{1}{2}at^2_c$$
Where $$v_{max}t_c=s_{max}$$
$$s_{max}=\frac{1}{2}at^2_c$$
$$[t_c=\frac{v_{max}}{a}]^2$$
$$t^2_c=\frac{v^2_{max}}{a^2}$$
After substituting the value of $$t^2_c$$ into the equation for $$s_{max}$$
$$s_{max}=\frac{1}{2}a\frac{v^2_{max}}{a^2}$$
$$s_{max}=\frac{1}{2}\frac{v^2_{max}}{a}$$
After solving for a:
$$a=\frac{v^2_{max}}{2s_{max}}$$
After substituting the value of a into Newton's Second Law, F=ma:
$$F=m\frac{v^2_{max}}{2s_{max}}$$
$$F\cdot s_{max}=\frac{1}{2}mv^2_{max}$$
Where $$F\cdot s_{max}=E_k$$
Then:
As we keep stating, this kinetic energy proof is not part of the 77 pages within The Yaldon Particle Theory. Here will be a direct link to The Yaldon Particle Theory: https://drive.google.com/file/d/0B-vl7kjDEZ8HV2lZN1dId3JpcUU/preview
When one fully understands this model, then one will appreciate the work done in order to develop it. Please do not continue to derail the topic of this thread away from the subject matter contained within this book. Thank you.
For example: A candy bar will have a certain amount of energy within it. Say that, we also have a stick of dynamite with the same amount of energy as the candy bar. The energy contained within the candy bar will be dispersed slowly, with a long period of time, after consumption. But, the energy contained within the stick of dynamite will be dispersed rapidly, over a shorter period of time, after it is ignited. As one can clearly see from this example, time is not a factor for the quantity of energy contained within a system.
Now once more, we will derive the kinetic energy formula, hopefully from an approach that you will prefer.
Starting from this equation:
$$v=at$$
The equation above can be plotted on a Cartesian graph, as seen in the following image: https://drive.google.com/file/d/0B-vl7kjDEZ8HXzByUWtrSUd5TDA/view
After multiplying the above equation by dt:
$$[v=at]\cdot dt$$
Performing integration:
$$v\int\limits_0^{t_c}dt=a\int\limits_0^{t_c}t~dt$$
$$v_{max}t_c=\frac{1}{2}at^2_c$$
Where $$v_{max}t_c=s_{max}$$
$$s_{max}=\frac{1}{2}at^2_c$$
$$[t_c=\frac{v_{max}}{a}]^2$$
$$t^2_c=\frac{v^2_{max}}{a^2}$$
After substituting the value of $$t^2_c$$ into the equation for $$s_{max}$$
$$s_{max}=\frac{1}{2}a\frac{v^2_{max}}{a^2}$$
$$s_{max}=\frac{1}{2}\frac{v^2_{max}}{a}$$
After solving for a:
$$a=\frac{v^2_{max}}{2s_{max}}$$
After substituting the value of a into Newton's Second Law, F=ma:
$$F=m\frac{v^2_{max}}{2s_{max}}$$
$$F\cdot s_{max}=\frac{1}{2}mv^2_{max}$$
Where $$F\cdot s_{max}=E_k$$
Then:
$$E_k=\frac{1}{2}mv^2_{max}$$
It is not a mere coincidence that we keep arriving to the same result for kinetic energy with all of these different approaches, by only using a precise language known as Mathematics. Are you really claiming that all of these proofs are wrong... and you are right? One has to be honest with the results, these proofs are not subjective.As we keep stating, this kinetic energy proof is not part of the 77 pages within The Yaldon Particle Theory. Here will be a direct link to The Yaldon Particle Theory: https://drive.google.com/file/d/0B-vl7kjDEZ8HV2lZN1dId3JpcUU/preview
When one fully understands this model, then one will appreciate the work done in order to develop it. Please do not continue to derail the topic of this thread away from the subject matter contained within this book. Thank you.
