Does this count as a proper derivation of the formula for work and kinetic energy? $v_{av}$ = average velocity,
$d$= distance,
$a$ = acceleration,
$m$ = mass
Given:
$$v_{av}*t=d$$
$$t=\frac{v_f-v_i}{a}$$
$$v_{av}=\frac{v_f+v_i}{2}$$
Then:
$$\frac{v_f+v_i}{2}*\frac{v_f-v_i}{a}=d$$
$$\frac{v_f^2-v_i^2}{2a}=d$$
$$\frac12(v_f^2-v_i^2)=ad$$
multiply by m to adjust for the amount of mass that accelerated
$$\frac12m(v_f^2-v_i^2)=mad$$
And if not, why is this not proper as a derivation of kinetic energy and work? What's incorrect, conceptually, about my process?
 A: This is not a proper derivation.
At a fundamental level, there are at least three important points that are not taken into account by this approach:


*

*as you consider a second mass point, it is somewhat difficult to adjust (in a non-arbitrary way) the derivation to obtain the correct energy term related to the angular momentum and/or rigid body rotation (think about a satellite in orbit around a planet, for example);

*this approach, as it is, cannot drive you to the corret answer if you have a force which act orthogonally to the direction of motion of mass; basically, it misses the point of vector calculus: would you use a scalar product, a vector product, or a combination of the two, and why? Without any further hypotesis, it is impossible to decide (think about a mass in uniform, circular motion; or a charge in motion in magnetic field);

*it is neglecting one of the most important issues: the fact that the position, velocities and potentials are in the most general case functions of time. Energy is not about average quantities. Try to write down the energy associated with a falling mass in a uniform gravitational field, derive it with respect to time, and see what happens... If you do the same using your approach, you will end up with meaningless results.

