If two forces are different (Force A < Force B) but the distance over which the forces are applied is the same, then by the work formula, the object that is pushed with force B should have more kinetic energy. Is there a way we can prove that?
(This is just so that I can understand the work formula better and get a better intuition as to why it is force × distance instead of anything else)
If you accept that the kinematic equations are true, then "force times distance" in the work-energy theorem follows directly from one of them: \begin{align*} v^2 - v_0^2 &= 2 a \Delta x \\ \frac{1}{2}v^2 - \frac{1}{2} v_0^2 &= a \Delta x & \text{(divide by 2)} \\ \frac{1}{2} m v^2 - \frac{1}{2} m v_0^2 &= (m a) \Delta x & \text{(multiply by $m$)} \\ \Delta (\mathrm{KE}) &= F \Delta x & \text{(Newton's Second Law)} \end{align*}
In words, what that first equation is telling you is that for a fixed distance $\Delta x$, an object that has a greater acceleration $a$ over that distance will change its speed (squared) by a greater amount. In particular, if two objects of the same mass start at rest, and one of them experiences a greater force, then it will have a greater acceleration, and it will end up moving faster (with more kinetic energy.)
