Physics

−0

Your analysis seems pretty good. To take an equivalent but more traditional example, imagine we throw a ball upwards. Ignoring air resistance and approximating the gravitational force as constant, the ball will accelerate downward with constant acceleration. If the initial (upwards) velocity is $v_{0}$ and the downward acceleration is $a = - g$ , the ball will reach the apex at time $t = v_{0} / a$ at which point it will have zero velocity and thus zero kinetic energy. It will have traveled a distance of $v_{0}^{2} / (2 a)$ . In general, work is force applied over a distance. In this case, the work done by the gravitational field is the (constant) force, $F = - m g$ , applied over this distance, $v_{0}^{2} / (2 a)$ , giving a work done of $m v_{0}^{2} / 2$ .

Moving more towards your analysis, once we fix a constant force (and constant mass), then we know work is directly proportional to the distance, $d$ , the mass travels, i.e. $W = F d$ . Your analysis shows that $d$ is quadratic in the initial velocity, thus work is quadratic in the initial velocity. I believe this was your goal. The slightly more detailed analysis in the previous paragraph will reproduce exactly the kinetic energy expression if you wanted more.