kinetic energy formula

Question

I cannot understand the derivation of the formula for potential energy ($\frac{1}{2}mv^2$). It is my understanding that mathematically the work done by a force $F$ should be the line integral of $F$ over a curve $\gamma$. So the force must be a funtion $F: \gamma \rightarrow \mathbb{R}^3$. Instead, my book and the wikipedia page treat $F$ as a function of time, which does not make sense to me. (I read the whole Analysis on manifolds by Munkres, so feel free to explain by means of manifolds and differential forms)

Answer 1

Treating $F$ as a function of time is simply defining $F$ in terms of 'when' instead of 'where', which is what you have suggested. This is better since $\gamma$ might cross itself, i.e. be at the same place at two different times, and you might experience two different forces at this position.

Answer 2

Let $F$ be the force vector and let $\gamma$ be the motion of the particle from time $t=a$ to $t=b$. The total work is $$ W=\int_a^b F(\gamma(t))\cdot \dot{\gamma(t)} dt $$ Force is the old mass time acceleration so $$ F(\gamma(t))=m\ddot{\gamma} $$ Plugging this into the above formula gives $$ W=\int_a^b m\ddot{\gamma}\cdot \dot{\gamma(t)} dt $$ Doing an integration by parts gives $$ \begin{align} \int_a^b m\ddot{\gamma}\cdot \dot{\gamma(t)} dt=(m|\dot{\gamma}|^2)\big|_a^b-\int_a^b m\ddot{\gamma}\cdot \dot{\gamma(t)} dt \end{align} $$ The same integral (the total work) appears on both sides of the equation so $$ 2\int_a^b m\ddot{\gamma}\cdot \dot{\gamma(t)} dt=(m|\dot{\gamma}|^2)\big|_a^b $$ Substituting the speed $v=|\dot{\gamma}|$ and dividing by 2 gives $$ W=\int_a^b m\ddot{\gamma}\cdot \dot{\gamma(t)} dt=\frac{1}{2}mv\big|_a^b=\frac{1}{2}m(v(b)^2-v(a)^2) $$ If we assume that the initial speed of the object was zero ($v(a)=0$) and let $V=v(b)$, we have that the total work from time $t=a$ to $t=b$ is $$ W=\frac{1}{2}mV^2. $$