Situación en la que la fuerza total que actúa sobre una partícula $k$ de un sistema de $N$ partículas sólo depende de las posiciones de éstas dentro del mismo:
$\vec{F}_k = \vec{F}_k (\vec{r}_1, \vec{r}_2, \dotsc, \vec{r}_N ), \quad k = 1, \dotsc, N$
En este caso es posible que exista, se considerará que es así, una función potencial $V (\vec{r}_1, \dotsc, \vec{r}_N )$ de $3N$ variables, $\vec{r}_k = (x_k, y_k, z_k )$, tal que:
$\begin{gather} F_{kx} = - \dfrac{\partial V}{\partial x_k}, \quad F_{ky} = - \dfrac{\partial V}{\partial y_k}, \quad F_{kz} = - \dfrac{\partial V}{\partial z_k} \\[1em] \vec{F}_k = \left(F_{kx}, F_{ky}, F_{kz} \right) = -\vec{\nabla}_k V \end{gather}$
Según la segunda ley de Newton:
$\begin{align} \vec{F}_k &= m_k \ddot{\vec{r}}_k = m_k \dfrac{d\vec{v}_k}{dt} = \\[1ex] &= \left(m_k \dfrac{dv_{kx}}{dt}, m_k \dfrac{dv_{ky}}{dt},m_k \dfrac{dv_{kz}}{dt} \right) \end{align}$
En donde:
$\begin{gather} \vec{v}_k = v_{kx}\vec{i} + v_{ky}\vec{j} + v_{kz}\vec{k} \\[1em] \begin{split} \dfrac{d\vec{v}_k}{dt} &= \dfrac{dv_{kx}}{dt}\vec{i} + \dfrac{dv_{ky}}{dt}\vec{j} + \dfrac{dv_{kz}}{dt}\vec{k} \\[1ex] &=\left(\dfrac{dv_{kx}}{dt}, \dfrac{dv_{ky}}{dt}, \dfrac{dv_{kz}}{dt} \right)\end{split}\end{gather}$
Por tanto:
$m_k \dfrac{dv_{kx}}{dt} = - \dfrac{\partial V}{\partial x_k}, \quad m_k \dfrac{dv_{ky}}{dt} = - \dfrac{\partial V}{\partial y_k},\quad m_k \dfrac{dv_{kz}}{dt} = - \dfrac{\partial V}{\partial z_k}$
Multiplicando las igualdades anteriores, respectivamente, por $v_{kx}$, $v_{ky}$, $v_{kz}$:
$\begin{array}{l} m_k v_{kx} \dfrac{dv_{kx}}{dt} = - \dfrac{\partial V}{\partial x_k} \underbrace{\dfrac{dx_k}{dt}}_{\displaystyle v_{kx}} \\[1ex] m_k v_{ky} \dfrac{dv_{ky}}{dt} = - \dfrac{\partial V}{\partial y_k} \underbrace{\dfrac{dy_k}{dt}}_{\displaystyle v_{ky}} \\[1ex] m_k v_{kz} \dfrac{dv_{kz}}{dt} = - \dfrac{\partial V}{\partial z_k} \underbrace{\dfrac{dz_k}{dt}}_{\displaystyle v_{kz}} \end{array}$
Sumándolas:
$m_k \left( v_{kx} \dfrac{dv_{kx}}{dt} + v_{ky} \dfrac{dv_{ky}}{dt} + v_{kz} \dfrac{dv_{kz}}{dt}\right) = - \left( \dfrac{\partial V}{\partial x_k} \dfrac{dx_k}{dt} + \dfrac{\partial V}{\partial y_k} \dfrac{dy_k}{dt} + \dfrac{\partial V}{\partial z_k} \dfrac{dz_k}{dt} \right)$
Siendo que:
$\begin{align} m_k \left( v_{kx} \dfrac{dv_{kx}}{dt} + v_{ky} \dfrac{dv_{ky}}{dt} + v_{kz} \dfrac{dv_{kz}}{dt}\right) &= m_k \vec{v}_k \cdot \dfrac{d\vec{v}_k}{dt} = \\[1ex] &= \dfrac{1}{2} m_k \dfrac{d}{dt}\left( \vec{v}_k \cdot \vec{v}_k \right) = \\[1ex] &= \dfrac{1}{2} m_k \dfrac{d}{dt}\left(v_k^2 \right) = \\[1ex] &= \dfrac{d}{dt}\left(\dfrac{1}{2} m_k v_k^2 \right)\end{align}$
Así pues:
$\dfrac{d}{dt}\left(\dfrac{1}{2} m_k v_k^2 \right) = - \left( \dfrac{\partial V}{\partial x_k} \dfrac{dx_k}{dt} + \dfrac{\partial V}{\partial y_k} \dfrac{dy_k}{dt} + \dfrac{\partial V}{\partial z_k} \dfrac{dz_k}{dt} \right), \quad k = 1, \dotsc, N$
Sumando, la anterior igualdad, para todas las partículas:
$\displaystyle \dfrac{d}{dt} \sum_{k=1}^N \left(\dfrac{1}{2} m_k v_k^2 \right) = - \sum_{k=1}^N \left( \dfrac{\partial V}{\partial x_k} \dfrac{dx_k}{dt} + \dfrac{\partial V}{\partial y_k} \dfrac{dy_k}{dt} + \dfrac{\partial V}{\partial z_k} \dfrac{dz_k}{dt} \right)$
Donde el término de la izquierda de la igualdad es la derivada de la energía cinética total:
$\dfrac{dE_{\smash[b]{\text{cin}}}}{dt} = \displaystyle \dfrac{d}{dt} \sum_{k=1}^N \left(\dfrac{1}{2} m_k v_k^2 \right)$
Además, aplicando la regla de la cadena, también:
$\begin{align} &V = V\left(x_1(t), y_1(t), z_1(t), \dotsc, x_N(t), y_N(t), z_N(t)\right) \\[1em] &\begin{split}\dfrac{dV}{dt} &= \dfrac{\partial V}{\partial x_1} \dfrac{dx_1}{dt} + \dfrac{\partial V}{\partial y_1} \dfrac{dy_1}{dt} + \dfrac{\partial V}{\partial z_1} \dfrac{dz_1}{dt} + {} \\[1ex] &+ \dotsb + \dfrac{\partial V}{\partial x_N} \dfrac{dx_N}{dt} + \dfrac{\partial V}{\partial y_N} \dfrac{dy_N}{dt} + \dfrac{\partial V}{\partial z_N} \dfrac{dz_N}{dt} = \\[1ex] &= \sum_{k=1}^N \left(\dfrac{\partial V}{\partial x_k} \dfrac{dx_k}{dt} + \dfrac{\partial V}{\partial y_k} \dfrac{dy_k}{dt} + \dfrac{\partial V}{\partial z_k} \dfrac{dz_k}{dt}\right) \end{split}\end{align}$
Por tanto:
$\begin{align} \dfrac{dE_{\smash[b]{\text{cin}}}}{dt} = - \dfrac{dV}{dt} &\Rightarrow \dfrac{dE_{\smash[b]{\text{cin}}}}{dt} + \dfrac{dV}{dt} = 0 \\[1ex] &\Rightarrow \dfrac{d}{dt}\left(E_{\text{cin}} + V \right) = 0\end{align}$
Así que:
$\boxed{ E_{\text{cin}} + V = E = \text{cte.}}$
En caso de que las fuerzas internas sí deriven de una función potencial $V$, pero no las fuerzas externas, entonces:
$\begin{align} &dW = \sum_{k=1}^N \vec{F}_k \cdot d\vec{r}_k = \sum_{k=1}^N \vec{F}_k \cdot \vec{v}_k\,dt \\[1ex] &\begin{split} \dfrac{dW}{dt} = \sum_{k=1}^N \vec{F}_k \cdot \vec{v}_k &= \sum_{k=1}^N \dfrac{d}{dt} \left( \dfrac{1}{2} m_k v_k^2 \right) = \\[1ex] &= \dfrac{d}{dt} \underbrace{\sum_{k=1}^N \dfrac{1}{2} m_k v_k^2}_{\displaystyle E_{\smash[b]{\text{cin}}}} = \\[1ex] &= \dfrac{dE_{\smash[b]{\text{cin}}}}{dt} \end{split} \\[1em] &dW = dW_{\text{int}} + dW_{\text{ext}} \\[1ex] &dW = \sum_{k=1}^N \vec{F}{}_k^i \cdot d\vec{r}_k + \sum_{k=1}^N \vec{F}{}_k^e \cdot d\vec{r}_k \\[1ex] &\begin{split} \dfrac{dW}{dt} &= \dfrac{\sum\limits_{k=1}^N \vec{F}{}_k^i \cdot d\vec{r}_k}{dt} + \dfrac{\sum\limits_{k=1}^N \vec{F}{}_k^e \cdot d\vec{r}_k}{dt} = \\[1ex] &= \dfrac{\overbrace{\color{blue}{\sum\limits_{k=1}^N -\vec{\nabla}_k V \cdot d\vec{r}_k}}^{\displaystyle \color{blue}{-dV}}}{dt} + \sum_{k=1}^N \vec{F}{}_k^e \cdot \dfrac{d\vec{r}_k}{dt} = \\[1ex] &= - \dfrac{dV}{dt} + \sum_{k=1}^N \vec{F}{}_k^e \cdot \vec{v}_k \end{split} \end{align}$
Por tanto:
$\begin{gather} \dfrac{dE_{\smash[b]{\text{cin}}}}{dt} = - \dfrac{dV}{dt} + \sum_{k=1}^N \vec{F}{}_k^e \cdot \vec{v}_k \\[1ex] \dfrac{d}{dt}\left(E_{\text{cin}} + V \right) = \sum_{k=1}^N \vec{F}{}_k^e \cdot \vec{v}_k \end{gather}$
Integrando se hallaría el trabajo:
$\displaystyle W_{\text{ext},1 \to 2} = \Delta E_{1 \to 2} = \int_{t_1}^{t_2} \negmedspace\sum_{k=1}^N \vec{F}{}_k^e \cdot \vec{v}_k \, dt$