A driven one-dimensional model with a time-dependent potential and a residual force: the mechanical-energy balance, the integrated work form and the normalization ambiguity computed on explicit data

Work in the force branch of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance with one label $N=1$, mass $m>0$, time window $I_0=\mathbb{R}$ and configuration domain $Q={\mathbb{R}}^3$. Write $q=(x,y,z)\in Q$ and $e_1=(1,0,0)$. Define $\mathbf{d}(t,q,v):=d(t,x,v)e_1$ and $\mathbf{F}(t,q,v):=F(t,x,v)e_1$ from the scalar expressions below, and set $U(t,q):=U(t,x)$, independent of $y,z$. Here the velocity argument of $d,F$ may be any $v\in {\mathbb{R}}^3$; these expressions are independent of it. Declare the potential presentation of Potential presentations of a force model, potential and mechanical energy, closed versus exact force fields, local potentials and equipotential level sets clause (a) $U(t,x):=\frac{1}{2}\lambda t{x}^2,d(t,x,v):=2ma+\lambda a{t}^3,$ where $a>0$ is a constant of dimension $\mathrm{\ell }{\tau }^{-2}$ and $\lambda >0$ a constant of dimension $\mu {\tau }^{-3}$, so that $U$ has dimension $\mu {\mathrm{\ell }}^2{\tau }^{-2}$ and $d$ has the dimension of a force (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework). Both maps are smooth; the force map of the model is $F(t,x,v)=-{\mathrm{\partial }}_{x}U(t,x)+d(t,x,v)=-\lambda tx+2ma+\lambda a{t}^3$, and the model satisfies $\mathbf{F}=-{\mathrm{\nabla }}_{q}U+\mathbf{d}$ on $\mathbb{R}\times {\mathbb{R}}^3\times {\mathbb{R}}^3$. On each compact time interval $\mathrm{\mid }t\mathrm{\mid }\le T$, its force has state-Lipschitz constant $\lambda T$; $\mathbf{d}$ is state-independent. Thus both satisfy the standing regularity. The one-dimensional computation below uses the rectilinear motion $q(t)=x(t)e_1$ and zero transverse coordinates and velocities. Each scalar force and residual-force expression denotes its $e_1$ component, $\mathrm{\parallel }q\mathrm{\prime }(t){\mathrm{\parallel }}_2^2=x\mathrm{\prime }(t{)}^2$, and $⟨\mathbf{d},q\mathrm{\prime }⟩=dx\mathrm{\prime }$; no constrained configuration domain or reaction force is assumed.

The curve $x(t):=a{t}^2(t\in \mathbb{R})$ is a motion of this model, and with $K$, $\mathcal{P}$ of Total mass, total momentum, centre of mass, angular momentum, torque, kinetic energy, power and work of a particle model in an inertial frame and $E=K+U(t,x(t))$:

For the numerical clauses below, put $s:=t\mathrm{/}(1\mathrm{s})$; displayed energy, power and work values include their SI units, and the interval $0\le s\le 1$ means $0\le t\le 1\mathrm{s}$.

1. on the interval $[0,1\mathrm{s}]$ the numerical data $m:=1\mathrm{k}\mathrm{g}$, $a:=1\mathrm{m}{\mathrm{s}}^{-2}$, $\lambda :=1\mathrm{k}\mathrm{g}{\mathrm{s}}^{-3}$ give $K(t)=2s^2\mathrm{J},U(t,x(t))=\frac{1}{2}{s}^5\mathrm{J},E(t)=(2s^2+\frac{1}{2}{s}^5)\mathrm{J},E^{\mathrm{\prime }}(t)=(4s+\frac{5}{2}{s}^4)\mathrm{W},$ and the balance of Mechanical energy balance for a potential presentation: $E^{\mathrm{\prime }}={\mathrm{\partial }}_{t}U$ plus the residual power, conservation for a static potential with no residual forces, and the integrated work form clause 1, $E^{\mathrm{\prime }}={\mathrm{\partial }}_{t}U(t,x(t))+d\dot{x}$, holds with ${\mathrm{\partial }}_{t}U(t,x(t))=\frac{1}{2}{s}^4\mathrm{W}$ and $d\dot{x}=(4s+2s^4)\mathrm{W}$;
2. the power is $\mathcal{P}(t)=4s\mathrm{W}$, the work over $[0,1\mathrm{s}]$ is $W([0,1\mathrm{s}])=2\mathrm{J}$, and the work-energy theorem $K(1\mathrm{s})-K(0)=W([0,1\mathrm{s}])$ holds; the integrated form of clause 3 of Mechanical energy balance for a potential presentation: $E^{\mathrm{\prime }}={\mathrm{\partial }}_{t}U$ plus the residual power, conservation for a static potential with no residual forces, and the integrated work form holds as $2\mathrm{J}=-\frac{1}{2}\mathrm{J}+(\frac{1}{10}+2+\frac{2}{5})\mathrm{J}$;
3. replacing $U$ by the equivalent potential $\tilde{U}(t,x)=U(t,x)+c(t)$ with $c(t):=-(1\mathrm{J})(t\mathrm{/}(1\mathrm{s}){)}^2$ (same force, different representative) gives $\tilde{E}(t)=E(t)+c(t)=(s^2+\frac{1}{2}{s}^5)\mathrm{J}$ and ${\tilde{E}}^{\mathrm{\prime }}(t)=(2s+\frac{5}{2}{s}^4)\mathrm{W}={\mathrm{\partial }}_t\tilde{U}(t,x(t))+d\dot{x}$, so the balance holds in the renormalized presentation with a different value and rate;
4. the model is driven and the energy is not conserved: $E(1\mathrm{s})-E(0)=\frac{5}{2}\mathrm{J}$ in the first normalization, while the balance attributes the change to the explicit time dependence, ${\int }_0^{1\mathrm{s}}{\mathrm{\partial }}_{t}Udt=\frac{1}{10}\mathrm{J}$, and the residual power, ${\int }_0^{1\mathrm{s}}d\dot{x}dt=\frac{12}{5}\mathrm{J}$, whose sum $\frac{1}{10}\mathrm{J}+\frac{12}{5}\mathrm{J}=\frac{5}{2}\mathrm{J}$ is the energy change $E(1\mathrm{s})-E(0)={\int }_0^{1\mathrm{s}}{E}^{\mathrm{\prime }}dt$.

• The three contributions are visible. On this data the balance reads $E^{\mathrm{\prime }}=(\frac{1}{2}{s}^4)\mathrm{W}+(4s+2s^4)\mathrm{W}$: the first term is the explicit time dependence of the potential and the remaining terms are the power of the residual force. Both are positive for $0<t\le 1\mathrm{s}$, so the model is driven; this is the behaviour clause 2 of Mechanical energy balance for a potential presentation: $E^{\mathrm{\prime }}={\mathrm{\partial }}_{t}U$ plus the residual power, conservation for a static potential with no residual forces, and the integrated work form excludes by hypothesis.

• The representative matters. The renormalized potential $\tilde{U}$ presents the same force and the same motion, yet $\tilde{E}$ differs from $E$ by $c(t)$ and its rate by $c^{\mathrm{\prime }}(t)$; a conservation statement must name the representative. The numerical data are declared model parameters, not measured values.