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The total work by the gas is the enclosed triangular area, $W=\frac{1}{2}{p}_0{V}_0$. The positive heat on the sloping leg isAdd $Q_{AB}=48p_0{V}_0/32$, excluding the emitting part of $BC$ and all of $CA$. HenceAs an independent conservation of energy check, the heat emitted on late $BC$ is $p_0{V}_0/32$ and on $CA$ is $80p_0{V}_0/32$; thus $Q_{\mathrm{in}}-Q_{\mathrm{out}}=16p_0{V}_0/32=W$. Using only the net heat on $BC$ would give an incorrect thermal efficiency.

The rod imposes a mechanical constraint, and its force is perpendicular to the wrecking ball's instantaneous velocity, so it does no work. In the constant parallel Newtonian gravitational field, the ball falls through the vertical distance $l-l\cos {\theta }_0$. Conservation of energy between $B$ and $W$ therefore givesand hence

Use $\mathbf{E}=-\nabla \varphi$ and Gauss's law $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$. The divergence theorem and the decay at infinity giveMultiplying by $1/2$ proves that the two expressions for the electrostatic energy agree.

The uniform volume charge density in the thick shell isSpherical symmetry and the integral form of Gauss's law giveThus $|\mathbf{E}|$ is zero inside, rises continuously through the charged region, reaches $Q/(4\pi {ϵ}_0{b}^2)$ at $b$, and then decays as $r^{-2}$. Taking $\varphi (\infty )=0$ and integrating $E=-d\varphi /dr$ givesThe electric potential is continuous, constant inside, decreases smoothly across the charge, and continues as $1/r$ outside.

As $b\to a$, the charge becomes a surface shell. ThenThe potential remains continuous, but the normal electric field jumps by $Q/(4\pi {ϵ}_0{a}^2)$ across the surface charge density.

The field-energy expression givesThe charge-potential expression gives the same result directly because the entire charge lies where the potential equals $\varphi (a)$:This is the electrostatic energy of a uniformly charged spherical shell.

Since $\varphi (a)=Q/(4\pi {ϵ}_{0}a)$ is proportional to $Q$,to first order. This is the work required to bring the additional charge $\delta Q$ from infinity to the potential created by the charge already present.

The work done along the trajectory isBy Newton's second law, $F=m\ddot{x}$, soThis is the work-energy theorem: the work equals the change in kinetic energy.

A force is conservative when its work between two points is independent of the path, equivalently when it is the negative gradient of a potential energy. Here a potential with $V(0)=0$ isConservation of energy for $x>0$ givesThus

If $v_0>v_e$, the particle escapes with asymptotic speed $\sqrt{v_0^2-v_e^2}$; the graph decreases from $v_0$ to this horizontal asymptote. At the escape velocity $v_0=v_e$, it decreases as $v_e{e}^{-x/(2\lambda )}$ and approaches zero only at infinity. If $v_0<v_e$, it reaches the turning pointwhere $v=0$, then returns on the negative branch of the same curve and oscillates symmetrically about the origin.

For oscillation put $w=1-v_0^2/v_e^2$. One quarter of the period iswhere $u=e^{-x/\lambda }$. Therefore

The first part of $\Delta U$ is the change of the quadratic energy associated with the active Ornstein–Uhlenbeck force. The second is minus the change in the particle's potential energy, measured in thermal-noise units. The time integral is the work done by active propulsion, again divided by its noise scale; the full log ratio is the trajectory's time-reversal asymmetry or entropy production.

In a stationary confining state the two endpoint terms remain $O(1)$ as $T\to \infty$ and have zero mean, while the mean active work and the mean log ratio grow proportionally to $T$. Endpoint-term distributions approach time-independent distributions with positive and negative fluctuations. Under mixing assumptions, the time-integrated work has a large-deviation distribution: its central part becomes approximately Gaussian with mean and variance proportional to $T$, while its far tails scale exponentially in $T$. The log-ratio distribution obeys the corresponding fluctuation theorem symmetry.

With an external force $\mathbf{F}$, the required forward and backward noise histories becomeRepeating the difference of squares and using $2\zeta /{\sigma }^2=\beta$ givesThe second term is $\beta$ times the work performed by the external force.

Consider a circular circumplanetary orbit well inside the Hill sphere, with dust speed $u^2=G{M}_p/a$ and unit plane normal $\hat{\mathbf{k}}$. During one dust orbital period, take the stellar direction $\hat{\mathbf{R}}$ and flux at the planet's stellar radius $a_p$ as constant. The velocity-dependent Poynting–Robertson drag has relative workThe constant terms have zero work over the closed circular orbit. Since $⟨u_i{u}_j⟩=(u^2/2)({\delta }_{ij}-k_i{k}_j)$,The coefficient is three for coplanar stellar and dust orbits; it is two for a dust plane normal to the instantaneous stellar direction. Averaging over a circular planetary year at fixed dust orbital inclination $I$ gives $3-{\sin }^{2}I/2$. Conservative perturbations must remain small enough for this circular-orbit average.

With $A=\beta G{M}_{\star }/c$ constant, average the drag work $\dot{\epsilon }=-A(2{\dot{r}}^2+v_{\theta }^2)/r^2$ over a bound elliptic orbit. Using $dt=r^{2}df/h$ gives the integral $\pi (2+3e^2){\mu }^2/h^2$, where $\mu$ is the conservative central parameter. Combining with $\dot{a}=2a^2\dot{\epsilon }/\mu$ givesThis is a weak-drag secular average. It also holds when the conservative parameter is $(1-\beta )G{M}_{\star }$, provided a bound orbit exists; the conservative parameter cancels but the drag amplitude remains $A$. For a circular Kepler orbit, $a^2(t)=a_0^2-4At$.

The net work on a particle equals its change in kinetic energy: