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With the reversed electric field, reflection about $x=d/2$ exchanges the electron and Positron. They therefore meet there with equal and opposite momenta. Each travels $d/2$, and the work-energy theorem gives its kinetic energy:Their combined kinetic energy is $1.000\mathrm{MeV}$, while each total relativistic energy is $\mathcal{E}=0.511+0.500=1.011\mathrm{MeV}$. The laboratory is the center-of-momentum frame of the colliding pair.

Local four-momentum conservation in the electron-positron annihilation excludes one real photon: the incoming total four-momentum is timelike, with zero spatial component and positive energy, while a single photon's four-momentum is null and has $|\mathbf{p}|=\mathcal{E}/c>0$. Two photons can balance their momenta by travelling in opposite directions, soThe equal photon energies follow from opposite equal momenta, not merely from sharing the incoming energy arbitrarily. Here $hc\simeq 1.23984\mathrm{MeV}\mathrm{pm}$. This is the free local annihilation idealization implicit in the question; significant momentum transfer to the external apparatus would change the kinematic system being considered.

For two equal point masses $m$ joined by a light tether of length $l$, the central moment of inertia is $m{l}^2/2$. With conserved angular momentum $L$, the rotational kinetic energy is $L^2/(m{l}^2)$. Slow, smooth contraction gives ${\dot{E}}_{\mathrm{rot}}=-T\dot{l}$ with the displayed leading tension. The exact radial correction is $E_{\mathrm{rad}}=m{\dot{l}}^2/4$ and $T=2L^2/(m{l}^3)-m\ddot{l}/2$, so the total work-energy theorem remains exact for non-quasi-static schedules too.

Place two equal point masses of $1\mathrm{kg}$ each on opposite sides of a fixed origin, at equal radius $r$, and let them share a signed angular velocity $\omega =\dot{\theta }$ in a plane. Their total mass is $M=2\mathrm{kg}$ and their moment of inertia about the perpendicular axis is $I=M{r}^2$. Radial pulling forces have zero torque about that origin, soTake an initial radius $r_i=1\mathrm{m}$ and angular velocity ${\omega }_i=1rad/s$, and a final radius $r_f=0.5\mathrm{m}$. ThenandThese are endpoint states with ${\dot{r}}_i={\dot{r}}_f=0$. During the pull, the full kinetic energy also contains radial motion: $K=M{\dot{r}}^2/2+K_{\mathrm{rot}}$. Assume the pulling apparatus supplies all net work to the two point masses; the work-energy theorem then fixes its net work at $+3\mathrm{J}$, including any radial acceleration and braking. Each mass spirals inward during the pull, allowing its inward force to do positive work. Conservation of angular momentum therefore predicts the final angular velocity without requiring kinetic energy conservation. For an apparatus included in a larger system, its stored energy supplies the increase consistently with conservation of energy. OpenStax uses the same mechanism to explain a skater pulling her arms inward. ^[ref]

Rotational kinetic energy is the kinetic energy of a rigid body's rotation. For pure rotation about an axis fixed in an inertial frame,Here $r_{\perp }$ is the perpendicular distance of a mass element $dm$ from that axis, $I$ is the moment of inertia about that same axis, and $\omega$ is the common angular speed, measured in radians per second. The rigid body assumption fixes the relative positions of the mass elements; they share angular velocity but have different speeds $v=\omega r_{\perp }$. Summing their kinetic energies $dm{v}^2/2$ gives the formula. This is a nonrelativistic model; deformation or internal motion can contribute additional kinetic energy.^[ref]

For a computed example, take a uniform disc and an ideal thin ring, each of mass $M=2\mathrm{kg}$ and radius $R=0.30\mathrm{m}$, rotating about its central perpendicular axis at angular speed $10rad/s$. The moment of inertia of a uniform disc is $M{R}^2/2=0.090\mathrm{kg}{\mathrm{m}}^2$, while the ring has moment of inertia $M{R}^2=0.180\mathrm{kg}{\mathrm{m}}^2$. Their rotational kinetic energies are therefore $4.5\mathrm{J}$ and $9.0\mathrm{J}$. The ring stores twice the energy at the same angular speed, because more mass lies farther from the axis. At the same nonzero rotational kinetic energy, its angular speed would instead be $1/\sqrt{2}$ times the disc's. This makes radial placement useful for storing energy in a rotating wheel, although a real design must also withstand the stresses of rotation.

For fixed-axis rotation with constant moment of inertia, write the signed rotation rate as $\Omega =\dot{\theta }$, with angle $\theta$ in radians. The axial component of net external torque obeys ${\tau }_{\mathrm{axis}}=I\dot{\Omega }$, so the work-energy theorem becomesAn opposing torque removes rotational kinetic energy; bearing friction must be included in the net torque. The signed $\Omega$ matters when the direction reverses. Rotation about a fixed axis need not be about a principal axis: the scalar energy formula still holds, but transverse bearing torques can be needed to maintain that axis.

For general rigid body motion, use the inertia tensor ${\mathsf{I}}_G$ about the centre of mass $G$:This is the kinetic-energy decomposition about the centre of mass. In a principal body frame, the rotational term is $\frac{1}{2}(I_1{\omega }_1^2+I_2{\omega }_2^2+I_3{\omega }_3^2)$, using the principal moments of inertia. Away from a principal axis, angular momentum and angular velocity need not be parallel. A freely rotating asymmetric rigid body can change the direction of angular velocity relative to the body even while rotational kinetic energy remains constant; Euler equations for a torque-free rigid body govern that motion. One constant axial moment of inertia is then insufficient.^[ref]

First test rest. With no other tangential forces, the required uphill static friction is $f_s=mg\sin \theta$. Its maximum is ${\mu }_{s}N$, where ${\mu }_s$ is the coefficient of static friction. ThusThe tangent function is increasing on this angle range, and its inverse tangent determines the last permitted resting angle ${\theta }_c$. At equality, the ideal model still permits limiting rest; a larger angle excludes it. Below the threshold, setting $f_s={\mu }_{s}N$ would incorrectly impose the maximum static friction even when less is required. Static friction on an inclined plane and inclined-plane measurement of static friction develop this onset criterion further.

For a block already sliding downhill, uphill kinetic friction has magnitude $f_k={\mu }_{k}N$, where ${\mu }_k$ is the coefficient of kinetic friction. Taking downhill as positive, Newton's second law becomesEvery term in the force balance scales with mass; division by $m>0$ removes it. Doubling mass doubles both gravitational weight and the contact forces, while leaving this acceleration unchanged if the coefficients remain unchanged. A mass-independent result is a prediction of these assumptions, rather than a universal property of every ramp.

For an original numerical model, stipulate $m=3\mathrm{kg}$, $g=9.8m/{\mathrm{s}}^2$, ${\mu }_s=0.45$ and ${\mu }_k=0.30$. The critical angle is ${\theta }_c\simeq 24.23^{\circ }$. At $20^{\circ }$, the normal force is $27.627\mathrm{N}$; rest requires $10.055\mathrm{N}$ of uphill static friction, below its $12.432\mathrm{N}$ bound. A resting block can stay there. If it is already sliding downhill, however, the kinetic friction is only $8.288\mathrm{N}$, giving downhill acceleration $a\simeq 0.5891m/{\mathrm{s}}^2$. Specifying the initial state matters.

At $30^{\circ }$, the downhill component of gravitational weight is $14.700\mathrm{N}$, while $N\simeq 25.461\mathrm{N}$ and the static friction bound is only $11.458\mathrm{N}$. Rest is impossible. During downhill sliding,Released from rest on a sufficiently long ramp, this ideal block covers its first $1.00\mathrm{m}$ in $t=\sqrt{2L/a}\simeq 0.9218\mathrm{s}$, reaching speed $v=\sqrt{2aL}\simeq 2.1697m/s$. As an independent energy calculation, the lost gravitational potential energy is $mgL\sin \theta =14.700\mathrm{J}$ and kinetic friction removes $f_{k}L\simeq 7.638\mathrm{J}$ of mechanical energy. The remaining $7.062\mathrm{J}$ equals $m{v}^2/2$ by the work-energy theorem.

At $\theta =\arctan {\mu }_k$, downhill sliding has zero acceleration and therefore constant velocity. Below that angle, a downhill-moving block slows; when it stops, test the static friction bound again. For uphill sliding, kinetic friction points downhill, so the signed downhill acceleration instead is $g(\sin \theta +{\mu }_k\cos \theta )$. The sign follows contact slip, and the downhill formula cannot be used unchanged after a reversal.

Work measures the energy transferred by a force as its point of application moves. For a point mass following a path $C$, add the contribution of the force along each small displacement:This line integral uses the actual path, and the dot product selects the component of force along the motion. A large force perpendicular to every small displacement does zero work; holding something still also does zero mechanical work on it, even if holding it feels tiring. The SI unit is $\mathrm{J}=\mathrm{N}\mathrm{m}$. ^[ref]

The usefulness of work is that it connects a description in terms of forces and distances to one in terms of energy. The work-energy theorem gives the change in kinetic energy from the total net work, often without solving for the entire motion. The work of one selected force equals that change only when the other forces contribute zero total work.

For a point mass of constant mass $m$ observed in an inertial frame, the net work along its actual path equals its change in kinetic energy:The middle expression is a line integral; the dot product selects the component of force along the displacement. Positive net work increases speed, negative net work decreases it, and zero net work leaves the endpoint speeds equal. Equal endpoint speeds do not imply equal velocities: a perpendicular force can continually change direction while doing zero work. ^[ref]

The bridge to Newton's second law is short: take the dot product of $m\dot{\mathbf{v}}={\mathbf{F}}_{\mathrm{net}}$ with velocity. The chain rule gives $d(m|\mathbf{v}{|}^2/2)/dt={\mathbf{F}}_{\mathrm{net}}\cdot \mathbf{v}$, equating the rate of change of kinetic energy to instantaneous power. Integrating gives the theorem. No conservative force assumption entered this argument; friction also contributes to the net work.

For the harmonic oscillator with spring stiffness $k=2N/m$, release a mass $m=0.5\mathrm{kg}$ from rest at $x=0.3\mathrm{m}$. The spring exerts $F=-kx$. Its work before the first passage through $x=0$ isThus $m{v}^2/2=0.09\mathrm{J}$ gives speed $0.6m/s$ there. The velocity is negative on this passage because the position is decreasing. The energy calculation alone determines its magnitude, so the direction still requires information about the motion.

For a time-independent potential energy $V$, the conservative force contributes work $-\Delta V$. Separating all other work as $W_{\mathrm{other}}$ gives $\Delta (T+V)=W_{\mathrm{other}}$. Consequently conservation of mechanical energy requires zero net other work; the work-energy theorem itself does not require that conservation. With damping, mechanical energy decreases even though the theorem remains valid. ^[ref]