Conservation of Mechanical Energy

In 1638 Galileo described a pendulum experiment anyone can repeat with a shoelace and a nut. Let the bob swing down from some height, and drive a nail into the wall so the string catches on it mid-swing. The bob's path kinks sharply — yet it climbs on the other side to exactly the height it started from, no matter where the nail sits. Something about the motion refuses to care about the route taken. That something is mechanical energy: the sum of kinetic energy $T$ and potential energy $U$ , which stays fixed whenever the only forces doing work are conservative. The principle is the great shortcut of mechanics. Instead of integrating Newton's second law along every twist of a roller-coaster track, you track two scalars and set their sum constant: height traded for speed at a fixed exchange rate. A bead on a curved wire, a planet on an ellipse, a pendulum at large amplitude — problems whose force equations are ugly all surrender to one line of energy bookkeeping. But touch a bicycle brake rim after a long descent and you meet the fine print: everything that does something ends up warm. Real machines leak organized energy into the disordered jiggling of molecules, and that leaked share never volunteers to come back. The full story of this article is therefore double — when the mechanical ledger balances exactly, and how to audit the books (with one number, the efficiency) when friction and drag start skimming from every transaction.

Key takeaways

When every force doing work on a system is conservative, the total mechanical energy $E = T + U$ is constant in time.
Kinetic energy $T = \tfrac{1}{2}mv^2$ measures motion; potential energy $U(\mathbf{r})$ stores configuration-dependent energy for forces derivable from $\mathbf{F} = -\nabla U$ .
The work-energy theorem $W_{\text{net}} = \Delta T$ plus the path-independence of conservative work yields $\Delta(T+U) = 0$ — the route never matters, only the endpoints.
Forces that do no work (normal forces, ideal string tension) leave the ledger untouched; friction and drag do negative work and drain it: $\Delta E = W_{\text{nc}}$ .
Nothing is ever destroyed — "lost" mechanical energy becomes thermal energy, and efficiency $\eta = E_{\text{useful}}/E_{\text{in}}$ measures how much survives in the organized form you wanted.
Energy conservation is the dynamical fingerprint of time-translation symmetry (Noether's theorem) and converts dynamics into algebra: $E$ alone fixes speed as a function of position.
Key idea: if every force that does work is conservative, then $T+U$ is a constant of the motion — energy conservation converts a differential problem in time into an algebraic relation between speed and position.

Where this lives today

The purest industrial use of equation (5) is the gravitational battery. Pumped-storage hydro plants — still the overwhelming majority of the world's grid storage — lift water hundreds of metres uphill when electricity is cheap and let it fall through turbines when it is scarce, banking terawatt-hours as $mgz$ and recovering 75–80% of it round trip. The missing 20% is equation (8) at work: turbulence, friction, and generator losses, all payable as heat. Electric vehicles run the same audit at street scale — regenerative braking routes what would have been brake-disc heat back into the battery, recovering the majority of the kinetic energy that a friction brake would simply degrade. The principle also polices modern computation. Molecular-dynamics simulations of proteins and materials integrate Newton's laws for millions of atoms interacting through conservative force fields; because the exact dynamics conserve $T+U$ , a slow drift of total energy is the standard diagnostic that the numerical integrator or time step is lying. When a simulation's ledger fails to balance, nobody blames the theorem — they fix the code. Three centuries on, that is perhaps the highest compliment mechanics pays to Galileo's nail.

The physical picture

Think of a frictionless ball rolling in a smooth bowl. As it descends it speeds up; as it climbs it slows; and it always returns to exactly its starting height — never higher, never lower. The bowl trades height for speed at a fixed exchange rate. The potential energy $U$ is a landscape, and the kinetic energy is the budget left over after paying for the current altitude. The picture becomes a genuine calculating tool the moment you draw the conserved energy $E$ as a horizontal line across a plot of $U(x)$ — like a flood level across a valley. The particle can only exist where the land lies below the water, $U(x)\le E$ , because the gap $T = E - U$ can never go negative. Where the line meets the curve, the particle momentarily stops and reverses: a turning point. Wells trap, hills repel, and the slope of the land is the force.

Figure

The flood-line picture — one conserved number organizes the whole motion

The same bookkeeping works along any track. Follow a cart over a frictionless hill and watch the two accounts: at every snapshot the kinetic and potential columns change, but their sum is the same stack of joules.

Figure

Height traded for speed at a fixed exchange rate — the two accounts always sum to E

Why should such a conserved sum exist at all? Because the laws of mechanics look the same today as tomorrow. By Noether's theorem, invariance under shifts in time forces a particular combination of motion and configuration — the energy — to stay constant. Friction "breaks" the mechanical version only because organized motion leaks into the microscopic disorder of heat: energy is still globally conserved, just no longer mechanical, and no longer recoverable. Keep that distinction in view; the last third of this article is about auditing exactly that leak.

The mathematics of the ledger

Everything begins with work — the mechanism by which a force changes a body's state of motion. Push with force $\mathbf{F}$ through a small displacement $\mathrm{d}\mathbf{r}$ and only the along-track component counts:

W = \int_{A}^{B}\mathbf{F}\cdot\mathrm{d}\mathbf{r} \qquad (1)

where: $W$ — work done by $\mathbf{F}$ from point $A$ to $B$ (joules); $\mathrm{d}\mathbf{r}$ — an infinitesimal step along the actual path. The dot product is the physics: a force perpendicular to the motion (a normal force, an ideal string tension) integrates to exactly zero work no matter how large it is. Feeding Newton's second law into (1) and integrating along the trajectory produces the work-energy theorem — the anchor to which every energy method is bolted:

W_{\text{net}} = \Delta T = \tfrac{1}{2}mv_B^2 - \tfrac{1}{2}mv_A^2 \qquad (2)

where: $W_{\text{net}}$ — the work done by the net force; $T = \tfrac12 mv^2$ — the kinetic energy. This holds for any forces whatsoever, conservative or not: it is nothing but $\mathbf{F}=m\mathbf{a}$ integrated once. All the leverage of this article comes from asking when the left side can be written as a change in something that depends on position only. That happens precisely for conservative forces — those whose work is path-independent, equivalently those that are the gradient of a scalar:

\mathbf{F} = -\nabla U \quad\Longleftrightarrow\quad \nabla\times\mathbf{F}=\mathbf{0} \qquad (3)

where: $U(\mathbf{r})$ — the potential energy; the curl condition is the practical test. Gravity and spring forces pass it; kinetic friction fails spectacularly — its direction depends on which way you slide, so a round trip does nonzero (negative) work. For a force passing the test, the work from $A$ to $B$ defines a single-valued function of position:

U_B - U_A = -\int_A^B \mathbf{F}\cdot\mathrm{d}\mathbf{r} \qquad (4)

where: the minus sign is a bookkeeping convention: doing work against the force loads the store. Only differences of $U$ are physical — you may put the zero wherever convenience suggests (the floor, the ceiling, infinity), and no measurable prediction changes. Combine (2) with (4): if every working force is conservative, $\Delta T = -\Delta U$ , and the sum is frozen. This is the headline:

E = T + U = \text{constant} \qquad (5)

where: $E$ — the total mechanical energy, set once by the initial conditions and then carried unchanged through the whole motion. Equation (5) is an integral of the motion: it converts "solve $m\ddot{\mathbf r} = \mathbf F$ along the whole path" into "evaluate two scalars at the two points you care about." The two potentials that cover most of introductory mechanics are worth carrying in your pocket:

U_{\text{grav}} = mgz,\qquad U_{\text{spring}} = \tfrac{1}{2}kx^2 \qquad (6)

where: $z$ — height above your chosen reference; $k$ — the spring constant; $x$ — extension from natural length. Near Earth's surface with $g \approx 9.8\ \mathrm{m/s^2}$ , every metre of height stores $9.8$ joules per kilogram — the exchange rate the bowl was using all along. In one dimension, (5) can be solved for the speed outright, which is the flood-line picture written as algebra:

v(x)=\sqrt{\frac{2\,[E-U(x)]}{m}} \qquad (7)

where: the square root exists only where $U \le E$ — the allowed region — and vanishes at the turning points $U(x_t)=E$ . No time-integration was ever performed: position in, speed out. Finally, the honest version for the real world. When non-conservative forces act, they do not invalidate the ledger — they appear as an explicit credit or (almost always) debit line:

\Delta E = W_{\text{nc}} \qquad (8)

where: $W_{\text{nc}}$ — the work done by all non-conservative forces. For friction and drag $W_{\text{nc}} < 0$ : mechanical energy drains monotonically, reappearing joule-for-joule as thermal energy. Dividing what survives by what you paid gives the efficiency,

\eta = \frac{E_{\text{useful}}}{E_{\text{in}}} \qquad (9)

where: $\eta$ — a pure number, $0 < \eta < 1$ for every real machine, often quoted as a percentage. Chained conversions multiply, $\eta_{\text{total}} = \eta_1\eta_2\eta_3\cdots$ , which is why a wall-plug-to-wheels electric drivetrain at $0.92 \times 0.94 \times 0.87 \approx 0.75$ beats a gasoline chain's $\approx 0.30$ — fewer, better stages.

Which equation answers which question

you want	use	notes
speed at a given position	$E$ conservation, eq. (5) or (7)	frictionless, or friction does no work
the work a force did	line integral, eq. (1)	perpendicular forces contribute zero
is a force conservative?	curl test, eq. (3)	friction and drag always fail
height/extension stored energy	standard potentials, eq. (6)	choose the zero anywhere convenient
problem with friction in it	leaky ledger, eq. (8)	$\Delta E$ equals friction work, negative
fraction of energy surviving	efficiency, eq. (9)	series stages multiply

Derivation AConservation of mechanical energy from Newton's second law

We are deriving the headline result (5): that $T+U$ is constant whenever the working forces are conservative. The strategy is to compute the rate of change of the kinetic energy directly from Newton's second law, substitute $\mathbf{F} = -\nabla U$ , and watch the right-hand side reassemble itself into minus the rate of change of the potential energy via the chain rule. The route is short but two hypotheses do real work and must be watched: the force must be a pure gradient (no friction hiding in $\mathbf{F}$ ), and $U$ must carry no explicit time dependence — a potential that changes shape under the particle pumps energy in or out.

Step 1. Differentiate the kinetic energy along the motion. With $T = \tfrac12 m\,\mathbf{v}\cdot\mathbf{v}$ , the product rule gives $\dot T = m\,\dot{\mathbf{v}}\cdot\mathbf{v}$ , and Newton's second law $m\dot{\mathbf{v}} = \mathbf{F}$ turns this into

\frac{\mathrm{d}T}{\mathrm{d}t} = \mathbf{F}\cdot\mathbf{v}

Why: this is the work-energy theorem in rate form — the net force feeds kinetic energy at the rate force-dot-velocity, the instantaneous power.

Step 2. Insert the conservative hypothesis $\mathbf{F} = -\nabla U$ :

\frac{\mathrm{d}T}{\mathrm{d}t} = -\nabla U\cdot\mathbf{v}

Step 3. Recognize the right side as a total time derivative. As the particle moves, $U(\mathbf{r}(t))$ changes at the rate given by the chain rule:

\frac{\mathrm{d}U}{\mathrm{d}t} = \nabla U\cdot\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} = \nabla U\cdot\mathbf{v}

Why the caveat matters: this chain-rule step is valid only if $U$ has no explicit time dependence, $\partial U/\partial t = 0$ . A moving wall or a spring whose stiffness is being cranked adds a $\partial U/\partial t$ term and breaks the conservation law — by exactly the power the external agent delivers.

Step 4. Add the two results: $\dot T = -\dot U$ , so

\frac{\mathrm{d}}{\mathrm{d}t}\bigl(T + U\bigr) = 0

and $E = T+U$ is a constant of the motion, fixed by the initial conditions. Quick check: for free fall with $U = mgz$ , $\nabla U = mg\hat z$ gives $\mathbf F = -mg\hat z$ , recovering the correct gravitational force; differentiating $\tfrac12 m\dot z^2 + mgz$ along the trajectory gives $\dot z(m\ddot z + mg) = 0$ , which vanishes since $\ddot z = -g$ — confirming the sum is constant.

Go deeper · Derivation B · 1 minWhy a conservative force admits a single-valued potential

Derivation BWhy a conservative force admits a single-valued potential

The claim buried in equation (4) is bolder than it looks: it says the integral defining $U$ gives the same answer along every path, so that a single number can be attached to each point of space. We prove this from the curl condition using Stokes' theorem, and the logic illuminates exactly why friction can never have a potential. The strategy: show that zero curl forces zero work around every closed loop, then split any loop into two paths between the same endpoints.

Step 1. Take any closed loop $C$ bounding a surface $S$ , and apply Stokes' theorem to a force with $\nabla\times\mathbf{F} = \mathbf{0}$ :

\oint_C \mathbf{F}\cdot\mathrm{d}\mathbf{r} = \int_S (\nabla\times\mathbf{F})\cdot\mathrm{d}\mathbf{S} = 0

Why: the circulation of a curl-free field vanishes around every closed loop — a round trip through a conservative force field always costs exactly nothing.

Step 2. Given two paths, path 1 and path 2, from $A$ to $B$ , run out along path 1 and back along path 2. That is a closed loop, so the total is zero, which forces

\int_{\text{path 1}}\mathbf{F}\cdot\mathrm{d}\mathbf{r} = \int_{\text{path 2}}\mathbf{F}\cdot\mathrm{d}\mathbf{r}

Why this is the whole ballgame: path-independence is precisely what lets $U_B - U_A \equiv -\int_A^B \mathbf F\cdot\mathrm d\mathbf r$ define a function of the endpoints alone. Galileo's nail-interrupted pendulum is this equation performed live — the kinked path and the free path bank the same energy.

Step 3. Contrast with kinetic friction, magnitude $f$ always opposing the velocity. Around a closed loop of length $L$ it does work $-fL \neq 0$ : the circulation test fails, no single-valued $U$ can exist, and the energy method must switch to the leaky-ledger form (8). This establishes the standard equivalence: conservative ⇔ curl-free ⇔ path-independent ⇔ zero loop work. For gravity $\mathbf F = -mg\hat z$ , any closed loop gives $\oint \mathbf F\cdot\mathrm d\mathbf r = -mg\oint \mathrm dz = 0$ , since the height returns to its starting value.

Go deeper · Derivation C · 1 minThe leaky ledger — dissipation, efficiency, and the bouncing ball

Derivation CThe leaky ledger — dissipation, efficiency, and the bouncing ball

Now we make equation (8) quantitative in the cleanest real system there is: a ball bouncing on a hard floor. The goal is three results — the energy lost per bounce, the geometric decay of bounce heights, and the finite total path length — using nothing but the ledger and one measured number. The strategy is to treat each impact as a "device" with a fixed efficiency, then compound. Watch how the mass drops out of every ratio: the bounce efficiency is a property of the ball-floor pairing, not of the ball alone.

Step 1. Split the forces on the ball into gravity (conservative, potential $mgh$ ) and the brief non-conservative squash at each impact. Between impacts, mechanical energy is conserved; at each impact, equation (8) says the loss equals the (negative) work of internal deformation forces.

Step 2. Define the coefficient of restitution $e$ as the ratio of rebound speed to impact speed. The mechanical energy at the top of each arc is pure potential, so the bounce efficiency is

\eta_{\text{bounce}} = \frac{mgh_1}{mgh_0} = \frac{h_1}{h_0} = e^2

Why the square: energy goes as the square of speed, so a speed ratio $e$ becomes an energy ratio $e^2$ . Note $m$ and $g$ cancel — measure two heights with a ruler and you have performed a precision efficiency experiment.

Step 3. Compound. The floor neither knows nor cares which bounce it is, so each impact keeps the same fraction $e^2$ of whatever arrives:

h_n = h_0\,e^{2n}

Why geometric decay appears: a fixed fractional tax applied repeatedly is exponential decay in disguise — the same mathematics as chained machine efficiencies $\eta_1\eta_2\eta_3\cdots$ in equation (9).

Step 4. Sum the path. The total distance is the first drop plus an up-and-down pair for every rebound:

d = h_0 + 2h_0e^2\left(1 + e^2 + e^4 + \cdots\right) = h_0\,\frac{1+e^2}{1-e^2}

Why it converges: the series ratio is $e^2 < 1$ . Infinitely many bounces, finite total distance — and every joule of the original $mgh_0$ ends up as heat and sound in the ball and floor; the resting ball holds none of it. *The algebra checks out: $h_0 + 2h_0e^2/(1-e^2) = h_0(1+e^2)/(1-e^2)$ ; for $e = 0.80$ and $h_0 = 2.00$ m this gives $2.00 \times 1.64/0.36 = 9.1$ m.*

Limiting cases

Case	Result	Meaning
constraint forces present	$\mathbf{N}\cdot\mathrm{d}\mathbf{r}=0$	normal forces and ideal string tensions act perpendicular to motion — zero work, conservation survives even though such forces are not "conservative"
pure dissipation	$\Delta E = -\!\int f\,\mathrm{d}s < 0$	kinetic friction and drag drain $E$ monotonically; the loss reappears exactly as heat
time-dependent potential	$\dfrac{\mathrm{d}E}{\mathrm{d}t} = \dfrac{\partial U}{\partial t} \neq 0$	a moving wall or varying spring breaks time-translation symmetry; the driver exchanges energy with the system
perfectly elastic bounce, $e \to 1$	$h_n = h_0$ forever	the ideal limit — mechanical energy conserved through the impact
perfectly inelastic, $e \to 0$	ball dead on first contact	all of $mgh_0$ degraded at once
closed, conservative system	$E = T+U$ exactly constant	the cleanest setting: isolated particles, conservative internal forces

Worked examples

1 · Loop-the-loop release height (confidence builder)

A cart starts from rest at height $h$ on a frictionless track and runs into a vertical circular loop of radius $R = 4.0\ \mathrm{m}$ . Find the minimum release height, and the cart's speed at the top and bottom of the loop when released from it. (Take $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. The dangerous point is the loop top, height $2R$ . There the track can only push inward, so the minimum-speed condition is gravity alone supplying the centripetal force: $mg = mv_{\text{top}}^2/R$ , i.e. $v_{\text{top}}^2 = gR = 9.8 \times 4.0 = 39.2\ \mathrm{m^2/s^2}$ .

Step 2. Energy conservation, eq. (5), from release to loop top: $mgh = \tfrac12 mv_{\text{top}}^2 + mg(2R)$ . Solve: $h = 2R + \tfrac{v_{\text{top}}^2}{2g} = 2R + \tfrac{R}{2} = \tfrac{5}{2}R$ .

Step 3. Numbers: $h_{\min} = 2.5 \times 4.0 = \boxed{10.0\ \mathrm{m}}$ , and $v_{\text{top}} = \sqrt{39.2} = \boxed{6.3\ \mathrm{m/s}}$ .

Step 4. Speed at the bottom, released from $10.0$ m: $v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 10.0} = \sqrt{196} = \boxed{14.0\ \mathrm{m/s}}$ . Notice what energy conservation did not need: the shape of the track between release and loop. Only heights entered.

2 · The friction patch (exam level)

A $2.0\ \mathrm{kg}$ block is released from rest at height $h = 3.0\ \mathrm{m}$ on a frictionless ramp. At the bottom it crosses a horizontal rough patch of length $d = 4.0\ \mathrm{m}$ with $\mu_k = 0.25$ , beyond which a second frictionless ramp rises. How high does the block climb the far ramp, and where does it finally come to rest?

Step 1. Energy at the bottom of the first ramp: $E_0 = mgh = 2.0 \times 9.8 \times 3.0 = 58.8\ \mathrm{J}$ .

Step 2. Friction toll per crossing, from eq. (8): $|W_{\text{nc}}| = \mu_k mg\,d = 0.25 \times 2.0 \times 9.8 \times 4.0 = 19.6\ \mathrm{J}$ .

Step 3. After the first crossing the ledger reads $58.8 - 19.6 = 39.2\ \mathrm{J}$ , so the far-ramp climb is $h' = \dfrac{39.2}{mg} = \dfrac{39.2}{19.6} = \boxed{2.0\ \mathrm{m}}$ .

Step 4. The block slides back down with $39.2$ J, pays another $19.6$ J on the return crossing, and climbs the first ramp to $1.0$ m. Coming back, the third crossing costs the final $19.6$ J.

Step 5. Bookkeeping shortcut for the final resting place: the block stops when friction has consumed everything, after a total sliding distance $D$ on the patch given by $mgh = \mu_k mg\,D$ , so $D = h/\mu_k = 3.0/0.25 = 12\ \mathrm{m}$ — exactly three crossings of the $4.0$ m patch. It comes to rest at the far edge of the patch, at the foot of the second ramp, with all $58.8$ J now living as heat in the two surfaces.

Origins · 1 minHermann von Helmholtz and how this was discovered

Hermann von Helmholtz

1821–1894 · Berlin, Königsberg, Heidelberg

Helmholtz came to the deepest law of physics from medicine. As a young Prussian army surgeon he was studying muscle metabolism, hunting down the "vital force" then believed to animate living tissue — and found instead that the heat and work a muscle produces are paid for, exactly, by the chemistry that fuels it. Generalizing ferociously, his 1847 memoir Über die Erhaltung der Kraft (On the Conservation of Force) argued that one conserved quantity — energy — flows unchanged through mechanics, heat, electricity, magnetism, and life itself, with the mechanical version $T+U$ as its cleanest special case. The physics establishment's leading journal rejected the paper as too speculative; Helmholtz published it as a pamphlet. He went on to invent the ophthalmoscope, measure the speed of nerve impulses his teachers had declared unmeasurable, and train a generation of physicists including Heinrich Hertz. The research society that ran much of German science for a century bore his name — a fitting ledger entry for the man who balanced the universe's books.

History

1638

— Galileo's interrupted pendulum: a bob catching on a nail still rises to its release height — energy remembers only the endpoints.

1673

— Christiaan Huygens shows that in elastic collisions the sum of

mv^2

is unchanged, isolating the quantity later called vis viva.

1740

— Émilie du Châtelet, drawing on 's Gravesande's brass-ball experiments, argues the "living force" goes as

v^2

, not

v

— settling the great vis viva controversy.

1843

— James Prescott Joule measures the mechanical equivalent of heat with paddle wheels: friction's "lost" work reappears, joule for joule, as warmth.

1847

— Hermann von Helmholtz unifies mechanics, heat, and electricity under one conservation law in Über die Erhaltung der Kraft.

1918

— Emmy Noether proves that energy conservation is the precise consequence of time-translation symmetry.

Going further

Texts. Morin, Introduction to Classical Mechanics, ch. 5 · Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., §1.2, §2.6 · Landau & Lifshitz, Mechanics, §6 · Taylor, Classical Mechanics, ch. 4 · Feynman, Lectures Vol. I, ch. 4 (the blocks-and-energy parable). Key reference results used in this article: $\mathbf{F}=-\nabla U\Leftrightarrow\nabla\times\mathbf{F}=0$ (via Stokes' theorem); loop-the-loop minimum height $h_{\min}=\tfrac{5}{2}R$ ; Earth escape speed $v_{\text{esc}}=\sqrt{2GM/R}\approx 11.2\,\text{km/s}$ ; harmonic-oscillator frequency $\omega=\sqrt{k/m}$ ; $g\approx 9.81\,\text{m/s}^2$ . Worked-example numbers ( $h_{\min} = 10.0$ m, $v_{\text{top}} = 6.3$ m/s, $v_{\text{bottom}} = 14.0$ m/s; friction-patch climb $2.0$ m and total patch distance $12$ m; bouncing-ball path $9.1$ m for $e=0.80$ , $h_0=2.00$ m) follow directly from the boxed formulas.

Builds on: Newton's Second Law · Velocity and Acceleration · Vectors in Kinematics and Coordinate Systems · Gravitational Force Near Earth's Surface. Leads to: Normal Force · Circular Motion and Centripetal Acceleration · Projectile Motion.

If energy is always conserved in the universe, why do we say friction "violates" conservation of mechanical energy?

Two different ledgers are in play. Total energy — mechanical plus thermal plus everything else — balances exactly, always; that is the universal law Helmholtz formulated. Mechanical energy, the sum $T+U$ this article tracks, is only the organized part, and friction is a one-way turnstile that converts it into disorganized thermal energy. So $T+U$ genuinely decreases (equation 8 quantifies the drain) while the grand total stays fixed. The practical upshot: when a problem contains friction, don't abandon energy methods — just add the friction work as an explicit debit line, $\Delta E = W_{\text{nc}} = -\mu_k N\,d$ for a patch of length $d$ .

Why can I put the zero of potential energy anywhere? Doesn't the energy have a real value?

Only differences of potential energy are ever measured. Equation (4) defines $U_B - U_A$ ; the value at a single point is a convention, like the zero of longitude. Shift every $U$ by a constant and $E$ shifts by the same constant, but every speed prediction from $v = \sqrt{2[E-U]/m}$ involves the difference $E-U$ , which is untouched. Choose the zero to kill as many terms as possible: floor level for a ramp problem, natural length for a spring, infinity for orbital mechanics (which is why bound orbits have negative $E$ — a perfectly healthy sign, not an error).

The normal force isn't conservative, and it isn't small. How exactly does it stay out of the energy equation?

Conservation needs the working forces to be conservative — forces that do no work at all never enter the books in the first place. A normal force is always perpendicular to the surface, and for a body sliding along that surface the velocity is parallel to it, so $\mathbf N\cdot\mathbf v = 0$ at every instant: zero power, zero work, zero ledger entry. The same holds for an ideal string's tension on a pendulum bob (always along the string, motion always perpendicular to it) and for the magnetic force on a charge. Loosely: forces that only steer are energy-invisible. This is why the loop-the-loop example never mentions the enormous normal force at the bottom of the loop.

Energy methods gave me the speed, but how do I find out how long the motion takes?

You generally can't — not from equation (5) alone, and that is by design. Energy conservation is one scalar relation between speed and position; it has already integrated out the time. If you need the schedule and not just the itinerary, take the one-dimensional relation $\dot x = \sqrt{2[E-U(x)]/m}$ , separate variables, and integrate $\mathrm{d}t = \mathrm{d}x/\dot x$ — doable in closed form for simple potentials, numerically otherwise. The interrupted-pendulum lesson holds here too: two tracks with the same endpoints give the same final speed but generally different travel times. (Finding the fastest track is the famous brachistochrone problem, and the answer is not the straight line.)

Where exactly does the kinetic energy go when a block skids to a stop?

Into microscopic mechanical energy of the two surfaces. At the sliding interface, asperities collide, weld, and tear billions of times per second; each event launches lattice vibrations that spread and randomize. Within microseconds the organized forward motion of $\sim 10^{24}$ atoms has become slightly-more-vigorous random jiggling of the same atoms — measurable as a temperature rise. Your hands after rubbing them, a drill bit after a hole, brake discs glowing after an Alpine descent: all are equation (8)'s debit column made visible. The reason you can't reclaim it is statistical: coaxing $10^{24}$ randomly-jiggling atoms to push the block forward in unison again is not forbidden by mechanics, just absurdly improbable — that asymmetry is the seed of the second law of thermodynamics.

A heat pump is advertised as 300 percent efficient. Doesn't that violate the ledger?

No law is broken — the number is a coefficient of performance, not an efficiency in the sense of equation (9). A heat pump does not convert its electrical input into heat; it uses the input work to move heat that already exists outdoors into your living room. Delivering 3 J of indoor heat per 1 J of electricity means 2 J were carried in from the cold outside air, and the full energy ledger (3 J delivered = 1 J work + 2 J transported) balances to the joule. The genuine bound is thermodynamic: the second law caps the COP at $T_h/(T_h - T_c)$ , which is why heat pumps look best in mild climates and lose their advantage as the outdoor temperature plummets.

Why does the minimum loop-the-loop speed come from gravity and not from energy conservation?

Because "makes it around the loop" is a force requirement, not an energy requirement. Energy conservation happily assigns a speed to any release height — even one for which the cart falls off the track. Staying on the track at the top demands a centripetal acceleration $v^2/R$ pointing downward, supplied by gravity plus the normal force; since the track can only push ( $N \ge 0$ ), the worst case is $N = 0$ , giving $v_{\text{top}}^2 \ge gR$ . Energy conservation then translates that speed requirement into the release height $h_{\min} = \tfrac52 R$ . The pairing is typical of good problems: dynamics sets the condition, energy converts it into geometry.

Is the vis viva argument just history, or does the factor of one-half in kinetic energy actually matter?

It matters, and the eighteenth century fought a forty-year war over it. Descartes' followers held that motion is measured by $mv$ ; Leibniz insisted on $mv^2$ . Both sides had valid experiments because they were tracking different conserved quantities — momentum and (twice the) kinetic energy. 's Gravesande's brass balls dropped into clay settled the energy side: doubling the impact speed quadrupled the dent depth, the $v^2$ signature, which Émilie du Châtelet drove home in her 1740 Institutions de physique. The modern factor $\tfrac12$ makes the bookkeeping exact: it is precisely what emerges when you integrate $\mathbf F = m\mathbf a$ along a path (Derivation A), and it is why $W = \Delta(\tfrac12 mv^2)$ with no stray factor of 2.