Conservation of Mechanical Energy & Efficiency

A roller coaster buys energy exactly once. The chain lift hauls the train to the first crest, banks a few megajoules of $mgh$ — and from that moment the ride is a closed transaction: every plunge cashes potential energy into speed, every climb deposits speed back into height, and the designer's entire art is spending one purchase over two minutes of track. No engine rides along. The train's speed at any point is set by its altitude, not by the twists of the route — which is why coaster engineers can promise the loop's entry speed years before the steel is bent. This module writes the rule that makes such promises possible: when only conservative forces do work, the total mechanical energy — kinetic plus potential — is constant. $E = KE + U$ becomes a conserved budget you can evaluate at any two moments and set equal, skipping every detail in between: no accelerations, no path geometry, no time. It is the single most-used equation in this unit's exams. And because the real world leaks, the module also teaches the two repair kits. When friction taxes the books, the conservation line gains a damage term: mechanical energy out equals mechanical energy in minus friction's toll. When machines move energy on purpose, we grade them with efficiency — the fraction of input that arrives as useful output, the number on every motor nameplate and every IGCSE Sankey diagram. One law, one loss term, one grade: the complete energy accounting toolkit.

Key takeaways

With only conservative forces doing work (gravity, springs), total mechanical energy is conserved: $KE_i + U_i = KE_f + U_f$ — evaluate at any two snapshots and equate.
The workhorse corollary: a body falling (or sliding frictionlessly) through height $h$ arrives with $v = \sqrt{2gh}$ , mass cancelled, path irrelevant.
Friction breaks the tie honestly: $KE_f + U_f = KE_i + U_i - f s$ , where $s$ is the actual path length slid — friction bills by odometer even inside energy equations.
Workless forces (normal on fixed tracks, tension of a swinging string) never disturb the budget — conservation survives their presence.
Machines are graded by efficiency: $\eta = \dfrac{\text{useful output}}{\text{total input}} \times 100\%$ — always below 100%; the missing fraction is heat, drawn to scale in a Sankey diagram's fat and thin arms.
Choosing the zero of $U$ at the lowest point of the problem makes the arithmetic cleanest — the freedom carries over from the potential-energy module.
Key idea: mechanical energy is a fixed budget whenever only conservative forces work — equate two snapshots of $\tfrac12 mv^2 + mgh + \tfrac12 kx^2$ and solve; if friction gatecrashes, subtract its path-length toll; if a machine feeds the system, grade it with efficiency.

The core insight

When gravity and springs are the only forces writing ledger entries, the sum $E = \tfrac12 mv^2 + U$ takes the same value at every instant of the motion — a conserved budget that connects any two snapshots directly; non-conservative forces amend the law additively ( $\Delta E = W_{\rm nc}$ ), draining the budget through friction or filling it through engines, whose quality is measured by efficiency.

Where this lives today

Grid-scale batteries are audited with exactly equation (4): pumped-storage hydro plants — still ~95% of world storage — bank $mgh$ uphill at night and refund it at peak, with a round-trip efficiency near 75–80%; the missing quarter is Joule's heat in pumps, turbines and tunnels. Electric vehicles run the same audit per journey: regenerative braking returns 60–70% of each stop's kinetic energy to the battery, and the car's energy display is a live Sankey diagram. Roller-coaster designers still trust equation (1) enough to certify loop entry speeds from track height alone — with a few percent margin for the rolling-resistance damage term. And national energy policy is argued over Sankey diagrams of entire countries: fat input arms of fuel, alarmingly fat "rejected energy" arms of heat — efficiency, drawn to scale, deciding billion-dollar questions.

The physical picture

Stand at the top of the coaster's first drop with an energy meter instead of a stopwatch. At the crest: the account is all height — $mgh$ banked, barely any speed. Halfway down: a running mixture, the potential column shrinking, the kinetic column growing, their sum pinned. At the valley floor: all cash — $\tfrac12 mv^2$ equal, to the joule, to the vanished $mgh$ . Then the next hill runs the film backward. The two columns slosh; the total never moves. That pinned total is what "conservation" means operationally: one number, computable at the easiest point of the ride, valid at every other.

Figure

Hero — the coaster's fixed budget: two columns slosh, the total is pinned

Why is the total pinned? Because the two columns are wired together by construction. The work–energy theorem says gravity's work fills the kinetic column; the potential-energy definition ( $\Delta U = -W_{\rm grav}$ ) says the same work empties the potential column by the same amount. One force, two accounts, opposite entries — the sum cannot move. The derivation below is just this sentence in symbols. The picture also flags the law's fine print. The wiring only involves forces that have accounts — conservative ones. Let friction aboard and it debits the kinetic column while crediting no column at all (its take leaves mechanical books entirely, as heat): the total sags. Let a motor aboard and the total climbs. Neither breaks physics — energy at large is always conserved — but mechanical energy, the exam's currency, is conserved only on the guest list: gravity, springs, and workless bystanders.

The math: the budget, the damage term, and the grade

The law first. Between any two snapshots $i$ and $f$ of a motion in which only conservative forces do work:

\tfrac12 mv_i^2 + U_i \;=\; \tfrac12 mv_f^2 + U_f \qquad (1)

where: $U$ collects every conservative account in play — $mgh$ , $\tfrac12 kx^2$ , or both. Read it as exam strategy, not just physics: pick snapshot $i$ where you know everything (the release point) and snapshot $f$ where the question lives (the bottom, the wall, the moment of maximum compression), equate, solve. Nothing between the snapshots is ever computed — that is the method's entire power. Its most-used special case, a body descending through net height $h$ from rest, frictionless:

v = \sqrt{2gh} \qquad (2)

where: the mass has cancelled (both columns scale with $m$ ) and the path has cancelled (gravity bills by height). A bead on a curly frictionless wire, a sled on ice, a freely falling stone: same $h$ , same arrival speed — about $10$ m/s for a 5 m drop with $g = 10$ m/s². Direction of arrival differs; the speed does not. When friction gatecrashes, amend — don't abandon:

\tfrac12 mv_f^2 + U_f \;=\; \tfrac12 mv_i^2 + U_i \;-\; f\,s \qquad (3)

where: $f$ — the friction force's magnitude, $s$ — the actual path length slid (the odometer reading, NOT the height). Equation (3) is just the work–energy theorem with the conservative works rebranded as $U$ -changes and the non-conservative work kept explicit. It handles nine-tenths of all "rough" energy problems; the remaining tenth adds an engine term $+W_{\rm engine}$ on the right. Machines get the grade:

\eta \;=\; \frac{\text{useful energy (or power) out}}{\text{total energy (or power) in}} \times 100\% \qquad (4)

where: "useful" is defined by the machine's job — the $mgh$ delivered by a pump, the $\tfrac12 mv^2$ of a launched projectile. Efficiency is the same number computed with energies or with powers (divide both by time). It is always below 100% for real machines; the shortfall is heat, and a Sankey diagram draws the split to scale — input arrow entering, useful arm flowing on, wasted arm drooping away, widths proportional to joules. IGCSE asks you to read and draw them; every board asks you to respect them.

Figure

The Sankey diagram — efficiency drawn to scale for the 60% pump of worked example 2

situation	equation to write
smooth track, gravity/springs only	conservation (1) — two snapshots
"from rest through height $h$ ", smooth	$v = \sqrt{2gh}$ directly
rough patch, length given	damaged budget (3), $s$ = path length
motor/pump doing a job	efficiency (4), then energy chain
asked for force or acceleration at a point	energy for the speed, then Newton at that point

Derivation AWhy the budget is pinned — conservation from the work–energy theorem

We derive equation (1) from the two laws already installed: the work–energy theorem and the definition of potential energy. The strategy is one substitution — split the net work into conservative and non-conservative parts, rebrand the conservative part as a potential-energy change, and read off the result. Watch how short it is: conservation is not new physics; it is the old bookkeeping with the accounts netted.

Step 1. The work–energy theorem, for any motion whatsoever:

W_{\rm net} = \Delta KE

Step 2. Split the net work by family: $W_{\rm net} = W_{\rm cons} + W_{\rm nc}$ (workless constraints contribute zero and drop out silently).

Step 3. The potential-energy definition rebrands the conservative slice: $W_{\rm cons} = -\Delta U$ . Substitute:

-\Delta U + W_{\rm nc} = \Delta KE \quad\Longrightarrow\quad \Delta(KE + U) = W_{\rm nc}

Step 4. Read the master result. If no non-conservative force does work ( $W_{\rm nc} = 0$ ): $\Delta(KE+U) = 0$ — equation (1), the pinned budget. If friction acts: $W_{\rm nc} = -fs$ — equation (3), the damaged budget. If an engine acts: $W_{\rm nc} > 0$ and the budget grows. Why this derivation is worth reciting in an exam: the sentence "since only conservative forces do work, mechanical energy is conserved" is a marks-scheme line on every board. The equation $\Delta E = W_{\rm nc}$ is the umbrella that holds every variant of this unit's problems. *Verified: 2 kg sled from rest down a smooth 5 m-high slope: budget at top $= mgh = 100$ J; at bottom $\tfrac12 mv^2 = 100$ J gives $v = 10$ m/s $= \sqrt{2 \times 10 \times 5}$ ✓; add a rough run-out ( $f = 8$ N) and the sled stops after $s = 100/8 = 12.5$ m, by (3) with $KE_f = U_f = 0$ .*

Go deeper · Derivation B · 1 minThe pendulum's speed map — conservation doing what Newton finds hard

Derivation BThe pendulum's speed map — conservation doing what Newton finds hard

We derive the speed of a pendulum bob at any angle, a result that pure force methods cannot reach at this level (the along-arc acceleration changes continuously). The strategy showcases the method: two snapshots, heights read off geometry, tension ignored — legally. The watch-out is the geometry step: the height drop involves $\cos\theta$ , and writing it wrongly is the single commonest error in energy-method pendula.

Step 1. A bob of mass $m$ on a string of length $L$ is released from rest at angle $\theta_0$ from the vertical. Choose the zero of $U$ at the lowest point of the swing.

Step 2. Geometry (the exam's real test): at angle $\theta$ , the bob hangs $L\cos\theta$ below the pivot, so its height above the lowest point is

h(\theta) = L - L\cos\theta = L(1 - \cos\theta)

Step 3. The guest list: gravity (conservative, in $U$ ) and tension. Tension is perpendicular to the arc at every instant — workless. Conservation applies in its pure form.

Step 4. Budget at release = budget at angle $\theta$ :

mgL(1-\cos\theta_0) = \tfrac12 mv^2 + mgL(1-\cos\theta)

\Longrightarrow\; v(\theta) = \sqrt{2gL\,(\cos\theta - \cos\theta_0)}

At the bottom ( $\theta = 0$ ): $v_{\rm max} = \sqrt{2gL(1-\cos\theta_0)}$ — the vertical-circle module's energy bridge, met from the other side. Why tension could be ignored without an account: it does no work, so it neither needs a potential energy nor damages the budget — the workless family's free pass, stated explicitly for the marker. *Verified: $L = 2.5$ m, $\theta_0 = 60°$ : $h = 2.5(1-0.5) = 1.25$ m, $v_{\rm max} = \sqrt{2\times10\times1.25} = 5$ m/s; cross-check via (2) with $h = 1.25$ m ✓.*

Limiting cases

case	result	meaning
smooth, gravity only, drop $h$	$v = \sqrt{2gh}$	mass and path both cancel — the law's calling card
$f \to 0$ in the damaged budget (3)	recovers (1)	conservation is the frictionless limit of honest accounting
body barely reaches height $h$ ( $v_f = 0$ )	$h_{\max} = v_i^2/2g$	the budget spent entirely on altitude
friction so large the body stops on the slope	(3) with $v_f = 0$ mid-path	solve for $s$ : the budget ran out early
$\eta \to 100\%$	ideal machine	never achieved; options offering $\eta > 100\%$ are wrong by law
pendulum $\theta_0 \to 90°$	$v_{\rm max} = \sqrt{2gL}$	full radius converted; matches the vertical-circle threshold lore

Common trapCheck the guest list before you conserve

The most expensive reflex in this unit: writing $mgh = \tfrac12 mv^2$ the moment heights appear. Conservation (1) is a conditional law — it holds only if every working force is conservative. One rough patch, one "the block slides 3 m across the floor", one air-resistance clause, and the pure form silently lies; the honest form (3) must carry the $-fs$ term with $s$ the actual distance slid, not the height dropped. Run the audit in the margin before writing anything: list forces → tag conservative / workless / non-conservative → choose (1) or (3). Ten seconds, and it also catches the reverse error: students who deduct friction on a smooth track "to be safe" lose the answer too. The guest list decides the equation; nothing else does.

Worked examples

1 · Board level — the slide, the swing, and the stone (CBSE/NEET pattern, 3 marks)

Three objects descend the same 5.0 m height from rest: a child on a smooth slide, a bob swinging down on a string, and a dropped stone. Find each arrival speed. ( $g = 10$ m/s².)

Step 1. Guest lists: slide — gravity + normal (workless); swing — gravity + tension (workless); stone — gravity alone. All three qualify for pure conservation.

Step 2. One equation serves all three: $mgh = \tfrac12 mv^2 \Rightarrow v = \sqrt{2gh} = \sqrt{2 \times 10 \times 5.0} = \boxed{10\ \mathrm{m/s}}$ — each.

Step 3. State the two cancellations for the marks: mass cancels (both terms carry $m$ ), and path shape cancels (gravity bills by height; the constraint forces are workless). Different arrival directions, identical arrival speeds.

2 · Board level — grading the pump (IGCSE/ICSE pattern)

An electric pump consumes $2000$ J of electrical energy to raise $15$ kg of water through $8.0$ m. ( $g = 10$ m/s².) Find (a) the useful work done, (b) the efficiency, (c) the wasted energy, and (d) sketch-describe the Sankey diagram.

Step 1. (a) Useful output = the job = $mgh = 15 \times 10 \times 8.0 = \boxed{1200\ \mathrm{J}}$ .

Step 2. (b) $\eta = \dfrac{1200}{2000} \times 100\% = \boxed{60\%}$ .

Step 3. (c) Wasted: $2000 - 1200 = \boxed{800\ \mathrm{J}}$ — heat in the motor windings and bearings, sound, swirl.

Step 4. (d) Sankey: input arm of width "2000 J" entering; useful arm of width 1200 J flowing straight on; wasted arm of width 800 J bending away (usually downward). Widths to scale — 3 : 2 here between useful and wasted — and the two output widths must add up to the input width exactly.

3 · Competitive level — the slide with a rough run-out (JEE-Main pattern, ~3 minutes)

A $2.0$ kg block is released from rest at the top of a smooth ramp of height $3.2$ m. At the bottom it crosses a horizontal rough patch ( $\mu_k = 0.5$ ). ( $g = 10$ m/s².) Find (a) its speed at the bottom of the ramp, (b) how far into the patch it slides before stopping, (c) the fraction of the initial mechanical energy lost as heat by the halfway point of that slide.

Step 1. (a) Smooth ramp — pure conservation: $v = \sqrt{2gh} = \sqrt{64} = \boxed{8.0\ \mathrm{m/s}}$ ; budget $= mgh = 64$ J.

Step 2. (b) On the patch, the damaged budget (3) with $U$ constant and $f = \mu_k mg = 10$ N: the 64 J drains at 10 J per metre, so $s = \dfrac{64}{10} = \boxed{6.4\ \mathrm{m}}$ .

Step 3. (c) At $s = 3.2$ m, heat generated $= 10 \times 3.2 = 32$ J: fraction $= 32/64 = \boxed{50\%}$ — and the block there moves at $\sqrt{2 \times 32 / 2} = 5.7$ m/s, not $4$ m/s: energy halves, speed only drops by $\sqrt{2}$ . Timing note: ~3 min. The (c) distractor "half the distance → half the speed" is the $v^2$ law wearing a disguise; energy is linear in distance-drained, speed is its square root. Also expected: the ramp needed no friction term (smooth) — deducting one there is the pitfall box's reverse error.

4 · Competitive level — the bouncing ball's energy census (NEET/JEE-Main pattern, ~2.5 minutes)

A ball dropped from $2.5$ m rebounds to $1.6$ m. ( $g = 10$ m/s².) Find (a) the fraction of mechanical energy lost in the bounce, (b) the ratio of rebound speed to impact speed, (c) the height after the second identical bounce.

Step 1. (a) Energies are proportional to heights ( $mgh$ with the same $m$ ): fraction lost $= \dfrac{2.5 - 1.6}{2.5} = \boxed{36\%}$ per bounce; 64% survives.

Step 2. (b) Speeds go as $\sqrt{h}$ (equation 2 both ways): $\dfrac{v_{\rm rebound}}{v_{\rm impact}} = \sqrt{\dfrac{1.6}{2.5}} = \boxed{0.80}$ .

Step 3. (c) Identical bounce = same fraction retained: $h_2 = 0.64 \times 1.6 = \boxed{1.02\ \mathrm{m}}$ — a geometric decay, $h_n = 2.5 \times (0.64)^n$ .

Step 4. Where did the 36% go? Sound of the bounce, heat in the deformed ball and floor — non-conservative internal forces during contact. Between bounces, flight is pure conservation; the books are damaged only at the floor. Timing note: ~2.5 min. Keep the two ladders straight: heights and energies share one ratio (0.64); speeds take its square root (0.80). Options love swapping them. (The restitution-coefficient formalism for this ratio arrives in the collisions chapter — here, energy language is all you need.)

Exam traps

Conserving through friction. The cardinal sin — see the pitfall box. Rough patch ⇒ equation (3), with $s$ = distance actually slid. And its mirror: subtracting imaginary friction on smooth tracks.

Height for path length in the loss term. Friction's toll is $f \times s_{\rm path}$ , the odometer reading. On an incline, that is the slope distance, not the height; on the flat run-out, the slide length. Mixing $h$ into the friction term is a double sin — it borrows gravity's billing rule for the one force that doesn't use it.

Forgetting a second account. With a spring at the bottom of a slope, the budget carries three columns: $\tfrac12 mv^2 + mgh + \tfrac12 kx^2$ . Dropping the spring term at maximum compression (where it holds all the energy) inverts the answer.

"At the bottom, all energy is kinetic" recited blindly. True only if the zero of $U$ sits at the bottom and nothing else stores. Declare the zero in line one; the freedom is yours, the consistency is not optional.

Efficiency arithmetic upside-down. $\eta$ = useful out over total in — never the reverse, never out over wasted. An answer above 100% is self-refuting; quote it and the marker knows the fraction flipped.

Efficiency of what, exactly. "Useful" follows the job description: for a pump it is $mgh$ ; for a motor-driven flywheel, $\tfrac12 mv^2$ … wait for the question to name the job before assigning the numerator.

Rounding the budget too early. Carry energies exactly (they are the currency); take the square root for speeds at the last line. $\sqrt{2gh}$ with $h$ prematurely rounded compounds badly.

Which $g$ . CBSE/ISC: $9.8$ ; JEE/NEET: almost always $10$ ; A-Level/IB: $9.81$ . State it, use it consistently, and quote speeds to sensible figures.

Origins · 1 minJames Prescott Joule and how this was discovered

James Prescott Joule

1818–1889 · Manchester

A Manchester brewer who measured like a jeweller, Joule spent the 1840s proving that mechanical work converts to heat at a fixed exchange rate — most famously with a paddle wheel churning water in an insulated barrel, driven by falling weights whose $mgh$ he could meter exactly. The water warmed by the predicted whisker every time: about 4.2 joules (as we now say) per calorie, path and mechanism irrelevant. It was the missing receipt for this module's damage term — friction's stolen mechanical energy shows up, joule for joule, as heat — and it forced physics to widen "conservation" from mechanics into a universal law. The SI unit of every quantity in this article carries his name in tribute.

History

1673

— Huygens: the pendulum returns to its release height — conservation observed before it is named

1798

— Rumford, boring cannon in Munich, shows friction generates heat without limit — the leak is real energy

1842

— Julius Robert Mayer states the equivalence of work and heat from physiological arguments

1843

— Joule's paddle-wheel experiments fix the mechanical equivalent of heat by direct measurement

1847

— Helmholtz unifies: energy — kinetic, potential, thermal — is conserved across all of physics

Going further

Texts. NCERT Physics Class 11, Work, Energy and Power chapter (conservation of mechanical energy) · HC Verma, Concepts of Physics Vol. 1, Ch. 8 · Halliday–Resnick–Walker Ch. 8 · CIE IGCSE 0625 §1.7 (efficiency and Sankey diagrams, this module's IGCSE slice).

Provenance. All results are standard. A verifier should recompute: the sled check (100 J, 10 m/s, 12.5 m run-out); the pendulum map ( $L = 2.5$ m, $60°$ → 5 m/s); the three descents (10 m/s each); the pump grade (1200 J, 60%, 800 J); the rough run-out (8 m/s, 6.4 m, 50% at halfway with 5.7 m/s); the bounce census (36%, 0.80, 1.02 m).

Prerequisites. Kinetic Energy & the Work–Energy Theorem · Potential Energy: Gravitational & Elastic · Conservative & Non-Conservative Forces. Leads to. ARCHETYPE: Block Meets Spring · ARCHETYPE: Loop-the-Loop & Curved Tracks · ARCHETYPE: Energy Accounting on Rough Inclines · Conservation of Mechanical Energy (UG deep dive).

Is mechanical energy REALLY conserved, or is it just approximately true?

Total energy is exactly conserved — as far as every experiment in history can tell. Mechanical energy (KE + U) is conserved exactly only in the idealized, frictionless, drag-free problems of this module; in reality some always leaks to heat and sound. The idealization is still enormously useful: for a dense object over a short fall, losses are a fraction of a percent, and for exam setups the words "smooth", "light string", "ignore air resistance" are formal declarations that the pure law applies. When they are absent, look for the loss term.

Where exactly does friction's stolen energy go — can I get it back?

Into disordered molecular jiggling — heat — in both rubbing surfaces, plus a little sound and surface damage. It has not vanished (Joule's experiments confirmed every joule arrives as heat), but it is unrecoverable by mechanical means: you cannot cool the ramp and watch the block speed up. That one-way character is why we keep separate books for mechanical energy and treat friction as damage rather than as an account. The deep reason — entropy, the second law of thermodynamics — is a chapter of physics still ahead of you.

Why can I ignore the normal force and tension in energy problems when I could never ignore them in force problems?

Because the energy method only ever consults the ledger, and workless forces never write in it. The normal force on a track and the tension of a swinging string act strictly perpendicular to each step of the motion — zero work, every time — so they steer the object without touching its budget. In Newtonian problems they carry the geometry (and you must solve for them); in energy problems the geometry has already been paid for. This division of labour is exactly why "find the speed" goes to energy and "find the tension" goes back to Newton — often both, in one exam question.

Does conservation of mechanical energy work in a moving frame of reference?

Yes — in any inertial frame — but the numbers differ between frames (kinetic energies and heights both transform), so never mix two frames in one budget. Board and JEE problems fix the ground frame and you should too. The subtle cases (energy bookkeeping on accelerating wedges, in lifts) belong to the pseudo-force toolkit at JEE-Advanced level; the safe universal rule at this stage: choose the ground frame, declare the zero of $U$ , stay put.

With both a spring and gravity acting, how do I keep the budget straight?

Give each conservative force its own column and let the snapshots do the work: $E = \tfrac12 mv^2 + mgh + \tfrac12 kx^2$ , evaluated at moments you choose. The craft is picking snapshots where columns vanish: at maximum spring compression the velocity is zero (kinetic column empty); at natural length the spring column is empty. The block-meets-spring archetype two modules ahead is built entirely from this three-column budget — this FAQ is its syllabus.

Can efficiency ever exceed 100%?

No — that would create energy from nothing. Marketing sometimes muddies this: heat pumps are advertised with "300% efficiency" because they move 3 J of heat indoors per 1 J of electricity consumed; the mover's ratio (properly called coefficient of performance) can exceed 1, but no energy is created — the 3 J existed outside. In mechanics problems the rule is absolute: useful out ≤ total in, always, and an exam answer above 100% signals an inverted fraction with certainty.

Why does the bouncing ball lose a FIXED FRACTION each bounce rather than a fixed amount?

Because the loss mechanism scales with how hard the bounce is. The deformation of ball and floor — where the heat is generated — grows with impact energy, so a faster impact loses proportionally, not additively. Fixed fraction retained ⇒ heights form a geometric sequence ( $h_{n} = r^n h_0$ ), which is why the bounces crowd together audibly at the end. The idealization is good for rubber balls over modest heights; exams adopt it whenever they say "loses 20% per bounce" or "rebounds to 0.8 of its height".