TECHNIQUE: Force Method or Energy Method?

A bead slides down a frictionless wire bent into some extravagant curve, released from height $h$ . Find its speed at the bottom. The force-method student draws a free-body diagram, discovers the normal force changes direction at every point of an unspecified curve, and dies in the algebra. The energy-method student writes $\tfrac12 mv^2 = mgh$ and is done in eight seconds. Now change one word — find the time to reach the bottom — and the tables turn completely: energy is silent about clocks, and no amount of $\tfrac12 mv^2$ will produce a second. Every mechanics problem you will meet in JEE, NEET or AP sits somewhere on this fork, and the single most valuable exam skill in the unit is choosing the right branch in the first ten seconds. You now own both toolkits: Newton's laws with free-body diagrams from the Laws of Motion chapter, and the energy ledger — work–energy theorem, conservation, the friction toll — from this one. What you may not yet own is the dispatcher: the quick triage that reads what a question asks and what the setup hides, and routes it to the cheap method instead of the expensive one. This module builds that dispatcher. The decision table, the tell-tale phrases ("find the speed at…" versus "find the time…" versus "find the tension at…"), the hybrid two-step that cracks half of JEE's circular-motion problems, the failure modes of each method — and drills where the same setup is solved both ways so you can feel the price difference directly.

Key takeaways

Energy methods answer "how fast / how far": speeds at positions, heights reached, distances to stop. They see only endpoints — path details, changing force directions, and the clock are invisible to them.
Force methods answer "how long / how hard": times, accelerations, tensions, normal forces. They see every instant — at the price of needing the geometry at every instant.
The tell in the question: time or any force value → Newton. Speed at a position with curved paths, springs, or variable forces → energy.
The hybrid two-step is JEE's favourite: energy for the speed at a point, then Newton at that single point for the tension or normal force (vertical circle, loop-the-loop, pendulum).
Friction is energy's toll booth: subtract $f \times (\text{path length})$ — the actual distance slid, never the displacement.
Energy is one scalar equation: it can fix ONE unknown. Two unknowns (e.g. two final speeds) need a second law — usually momentum. That triage is the next technique module.
Key idea: read what the question asks before touching the physics — clocks and forces summon Newton, speeds-at-positions summon energy, and "force at a point on a curved ride" summons energy first, then Newton at that point.

The core insight

Force methods track every instant (so they can answer time, acceleration, tension and normal-force questions but need the full geometry), while energy methods compare only endpoints (so they crush speed-at-position questions on any path but are blind to time and to individual forces) — and the highest-yield exam questions are hybrids: energy conservation to the point of interest, then Newton's second law at that single point.

Where this lives today

Professional mechanics is dispatch all the way up. Crash-safety engineers run the energy books first — a car's kinetic energy must be absorbed across the crumple distance, which sets the average force — and only then simulate Newton instant-by-instant to find the peak forces that injure. Roller-coaster designers size hills and loops entirely by energy (speed at every height, one line each) and then audit the g-forces pointwise with Newton, exactly this module's two-step at industrial scale. In advanced physics the fork deepens into a formalism: Lagrangian mechanics rebuilds all of dynamics from energies precisely because constraint forces (this chapter's workless $N$ and $T$ ) drop out of the books automatically — the energy method's superpower, promoted to a first principle. And every competitive-exam hall on Earth remains the technique's natural habitat: the candidates who finish JEE mechanics are rarely faster at algebra — they are faster at choosing, spending ten seconds on dispatch to save ten minutes on regret.

The physical picture

Think of the two methods as two kinds of surveillance on the same journey. The force method is a dashcam: it records everything, instant by instant — where the object is, which way each force points, how fast the velocity is turning. From the full recording you can extract anything: the time stamp of any event, the force at any moment. But you must watch the whole tape, and if the road curves in complicated ways, the tape is long and the geometry brutal. The energy method is a pair of border checkpoints. It photographs the object twice — once at the start, once at the end — and compares the two photographs using an accounting identity: kinetic energy plus banked potential energy, minus the friction toll paid en route, must balance. It never watches the journey. That blindness is its superpower — the wildest curve between checkpoints costs nothing, because only the photographs matter — and its fatal limitation: no clock appears in either photograph, and no individual force does either.

Figure

Hero — the dispatcher: what does the question ask?

Why does this triage matter so much under time pressure? Because the same physical setup can be a thirty-second problem or a ten-minute problem depending on the route, and papers are engineered around that gap. The setter who asks for the speed at the bottom of a curved slide is testing whether you know Newton is the wrong tool; the setter who asks for the time down a straight incline is testing the reverse. And the setter who asks for the normal force at the top of a loop is testing whether you can chain the methods — most of the marks are lost by students who stop after the energy step, presenting a speed to a question that asked for a force.

The toolkit: the dispatch table

the question asks for	the setup contains	method	why
speed at a position, height reached	curved track, swinging string, any shape	energy	endpoints only; geometry irrelevant
speed after compression/stretch	a spring	energy	variable force = area/ledger territory
distance to stop	friction	energy	one unknown $d$ in the toll $f\,d$
time, or anything with "when"	constant forces, straight paths	force (F = ma + suvat)	only Newton owns the clock
acceleration at an instant	anything	force	$a$ is Newton's native output
tension / normal force at a point	circular or curved motion	hybrid	energy → $v^2$ there, Newton → the force
"minimum speed to complete the loop"	vertical circle	hybrid	Newton at top ( $N$ or $T$ = 0) sets $v_{\rm top}$ , energy carries it elsewhere
two unknown final speeds	collision, two bodies	energy alone FAILS — add momentum	one scalar equation fixes one unknown

The ten-second triage. Move 1: underline what is asked (a speed? a time? a force?). Move 2: scan the setup for energy's friends (springs, curves, "smooth") and enemies (a demanded clock, a demanded force). Move 3: if a force is asked on a curved ride, plan the two-step: energy to the point, Newton at the point. Move 4: count unknowns — energy is one equation; if two survive, recruit momentum or Newton. Move 5: only then start writing.

Derivation ASame hill, both routes — feeling the price difference

We solve one problem twice — a block sliding from rest down a smooth incline of height $h = 5$ m and angle $\theta = 30°$ , asking for the speed at the bottom — first by force method, then by energy, to expose exactly where the routes converge and what each one had to pay. The strategy: run each method honestly and completely, then note which pieces of work the energy route never performed. Watch the angle $\theta$ do something remarkable in the second route: vanish.

Step 1. Force route. FBD: weight $mg$ down, normal $N$ perpendicular to the slope. Resolve along the slope: $ma = mg\sin\theta$ , so $a = g\sin\theta = 10 \times 0.5 = 5\ \mathrm{m/s^2}$ (taking $g = 10$ ).

Step 2. Kinematics along the slope. The slope length is $s = h/\sin\theta = 5/0.5 = 10$ m. Then $v^2 = 2as = 2(5)(10) = 100$ , so $v = 10$ m/s.

Step 3. Energy route. Two snapshots: top (at rest, height $h$ ), bottom (speed $v$ , height 0). Smooth slope, normal force perpendicular to motion — does no work. Ledger:

\tfrac12 mv^2 = mgh \;\;\Rightarrow\;\; v = \sqrt{2gh} = \sqrt{100} = 10\ \mathrm{m/s}

Why the angle vanished: the energy ledger only reads the photographs — height lost, speed gained. The slope's steepness affects HOW LONG the trip takes and WHAT the acceleration was, but not the arrival speed. The force route had to compute both of those intermediate quantities and then watch them cancel: $v^2 = 2(g\sin\theta)(h/\sin\theta) = 2gh$ .

Step 4. Now perturb the problem and watch the routes separate. Make the slope a curved slide of the same height: the force route collapses ( $\theta$ changes at every point; suvat is illegal under changing acceleration) while the energy route is untouched — still $v = \sqrt{2gh} = 10$ m/s. Instead ask for the time to reach the bottom of the original straight incline: the energy route collapses (no clock in the ledger) while the force route finishes cheerfully: $t = v/a = 10/5 = 2$ s. Verified: both routes give $v = 10$ m/s on the straight incline; the cancellation $2(g\sin\theta)(h/\sin\theta) = 2gh$ is exact for every angle, which is precisely the path-independence the energy method banks on.

Go deeper · Derivation B · 1 minThe hybrid two-step — tension at the bottom of a swing

Derivation BThe hybrid two-step — tension at the bottom of a swing

We derive the tension in a pendulum string at the lowest point of its swing — released from rest with the string horizontal, length $L$ — the canonical problem that neither method can finish alone. The strategy: energy owns the journey (a quarter-circle of continuously turning force directions — Newton would suffer), Newton owns the instant (tension is a force — energy is blind to it). The handoff variable is $v^2$ at the bottom: energy produces it, Newton consumes it. Watch the final answer exceed $mg$ by a term that exists only because the bob is turning.

Step 1. Energy leg, from release (rest, height $L$ above the bottom) to the lowest point (speed $v$ ): the tension is perpendicular to the motion at every instant, so it does no work — the ledger sees only gravity:

\tfrac12 mv^2 = mgL \;\;\Rightarrow\;\; v^2 = 2gL

Step 2. Newton leg, at the single instant of interest. At the lowest point the bob moves in a circle of radius $L$ , so its acceleration points up (centripetal), magnitude $v^2/L$ . FBD there: tension $T$ up, weight $mg$ down:

T - mg = \frac{mv^2}{L}

Why Newton is legal here even though the motion is complicated: the second law is instantaneous — it needs no history, only the forces and the acceleration NOW. We are using each method exactly on its home turf.

Step 3. The handoff. Substitute $v^2 = 2gL$ from the energy leg:

T = mg + \frac{m(2gL)}{L} = 3mg

Why more than $mg$ : the string must both support the weight AND bend the velocity into a circle. Students who answer $mg$ have forgotten the bob is turning; students who answer $2mg$ dropped one of the two jobs. *Verified: the celebrated result — a pendulum released from the horizontal pulls with exactly three times the bob's weight at the bottom, independent of $L$ and $m$ 's value. Limits: released from rest AT the bottom ( $h = 0$ ): $v^2 = 0$ , $T = mg$ ✓ (just hanging); the vertical-circle module's bottom-of-loop $T = 6mg$ for $h = 2L$ … wait, for full-loop entry $v^2 = 5gL$ gives $T = mg + 5mg = 6mg$ ✓ — the same two-step, harder entry speed.*

Figure

Same hill, two surveillance styles — the dashcam pays for every instant, the checkpoints pay for two

When it backfires

Energy cannot see the clock — ever. No rearrangement of $\tfrac12mv^2 = mgh$ produces a time. If "how long" appears, you need Newton (constant acceleration) or, later, calculus/impulse. The most common misuse of energy in scripts is a correct speed presented as the answer to a time question.

Energy cannot name individual forces. The ledger deals in work totals, and constraint forces (normal, tension) that act perpendicular to motion never even enter it. Any asked force routes through Newton at a point — usually after energy has delivered $v^2$ there.

Friction charges by the path, not the displacement. The toll is $f \times$ (distance actually slid). On a there-and-back trip of displacement zero, the toll is $f \times$ (total path) — the ledger must charge both legs. Forgetting this converts every oscillating-with-friction problem into nonsense.

One equation, one unknown. Energy conservation is a single scalar statement. A collision with two unknown outgoing speeds, a two-block-and-spring system with both blocks moving — energy alone leaves you one equation short. The reinforcement arrives from momentum conservation; dispatching between those two laws is the next technique module (Which Conservation Law?).

"Smooth" is load-bearing. The clean ledger $\tfrac12mv^2 = mgh$ assumes no friction toll. On rough tracks you must know the path length — which, on a curved slide, the problem may deliberately withhold. If friction acts on an unknown curve, neither elementary method finishes: that is the setter telling you the question wants something else (often "heat generated = energy gap", computed from the two photographs).

Force methods die on curves and variable forces. suvat requires constant acceleration. A spring's force changes with $x$ ; a curve's normal force changes direction continuously. Using $v^2 = u^2 + 2as$ with a spring's "average-ish" force is the classic manufactured wrong answer.

Common trapEnergy is the advanced method, so it should be tried first everywhere

Half true, and the false half is expensive. Energy feels superior because it demolishes problems Newton finds impossible (curved tracks, springs) — but it answers only two kinds of question: how fast at a place, and how far until the ledger runs dry. Exam setters weaponize the overconfidence: they dress a time question in energy costume (a curved slide, a released pendulum… "how long until it reaches the bottom?") and harvest the students who autopilot into $\tfrac12mv^2 = mgh$ , stare at it, and burn four minutes discovering it cannot produce seconds. The counterpart trap catches force-method loyalists: asked for a speed after a spring launch, they attempt $F = ma$ with the spring's maximum force as if it were constant, earning a wrong answer with confident working. Neither method is "first"; the question is first. Read what is asked, then dispatch — the ten seconds of triage are the highest-paid seconds on the paper.

Worked examples — the drill ladder

Drill 1 · Board level — the curved slide (dispatch: energy)

A child of mass $30\ \mathrm{kg}$ starts from rest at the top of a frictionless playground slide of height $3.2\ \mathrm{m}$ whose shape is an irregular curve. Find her speed at the bottom. ( $g = 10\ \mathrm{m/s^2}$ .)

Step 1. Triage: speed at a position + curved smooth track → energy. (Force method is not merely slow here — without the curve's equation it is impossible.)

Step 2. Two photographs: top (rest, $h = 3.2$ m), bottom (speed $v$ , $h = 0$ ). Normal force ⊥ motion throughout — workless.

Step 3. Ledger: $\tfrac12 mv^2 = mgh \Rightarrow v = \sqrt{2 \times 10 \times 3.2} = \boxed{8.0\ \mathrm{m/s}}$ . The mass and the curve's shape never entered — say so; it earns the reasoning mark.

Drill 2 · Board level — one word changes the method

A block slides from rest down a smooth straight incline, height $3.2\ \mathrm{m}$ , angle $30°$ . Find the TIME to reach the bottom. ( $g = 10\ \mathrm{m/s^2}$ .)

Step 1. Triage: time asked → force method, and the straight incline makes it legal (constant $a$ ).

Step 2. Along-slope: $a = g\sin\theta = 5\ \mathrm{m/s^2}$ ; slope length $s = h/\sin\theta = 6.4$ m.

Step 3. suvat from rest: $s = \tfrac12 at^2 \Rightarrow t = \sqrt{2s/a} = \sqrt{2.56} = \boxed{1.6\ \mathrm{s}}$ .

Step 4. Note the dispatcher's fine print: had the slide been curved, this question would be unanswerable at board level — which is exactly why curved-track problems never ask for time.

Drill 3 · Board level — the spring launcher (dispatch: energy; force method illegal)

A spring of constant $k = 500\ \mathrm{N/m}$ , compressed $0.20\ \mathrm{m}$ , launches a $0.50\ \mathrm{kg}$ trolley on a smooth horizontal track. Find the launch speed — and state why $v^2 = u^2 + 2as$ must not be used.

Step 1. Triage: speed after a spring → energy. The spring force varies from $kx = 100$ N to zero across the launch: no constant $a$ , so suvat is illegal.

Step 2. Ledger: $\tfrac12 kx^2 = \tfrac12 mv^2 \Rightarrow v = x\sqrt{k/m} = 0.20\sqrt{1000} = \boxed{6.3\ \mathrm{m/s}}$ .

Step 3. The illegal route, exposed: using the maximum force in suvat gives $v = \sqrt{2(200)(0.2)} = 8.9$ m/s — a 40% overestimate, and a distractor you will meet as an MCQ option. (It double-counts: the force is $100$ N only at the first instant.)

Drill 4 · Competitive level — the loop's normal force, two-step (JEE-Main pattern, ~3 minutes)

A block is released from rest at height $h = 3R$ on a smooth track that ends in a vertical circular loop of radius $R$ . Find the normal force from the track at the TOP of the loop, in units of $mg$ .

Step 1. Triage: a force at a point on a curved ride → hybrid. Energy to the top of the loop, Newton at the top.

Step 2. Energy leg (smooth): from $h = 3R$ to the loop top at height $2R$ : $\tfrac12 mv_{\rm top}^2 = mg(3R - 2R) \Rightarrow v_{\rm top}^2 = 2gR$ .

Step 3. Newton leg at the top: both $N$ and $mg$ point toward the centre (downward), and together they supply the centripetal demand: $N + mg = \dfrac{mv_{\rm top}^2}{R} = 2mg$ .

Step 4. $N = \boxed{mg}$ . Audits: $v_{\rm top}^2 = 2gR > gR$ ✓ (above the minimum — contact maintained, the Newton step was legal); release from exactly $h = 5R/2$ would give $N = 0$ ✓ (the critical case). Timing note: ~3 min, of which the physics is 90 seconds — the rest is the legality audit in Step 4, which JEE multi-options are built to test. Never run the Newton step at the top without first checking $v_{\rm top}^2 \ge gR$ : below that, the block has already left the track and the FBD you drew doesn't exist.

Drill 5 · Competitive level — friction makes it a quadratic; the dispatcher chains three tools (~4 minutes)

A $2.0\ \mathrm{kg}$ block moving at $6.0\ \mathrm{m/s}$ on a rough floor ( $\mu = 0.25$ ) hits a spring of constant $k = 100\ \mathrm{N/m}$ . Find (a) the maximum compression, (b) the time the block takes to first come to rest — or explain why (b) cannot be answered with this chapter's tools. ( $g = 10\ \mathrm{m/s^2}$ .)

Step 1. Triage (a): distance-flavoured unknown with a spring and friction → energy with the toll. Friction force $f = \mu mg = 5.0$ N.

Step 2. Ledger to maximum squash: $\tfrac12 mv^2 = \tfrac12 kx^2 + f x$ , i.e. $36 = 50x^2 + 5x$ , so $50x^2 + 5x - 36 = 0$ .

Step 3. Positive root: $x = \dfrac{-5 + \sqrt{25 + 7200}}{100} = \dfrac{-5 + 85}{100} = \boxed{0.80\ \mathrm{m}}$ . (Engineered discriminant: $\sqrt{7225} = 85$ .)

Step 4. Triage (b): a time, under a force that varies with position (spring + friction) — not constant acceleration, so suvat is illegal, and energy is clock-blind. State it: "the time cannot be found by suvat or energy methods; it requires the oscillation toolkit" (it is a quarter-period-like quantity of damped SHM — the Oscillations chapter's business). Recognizing a question your current tools cannot answer is the tested skill in multi-part JEE items; guessing $t = x/v_{\rm avg}$ earns the designed zero. Timing note: ~4 min. Part (a) is the rough-spring quadratic from Block Meets Spring — 2.5 minutes with the discriminant engineered clean. Part (b) is a 30-second sentence IF the dispatcher is awake; it is a ten-minute derailment if not. Learn the smell of "time + variable force": it is either an SHM question in disguise or a trap.

Exam traps

Answering a speed to a time question. The flagship dispatch failure — energy runs happily, produces $v$ , and the mark scheme wanted $t$ . Underline the asked quantity before writing anything.

suvat with a spring. $v^2 = u^2 + 2as$ demands constant acceleration; springs never provide it. The "maximum force" and "average force" substitutions both manufacture wrong numbers (the max-force version overshoots by $\sqrt2$ on a clean launch).

Stopping after the energy step in hybrid problems. The loop/pendulum/vertical-circle family asks for $N$ or $T$ ; energy only delivers $v^2$ . Budget the second step — it is where the marks are.

Newton at the top of a loop without the contact audit. If $v_{\rm top}^2 < gR$ the block left the track earlier and your FBD describes a ghost. Check legality before computing.

Friction charged on displacement instead of path. Round trips and oscillations: the toll is $f \times$ total distance slid. An exam's "block slides up, comes back down" charges friction twice.

Charging the normal force work on a curve. On tracks and strings, $N$ and $T$ are perpendicular to motion: zero work, and that is why the clean ledger works at all. (Exception territory — moving pulleys, accelerating wedges — is beyond this chapter and flagged as such in JEE-Advanced.)

Using energy with two unknowns. One scalar equation fixes one unknown. If two final speeds survive your snapshot, you are one law short — recruit momentum (next module) instead of inventing a second energy equation.

Method mismatch under time pressure. If your chosen route generates a page of trigonometry for a 2-mark question, stop — the setter priced it for the other method. Marks per minute is the objective function.

Origins · 1 minÉmilie du Châtelet and how this was discovered

ÉD

Émilie du Châtelet

1706–1749 · Cirey & Paris

The force-versus-energy fork is old enough to have had a war named after it: the vis viva controversy. Descartes and Newton's followers measured motion by $mv$ ; Leibniz insisted the true measure was $mv^2$ , his "living force". Émilie du Châtelet — mathematician, physicist, and the mind that gave France its still-definitive translation of Newton's Principia — settled the physics in her 1740 Institutions de Physique. She championed 's Gravesande's brutal little experiment: brass balls dropped into soft clay from different heights dent it in proportion to $v^2$ , not $v$ — the clay is an energy ledger, not a momentum one. Her synthesis held that both quantities matter, each conserved in its own arena — precisely this module's dispatch table, written 280 years early. She completed the Principia translation while pregnant at 42, working eighteen-hour days against a premonition she would not survive the birth. She was right; the book appeared posthumously, and d'Alembert's generation finally conceded that the "war" had two victors.

History

1686

— Leibniz attacks the Cartesian

mv

as the measure of motion, proposing vis viva

mv^2

: the fork is born.

1722

— 's Gravesande's clay experiments show dent depth ∝ drop height ∝

v^2

: nature keeps an energy ledger too.

1740

— du Châtelet's Institutions de Physique argues both measures are real and separately conserved.

1743

— d'Alembert's Traité de dynamique declares the controversy a dispute over words: use each where it answers.

1847

— Helmholtz's conservation of energy completes the ledger; physics now runs permanently on both books.

Going further

Texts. HC Verma, Concepts of Physics Vol. 1 — compare the worked examples of Ch. 5–6 (force) with Ch. 8 (energy) on the same setups · Morin, Introduction to Classical Mechanics, Ch. 5's problems, many solved both ways · NCERT Class 11 Ch. 6, whose examples quietly model the dispatch.

Provenance. All results are standard chapter material re-audited: Drill 1's $\sqrt{2gh} = 8.0$ m/s; Drill 2's $t = 1.6$ s; Drill 3's $6.3$ m/s and the illegal-suvat $8.9$ m/s; Drill 4's $N = mg$ at $h = 3R$ (and $N = 0$ at $h = 5R/2$ ); Drill 5's quadratic $50x^2 + 5x - 36 = 0$ with root $0.80$ m; the pendulum's $3mg$ . A verifier should recompute each and confirm the two-route agreement in derivation A.

Prerequisites. Kinetic Energy & the Work–Energy Theorem · Conservation of Mechanical Energy · Free-Body Diagrams (and the NEWT archetype gym for the corners). Leads to. Which Conservation Law? (technique) · Option Elimination & Answer Checking (track capstone) · Limiting & Extreme Cases (the companion audit).

How do I decide fast — is there a literal checklist?

Underline the asked quantity, then run three questions. (1) Is it a time, or an acceleration, or a named force? → Newton owns it (possibly after an energy step for $v^2$ ). (2) Is it a speed-at-a-place, a height, or a stopping distance? → energy owns it. (3) Does the setup contain a spring, a curve, or the word "smooth"? → energy is being invited; does it contain "how long" or a straight track with constant forces? → Newton is being invited. Conflicts resolve toward the asked quantity, not the setup: a curved track with a time question is not solvable by either elementary route — which, in an exam, means you misread something or the question wants a statement, not a number.

Why exactly can't the energy method find times?

Because the ledger's entries — $\tfrac12mv^2$ , $mgh$ , $\tfrac12kx^2$ , $f d$ — contain positions and speeds but no $t$ anywhere. Two journeys between the same photographs, one leisurely and one violent, post identical ledgers. The information was never recorded, so no algebra can extract it. Newton's law, by contrast, is a statement about rates — force determines how velocity changes per second — so the clock is built into its grammar. Deeper down, energy conservation is what survives of Newton's law after the time variable is deliberately integrated away; asking it for a time is asking for the thing it discarded to become powerful.

In the hybrid, why is Newton's law valid at a single instant when the motion is so complicated?

Because $F = ma$ is a pointwise law: it relates the forces acting now to the acceleration now, with no memory of the path. Complicated history is irrelevant — what makes curved problems hard for the pure force method is not that Newton fails but that you'd have to apply him at every point and stitch the answers together with calculus. The hybrid avoids the stitching: energy (which loves whole journeys) carries $v^2$ to the point of interest, and Newton is invoked exactly once, where the question lives. Each law works its home turf; the handoff variable is $v^2$ .

Does the work–energy theorem count as the force method or the energy method?

It is the bridge — literally the force method with the clock integrated out. Net work equals $\Delta$ KE is derived from $F = ma$ , and it holds for every force, conservative or not, which makes it the most general single line in the unit. In dispatch terms it behaves as an energy tool (two snapshots, no clock) but with Newton's bookkeeping (you must count the work of every force, including friction — nothing is "banked" as PE unless you choose to). When friction or applied forces muddy a problem, WET is usually the safest ledger to open: conservation-of-mechanical-energy is WET with the no-toll clause activated.

What about momentum — where does IT fit in the dispatch?

Momentum is the third counter at the same bank, and it opens when interactions between bodies dominate: collisions, explosions, recoil — events where internal forces are violent, brief and unknowable, exactly where both force and energy methods choke (forces unmeasurable; kinetic energy not conserved). Its tell-tale phrases: "collide", "stick", "explode", "recoil", "fragments". The full three-way dispatch — energy vs momentum vs both — is the next technique module (Which Conservation Law?), built on the momentum chapter this one precedes. For now, the amber flag to plant: two unknown final velocities means energy alone is one equation short.

Is one method more "correct" or more fundamental than the other?

They are the same physics wearing different bookkeeping. Energy conservation for mechanics is derivable from Newton's laws (that derivation is the work–energy theorem), so no experiment can ever separate them — du Châtelet's resolution of the vis-viva war, in modern dress. "Which is fundamental" flips at higher levels: advanced mechanics (Lagrangian, Hamiltonian) and quantum theory take energy as primary and derive forces, because energy survives into theories where "force" does not. For exam purposes the equality is the point: any problem solvable both ways must give the same answer, which is a free cross-check — Drill 1's slide can be audited by the incline special case in thirty seconds.