ARCHETYPE: Block Meets Spring

At the end of every railway terminus stands a squat steel frame with two enormous springs: the buffer stop. Its job description is a physics problem — a mass arrives with kinetic energy; convert all of it to spring energy before the compression runs out — and its design equation is the one this archetype drills. A block slides into a spring: how far does the spring squash? How does friction change the answer? Does the block bounce back with the speed it brought, and can it end up stuck? Exam setters adore this setup because it forces the whole unit through one narrow doorway. The clean version is two lines of energy conservation. The rough-floor version adds friction's toll and produces a genuine quadratic — one of the few places a board exam makes you solve one. The rebound version tests whether you know that a spring pushes but cannot pull, and hides a wicked follow-up: the block's maximum speed does not occur where it first touches the spring, and its momentary halt at maximum compression is not equilibrium — the force there is at its largest. The archetype's punchlines: maximum compression from a three-column energy budget, the quadratic discipline for rough floors, the equilibrium-versus-extremes map along the spring, and the rebound audit that decides between "returns at full speed," "limps home slower," and "parks against the spring forever."

Key takeaways

Smooth floor: all kinetic energy becomes spring energy at maximum compression: $\tfrac12 mv^2 = \tfrac12 kx_{\max}^2$ , so $x_{\max} = v\sqrt{m/k}$ .
Maximum compression is where $v = 0$ — and where the force (and deceleration $kx_{\max}/m$ ) is LARGEST. It is the opposite of equilibrium.
Rough floor: friction taxes the journey — $\tfrac12 mv^2 = \tfrac12 kx_{\max}^2 + f\,x_{\max}$ — a quadratic in $x_{\max}$ ; take the positive root.
The block's maximum speed occurs after contact, at the point where spring force balances the other along-track forces ( $kx = f$ on a rough floor; $kx = mg\sin\theta$ on an incline) — not at first touch.
Rebound: an unattached spring pushes the block back and releases it at natural length — with the arrival speed on a smooth floor, slower on a rough one (friction charges both legs), possibly not at all if the budget dies en route.
On an incline, the descent distance includes the compression itself: $mg\sin\theta\,(d + x_{\max}) = \tfrac12 kx_{\max}^2$ for a smooth slope.
Key idea: run the three-column budget ( $\tfrac12 mv^2$ , $\tfrac12 kx^2$ , friction/gravity tolls) between snapshots — first touch, maximum squash, release — and every question in the family falls to the same audit.

The core insight

When a moving block meets a spring, energy bookkeeping between three snapshots — contact, maximum compression (all motion banked in $\tfrac12 kx^2$ , force maximal, equilibrium nowhere near), and release at natural length — answers every variant: equate the budget for the smooth case, add friction's path-length toll (a quadratic) for the rough case, and audit the return leg separately to settle the rebound.

Where this lives today

The archetype is safety engineering's founding calculation. Railway buffer stops are sized by equation (1) run backward: given the heaviest permitted train and a survivable deceleration, solve for $k$ and the travel $x$ — modern friction-hydraulic stops deliberately add a large "friction" term so the energy is dissipated rather than refunded (a rebounding train is its own hazard: engineers want the parked ending). Highway crash attenuators — the yellow barrel arrays before bridge pillars — are tuned spring-and-crush budgets stretching the stop over metres to keep $a_{\max}$ under about 20g. Elevator shafts keep code-mandated buffer springs at the pit bottom. And at the playful end, every pinball launcher and Nerf gun is worked example 2: a latched $\tfrac12 kx^2$ , a released puck, and a speed you can now predict before pulling the trigger.

The setup

Geometry first, because in an archetype the geometry is the physics. A block of mass $m$ slides along a floor at speed $v$ toward a light spring of constant $k$ , one end fixed to a wall, the other end free. The block is not attached — it can press the spring but the spring can only push back. Three snapshots organize everything: (i) first touch, spring at natural length, block at full speed; (ii) maximum compression $x_{\max}$ , block momentarily at rest, spring force at its peak; (iii) release, the spring back at natural length shoving the block away, after which the spring stands still and the block coasts (or doesn't).

Figure

The archetype in three snapshots — touch, maximum squash, release

Two readings of the picture carry the archetype. First, the spring is an energy trap with a spring-loaded door: it absorbs the block's kinetic energy quadratically ( $\tfrac12 kx^2$ — the second centimetre of squash costs triple), holds all of it for one instant at $x_{\max}$ , then hands it back on the return leg. On a smooth floor the refund is perfect — conservative force, clean books. Second, the drama is force, not just energy: the block decelerates hardest exactly when it moves slowest, and the instant of rest at $x_{\max}$ is the instant of maximum net force $kx_{\max}$ . Every "the block is in equilibrium at maximum compression" option is bait.

The master solution pattern

Every variant runs the same audit. Choose two snapshots, write the three-column budget — kinetic $\tfrac12 mv^2$ , spring $\tfrac12 kx^2$ , plus gravity's column on slopes — and charge friction's toll on the path actually slid. Smooth floor, snapshots (i)→(ii):

\tfrac12 mv^2 = \tfrac12 k\,x_{\max}^2 \qquad\Longrightarrow\qquad x_{\max} = v\sqrt{\frac{m}{k}} \qquad (1)

where: the scalings do exam duty on their own — doubling the speed doubles the compression (not quadruples: $x \propto v$ , since both sides are quadratic); a stiffer spring squashes as $1/\sqrt{k}$ ; a heavier block as $\sqrt{m}$ . Read the buffer-stop design brief in it: to halve the stopping distance you need four times the spring constant. Rough floor, same snapshots, friction charging the compression distance:

\tfrac12 mv^2 = \tfrac12 k\,x_{\max}^2 + f\,x_{\max} \qquad (2)

where: $f = \mu_k mg$ — and the equation is now a genuine quadratic, $\tfrac12 kx^2 + fx - \tfrac12 mv^2 = 0$ , solved by formula with the positive root (the negative root is the algebra's ghost, not a compression). This is one of the few places boards make you run the quadratic formula under time pressure; example 3 drills it with clean numbers. The force map along the track (rough floor) — where the block actually speeds and slows:

\text{max speed at } kx = f \qquad\text{(the moving "equilibrium" point)}\qquad (3)

where: before that point the spring's push is weaker than the help... careful — on the approach, both spring force and friction oppose the block, so it decelerates from first touch; equation (3)'s balance matters on the return leg, where the spring pushes forward and friction drags back: the block gains speed until $kx$ falls to $f$ , then loses it. On a smooth floor ( $f = 0$ ) the maximum return speed is at natural length — the release point. On an incline (approach), gravity helps the block: it accelerates into the spring until $kx = mg\sin\theta$ , and only then begins slowing — the counter-intuitive beat that JEE loves.

Derivation AThe smooth case — and why maximum compression is maximum force

We derive equation (1) and pin the archetype's central trap: the state of the block at maximum compression. Strategy: two snapshots of the pure budget, then one glance at Newton for the force. Watch how little algebra there is — the smooth case is deliberately cheap so the ideas stay visible.

Step 1. Guest list between snapshots (i) and (ii): spring force (conservative — banked as $\tfrac12 kx^2$ ), gravity and normal (perpendicular to motion — workless). Conservation applies in pure form.

Step 2. Budget: $\tfrac12 mv^2 + 0 = 0 + \tfrac12 kx_{\max}^2$ , giving $x_{\max} = v\sqrt{m/k}$ .

Step 3. The state at (ii): velocity zero — but Newton's second law reads the force, and the spring pushes with its peak value $kx_{\max}$ . The block's acceleration there is

a_{\max} = \frac{k\,x_{\max}}{m} = v\sqrt{\frac{k}{m}}

— its largest of the whole encounter. The block is at rest and maximally shoved: the perfect anti-equilibrium.

Step 4. Partial-compression speeds come from the same budget: at compression $x < x_{\max}$ , $\tfrac12 mv_x^2 = \tfrac12 mv^2 - \tfrac12 kx^2$ . At half the maximum squash, $v_x = v\sqrt{1 - \tfrac14} = 0.87v$ — the block still carries 75% of its energy; the spring's quadratic bite is gentle early and savage late. Why the block leaves at natural length with speed $v$ exactly: the return leg replays the budget in reverse — conservative forces refund in full. The spring cannot pull, so beyond natural length it simply disengages; block coasts at $v$ , spring stays put. Attached springs (bolted blocks) run past natural length into stretch and oscillate — a different problem, flagged by the word "attached." *Verified: $m = 2$ kg, $v = 4$ m/s, $k = 800$ N/m: $x_{\max} = 4\sqrt{2/800} = 0.2$ m; check $\tfrac12 \times 800 \times 0.04 = 16$ J $= \tfrac12 \times 2 \times 16$ ✓; $a_{\max} = 800 \times 0.2/2 = 80$ m/s² — eight g's, the buffer-stop's argument for softer springs.*

Go deeper · Derivation B · 1 minThe rough case — the quadratic, the return audit, and the stuck verdict

Derivation BThe rough case — the quadratic, the return audit, and the stuck verdict

We derive the rough-floor compression (2), then run the return leg to see what the block keeps — and when it never comes home at all. Strategy: same snapshots, friction charging the odometer both ways; the return is a fresh audit from snapshot (ii). Watch the stuck check at the turnaround — a static-friction question hiding inside an energy problem.

Step 1. Approach, (i)→(ii): friction $f = \mu_k mg$ opposes the motion over the distance $x_{\max}$ :

\tfrac12 mv^2 = \tfrac12 kx_{\max}^2 + f\,x_{\max}

Rearranged: $\tfrac12 kx^2 + fx - \tfrac12 mv^2 = 0$ , a quadratic in $x$ ; the physical root is

x_{\max} = \frac{-f + \sqrt{f^2 + km v^2}}{k}

Step 2. The turnaround interrogation. At rest at $x_{\max}$ , does the block move again? It does only if the spring's push beats static friction's ceiling:

k\,x_{\max} > \mu_s m g \quad\Rightarrow\quad \text{rebounds};\qquad k\,x_{\max} \le \mu_s mg \quad\Rightarrow\quad \text{parked forever, spring still compressed}

The parked case is a legal, stable ending — the spring holds a permanent compression, friction holds the spring. Exam setters spring it (so to speak) on students who believe "springs always bounce things back."

Step 3. Return leg, (ii)→(iii), fresh audit: the spring pays out $\tfrac12 kx_{\max}^2$ , friction charges $f\,x_{\max}$ again (it flipped with the motion):

\tfrac12 m v'^2 = \tfrac12 kx_{\max}^2 - f\,x_{\max}

The block leaves slower than it arrived — the round trip cost $2f x_{\max}$ , friction's double toll from the conservative-forces module, cashed here.

Step 4. The block can also die on the way out: if $\tfrac12 kx_{\max}^2 \le f\,x_{\max}$ (i.e. $kx_{\max} \le 2f$ ), the budget zeroes before natural length… combined with Step 2's ceiling this brackets three endings: full rebound, partial crawl-out, permanent park. Read $kx_{\max}$ against $\mu_s mg$ and $2f$ and the story picks itself. Verified: the numbers of worked example 3 — $m = 2$ kg, $v = 2$ m/s, $\mu = 0.5$ , $k = 100$ N/m: budget 4 J = $50x^2 + 10x$ → $x_{\max} = 0.2$ m; turnaround: $kx_{\max} = 20$ N $> \mu mg = 10$ N → rebounds; return: spring 2 J minus friction 2 J = 0 → the block stops exactly at natural length. All three phases in one problem.

The variants examiners actually pose

The incline feed. A block released a distance $d$ up a smooth slope slides into a spring at the bottom: the budget gains gravity's column, and — the classic catch — the block keeps descending while compressing, so gravity pays over $d + x_{\max}$ , not $d$ : $mg\sin\theta\,(d + x_{\max}) = \tfrac12 kx_{\max}^2$ . Another quadratic; example 4 runs it. Bonus beat: maximum speed occurs after contact, at $kx = mg\sin\theta$ .

The half-compression interrogation. "Find the speed when the spring is compressed to half the maximum": budget with $x = x_{\max}/2$ — kinetic energy is $\tfrac34$ of the arrival value on a smooth floor, never half. The quadratic storage is the whole point.

The two-spring tandem and the pre-compressed launcher. Springs in series/parallel replace $k$ (Hooke's-law module rules); a spring pre-compressed by a latch and released is the same budget run backward — launcher speed $v = x\sqrt{k/m}$ .

The graph version. AP and IB dress the archetype as an $F$ – $x$ graph: compression force line of slope $k$ ; the stopping compression is where the accumulated area equals the arrival kinetic energy.

Figure

The speed map along the spring — where the block is fastest, and why it isn't at first touch

Limiting cases

case	result	meaning
$k \to \infty$ (rigid wall)	$x_{\max} \to 0$ , $a_{\max} \to \infty$	rigid stops are violent — the buffer-stop's reason to exist
$k \to 0$ (marshmallow spring)	$x_{\max} \to$ huge	soft springs need room; stopping distance trades against force
$v \to 2v$ , smooth	$x_{\max}$ doubles, $a_{\max}$ doubles	linear, not quadratic — both budget columns are squares
$f$ huge	block stops before or barely into the spring	friction can pre-empt the archetype entirely
$\mu_s$ large at turnaround	$kx_{\max} \le \mu_s mg$ : parked	the spring-held, friction-locked ending
$\theta \to 0$ on the incline variant	recovers the flat smooth case	the $(d + x)$ subtlety gracefully disappears

Common trapAt maximum compression the block is in the LEAST equilibrium of its life

" $v = 0$ , so the block is in equilibrium" — the archetype's flagship error. Equilibrium is about force, not speed: at $x_{\max}$ the spring pushes with its largest value $kx_{\max}$ , the acceleration $kx_{\max}/m$ peaks, and the block is about to be hurled back. Rest and equilibrium coincide only where zero velocity meets zero net force — and along this track the net-force zero lives elsewhere entirely (at $kx = f$ on a rough return, at $kx = mg\sin\theta$ on an incline — the points of maximum SPEED, where the block is moving fastest and most "in balance" at once). Bank the inversion as a slogan: maximum squash = maximum force; balanced force = maximum speed. Options pairing "momentarily at rest" with "net force zero" are testing exactly this line.

Worked examples

1 · Board level — the buffer stop, smooth (CBSE/ISC pattern, 3 marks)

A $2.0$ kg trolley moving at $4.0$ m/s on a smooth track runs into a light buffer spring of constant $k = 800$ N/m. Find (a) the maximum compression, (b) the maximum force on the trolley, (c) the trolley's speed when the compression is $0.10$ m.

Step 1. (a) Budget, touch → max squash: $\tfrac12 \times 2 \times 16 = \tfrac12 \times 800 \times x^2 \Rightarrow x_{\max} = \boxed{0.20\ \mathrm{m}}$ .

Step 2. (b) Peak force at peak squash: $F = kx_{\max} = 800 \times 0.2 = \boxed{160\ \mathrm{N}}$ (deceleration $80$ m/s² — quote it and note the trolley is not in equilibrium there).

Step 3. (c) Partial budget: $\tfrac12 \times 2 \times v^2 = 16 - \tfrac12 \times 800 \times 0.01 = 12$ J $\Rightarrow v = \boxed{3.5\ \mathrm{m/s}}$ — half the squash has taken only a quarter of the energy.

2 · Board level — the launcher runs the film backward (ICSE/A-Level pattern)

A spring of constant $k = 600$ N/m is compressed $10$ cm by a latch behind a $0.60$ kg puck on a smooth surface. The latch is released. Find (a) the puck's launch speed, (b) its speed when the spring still holds $25\%$ of the stored energy.

Step 1. (a) Stored: $\tfrac12 \times 600 \times 0.01 = 3.0$ J. All of it becomes kinetic at natural length: $v = \sqrt{2 \times 3.0/0.6} = \boxed{3.2\ \mathrm{m/s}}$ (√10, 2 s.f.).

Step 2. (b) Kinetic share $= 75\%$ of 3.0 J $= 2.25$ J: $v = \sqrt{2 \times 2.25/0.6} = \boxed{2.7\ \mathrm{m/s}}$ .

Step 3. Note the checkpoint: 25% of the energy left in the spring means the compression is at half its initial value ( $U \propto x^2$ ) — the quadratic account read in reverse.

3 · Competitive level — the rough floor's quadratic, with all three endings audited (JEE-Main pattern, ~3.5 minutes)

A $2.0$ kg block slides at $2.0$ m/s into a spring of constant $k = 100$ N/m on a floor with $\mu_s = \mu_k = 0.5$ . ( $g = 10$ m/s².) Find (a) the maximum compression, (b) whether the block rebounds, (c) its speed back at the spring's natural length.

Step 1. (a) Budget with friction's toll ( $f = \mu mg = 10$ N): $\tfrac12 \times 2 \times 4 = \tfrac12 \times 100\,x^2 + 10x$ , i.e. $50x^2 + 10x - 4 = 0$ . Quadratic formula: $x = \dfrac{-10 + \sqrt{100 + 800}}{100} = \dfrac{-10 + 30}{100} = \boxed{0.20\ \mathrm{m}}$ (positive root only).

Step 2. (b) Turnaround interrogation: spring push $kx_{\max} = 20$ N versus static ceiling $\mu_s mg = 10$ N. $20 > 10$ : $\boxed{\text{it rebounds}}$ .

Step 3. (c) Return audit: $\tfrac12 mv'^2 = \tfrac12 kx_{\max}^2 - f x_{\max} = 2.0 - 2.0 = 0$ → $v' = \boxed{0}$ . The block crawls back and stops exactly at natural length — 2 J banked at max squash, 2 J burned crawling home. Of the original 4 J, friction took everything, half on each leg. Timing note: ~3.5 min. The quadratic's discriminant ( $f^2 + kmv^2 = 900$ ) is engineered clean — recompute it symbolically before plugging. Designed errors: dropping friction from the approach (gives 0.28 m), forgetting the positive-root discipline, and skipping the Step-2 interrogation (a stiffer $\mu_s$ would have parked the block at 0.2 m forever — always check before computing a return).

4 · Competitive level — down the incline into the spring (JEE-Advanced flavour, ~4 minutes)

A $1.0$ kg block is released from rest on a smooth $30°$ incline, a distance $1.2$ m (along the slope) above the free end of a spring ( $k = 100$ N/m) fixed along the incline. ( $g = 10$ m/s².) Find (a) the maximum compression, (b) the compression at which the block moves fastest, and (c) that maximum speed.

Step 1. (a) Gravity pays over the whole descent, $d + x$ : $mg\sin 30°\,(1.2 + x) = \tfrac12 kx^2$ , i.e. $5(1.2 + x) = 50x^2 \Rightarrow 50x^2 - 5x - 6 = 0$ . Discriminant $25 + 1200 = 1225 = 35^2$ : $x_{\max} = \dfrac{5 + 35}{100} = \boxed{0.40\ \mathrm{m}}$ .

Step 2. (b) After contact the block still accelerates while gravity's pull beats the spring's push; the speed peaks where they balance: $kx = mg\sin 30° \Rightarrow x = \dfrac{5}{100} = \boxed{0.05\ \mathrm{m}}$ — barely into the spring, and emphatically not at first touch.

Step 3. (c) Budget from release to $x = 0.05$ m: $\tfrac12 mv^2 = mg\sin\theta (1.2 + 0.05) - \tfrac12 k (0.05)^2 = 5 \times 1.25 - 0.125 = 6.125$ J → $v = \boxed{3.5\ \mathrm{m/s}}$ .

Step 4. Audit: at max compression the budget closes — $5 \times 1.6 = 8$ J deposited by gravity, $\tfrac12 \times 100 \times 0.16 = 8$ J in the spring ✓. Timing note: ~4 min. Two designed errors carry the marks: charging gravity over $d$ instead of $d + x$ (yields the wrong quadratic $50x^2 = 6$ ), and answering (b) with "at first touch" or "at max compression". The balance point $kx = mg\sin\theta$ is also where an attached block would eventually oscillate about — a free preview of SHM.

Exam traps

Equilibrium at max compression. The flagship trap — see the pitfall box. Max squash = max force; the net-force zero (max speed) lives at $kx = f$ or $kx = mg\sin\theta$ .

Friction charged one way. The toll applies on approach AND return, each over the distance actually slid. Round-trip losses double; example 3's block lost half its energy on each leg.

Gravity's distance on inclines. The block descends $d + x_{\max}$ , not $d$ . Missing the $+x$ shrinks the quadratic's constant term and every root after it.

Peak force used for energy. $W \ne kx_{\max} \cdot x_{\max}$ ; the spring's storage is the triangle $\tfrac12 kx^2$ . The factor-2 ghost from the variable-force module haunts this archetype specifically.

Negative root kept, or discriminant botched. The quadratic's negative root is unphysical. Compute the discriminant symbolically ( $f^2 + kmv^2$ for the flat case) before substituting — the numbers are usually engineered to come out square.

The spring that pulls. Unattached springs release the block at natural length, full stop. Only the word "attached" (or "fastened", "connected") licenses stretch, oscillation, or negative spring displacement.

Rebound assumed. At the turnaround, interrogate static friction: $kx_{\max}$ versus $\mu_s mg$ . Parked-forever is a real answer, and "find the final resting position" questions expect you to consider it.

Linear-speed intuition on partial compressions. At half of $x_{\max}$ the block retains $\sqrt{3}/2 \approx 87\%$ of its speed, not half. The spring's early centimetres are cheap — quadratic storage, always.

Origins · 1 minRobert Hooke and how this was discovered

Robert Hooke

1635–1703 · Gresham College, London

Hooke published the spring law that runs this archetype as a 1660 anagram — ceiiinosssttuv, unscrambled in 1678 to ut tensio, sic vis, "as the extension, so the force" — staking priority while he kept experimenting. Curator of Experiments to the Royal Society, he weighed air, drew the first microscopic cell, rebuilt London after the Great Fire alongside Wren, and feuded with Newton over almost everything. The linearity he codified is why the buffer-stop budget is a clean triangle ( $\tfrac12 kx^2$ ) and why the block-meets-spring quadratic has constant coefficients: every result in this module leans on his one sentence.

History

1660

— Hooke poses his spring law as an anagram; the solution "as the extension, so the force" follows in 1678

1784

— Coulomb's torsion balance turns spring energy storage into precision measurement

1846

— Railway buffer stops standardize the moving-mass-meets-spring design brief

1930

s — Highway crash barriers adopt controlled compression to cap deceleration

1959

— Volvo's three-point era begins the modern science of stretching stopping distances to shrink forces

Going further

Texts. NCERT Physics Class 11, Work, Energy and Power chapter (the block–spring worked example) · HC Verma, Concepts of Physics Vol. 1, Ch. 8 · Halliday–Resnick–Walker Ch. 8 (spring–block energy problems).

Provenance. All results are standard. A verifier should recompute: derivation A's anchor ( $x_{\max} = 0.2$ m, $a_{\max} = 80$ m/s²); derivation B / example 3's triple audit ( $x_{\max} = 0.2$ m, rebound verdict 20 N > 10 N, return speed exactly 0); example 2's launcher (3.0 J, 3.2 then 2.7 m/s); example 4's quadratic ( $1225 = 35^2$ , $x_{\max} = 0.4$ m, peak speed 3.5 m/s at 5 cm).

Prerequisites. Spring Force & Hooke's Law · Conservation of Mechanical Energy & Efficiency · Work by a Variable Force & F–x Graphs. Leads to. ARCHETYPE: Mass Dropped on a Vertical Spring (ahead in this unit) · TECHNIQUE: Force Method or Energy Method? · Conservation of Mechanical Energy (UG deep dive).

Why does maximum speed happen AFTER the block touches the spring on an incline?

Because touching is not braking — the spring's push starts at zero (natural length exerts no force) while gravity's along-slope pull is constant. Just after contact the net force is still downhill: the block keeps gaining speed. Only when the compression reaches $x = mg\sin\theta/k$ does the spring's push match gravity, and beyond it the net force flips uphill and braking begins. That balance point is the speed maximum. The same beat plays on a rough floor's return leg at $kx = f$ . First touch is where the plot changes, not where the speed peaks.

The spring did negative work on the block while compressing — where is that energy during the pause at x_max?

In the spring's bent coils: elastic potential energy, $\tfrac12 kx_{\max}^2$ , held in the deformed bonds of the steel. The pause is only kinetic rest — the energy never stopped existing, it changed columns. On a smooth floor the return leg empties the coil back into motion, joule for joule. This is precisely what "conservative force" purchased in the earlier module: a storage device with a full-refund policy. Friction, meanwhile, has no coil — its take during both legs is already heat.

What changes if the block is ATTACHED to the spring?

Everything after natural length. An attached block sails past natural length, stretches the spring, and is pulled back — the system oscillates about the equilibrium point instead of disengaging. On a smooth floor the result is perpetual oscillation (simple harmonic motion, next chapter); with friction, a shrinking oscillation that parks somewhere the spring force can no longer beat static friction. The unattached block of this archetype gets exactly one compression and one push-off — read the problem's verb ("rests against", "strikes" vs "is fastened to") before writing anything.

In the rough-floor problem, why does friction appear in the energy equation but static friction decide the rebound?

Two different jobs at two different moments. While the block slides, kinetic friction acts and bills energy by distance — that is the $fx$ term in the budget. At the instant of rest at $x_{\max}$ , nothing slides: the question "does it move again?" is a force standoff between the spring's push $kx_{\max}$ and static friction's ceiling $\mu_s mg$ . Energy methods find where things stop; force methods decide whether stopped things restart. The archetype needs both, in that order.

Could the block bounce back and forth on a rough floor several times before stopping?

Against an unattached spring — no second bounce is available from the spring side: after release the block travels away, and friction only slows it; it never returns (nothing pulls it back). It either escapes the region or grinds to rest on open floor. An attached block, or a block shuttling between TWO springs (a JEE-Advanced staple), does oscillate with shrinking amplitude, losing $2f \times$ (travel) per round trip; the final resting place is wherever the remaining spring force first fails to beat static friction. The audit is the same budget, applied leg by leg — tedious by design, which is why setters usually ask for just the first or second turnaround.

How do I decide between energy methods and F = ma for spring-impact problems?

Energy for distances and speeds; Newton for forces and instants. "Maximum compression", "speed at compression $x$ ", "how far up does it rebound" — budgets, every time (the spring's force varies with position, so kinematics equations are illegal anyway). "Maximum force", "acceleration at $x_{\max}$ ", "does it move again" — Newton at a snapshot. The archetype's standard full question walks you through both on purpose; answering force questions with energy (or vice versa) is the classic way to stall. The decision tree gets its own technique module at the end of this unit.