Spring Force & Hooke's Law

In 1676, Robert Hooke published the most important discovery of his life as a scrambled anagram — ceiiinosssttuv — so he could claim priority without telling rivals what he had found. Two years later he unscrambled it: ut tensio, sic vis. "As the extension, so the force." Stretch a spring twice as far and it pulls back exactly twice as hard. That one proportionality is the reason your bathroom scale, your car's suspension, the balance wheel in a mechanical watch, and the force sensor in an atomic force microscope all work. The law earns its place in the Laws of Motion chapter because a spring is a force you can read with a ruler. A string's tension adjusts invisibly to whatever the problem demands; a spring's force is written in its geometry — measure the stretch $x$ , multiply by the stiffness $k$ , done. That makes springs the measuring instruments of mechanics (every "spring balance" question runs on it) and also the exam's favourite twist on Newton's-law problems: unlike a string, a spring can push as well as pull, and unlike a string, it cannot change its force instantly. Every board tests this law — IGCSE through load–extension graphs and the limit of proportionality, CBSE/ISC and NEET through spring balances and vertical setups, JEE through series/parallel combinations and the treacherous "spring is suddenly cut" instant. This article builds the law, the graph, the combinations, and the reflexes: where to measure $x$ from, which way the force points, and why the spring is the honest witness in any FBD it joins.

Key takeaways

A spring resists changes to its natural length with a restoring force $F = -kx$ , where $x$ is the extension or compression measured from natural length and $k$ is the spring constant (stiffness) in $\mathrm{N/m}$ .
The minus sign means restoring: stretched, the spring pulls back; compressed, it pushes out — always toward natural length. Springs push AND pull; strings only pull.
The law is linear only up to the limit of proportionality; beyond the elastic limit the spring deforms permanently — the load–extension graph tells the whole story.
A mass hanging in equilibrium stretches a vertical spring by $x_0 = mg/k$ — the working principle of every spring balance.
Combinations: springs in series share the force and add flexibility ( $1/k_s = 1/k_1 + 1/k_2$ — softer); springs in parallel share the load and add stiffness ( $k_p = k_1 + k_2$ — stiffer).
Key idea: a spring's force is written in its geometry — read the stretch, multiply by $k$ , point the force toward natural length: $F = -kx$ .

The core insight

Within its elastic limit a spring exerts a restoring force proportional to its deformation, $F = -kx$ , directed back toward natural length — so the force is determined by the visible stretch, making the spring both a force-maker and a force-meter.

Where this lives today

Your phone knows which way you're holding it because of Hooke's law. Inside every MEMS accelerometer a microscopic silicon proof mass hangs on silicon flexures — springs etched thinner than a hair — and acceleration is read off their deflection, $x = ma/k$ , thousands of times a second. The atomic force microscope runs the law at its most delicate: a cantilever of $k \sim 0.1\text{–}10\ \mathrm{N/m}$ rides over a surface, and sub-nanonewton forces map individual atoms through the deflection of that one calibrated spring. At the other end of the scale, vehicle suspensions tune $k$ against passenger comfort and cornering, and seismic isolation for gravitational-wave detectors stacks soft springs in series — deliberately exploiting $1/k$ -addition — to make the world's stillest platforms. Wherever engineering needs to measure a force, it still does what Hooke did: let the force stretch something honest, and read the ruler.

The physical picture

Take a spring hanging free and call its length the natural length $L_0$ . Everything in this article is measured from there. Pull the end down by $x$ and the spring pulls up on your hand; push it up by $x$ and it pushes back down. Both responses point toward the natural length — the spring always fights to undo whatever you did. That is the meaning of the minus sign in $F = -kx$ : not "the force is negative," but "the force opposes the displacement."

Figure

One spring, three states — the force always points back toward natural length

Contrast this with the string you just mastered. A string's tension is a constraint force: it takes whatever value keeps the string's length fixed, and you discover that value from Newton's second law. A spring's force is a law force: the geometry dictates it — $kx$ , no negotiation. This difference has two exam-grade consequences. First, a spring can push (a compressed spring is a loaded launcher), while a string never can. Second, a spring's force cannot change instantly: to change $F$ you must change $x$ , and the ends of a spring cannot teleport. Cut a string above a hanging pair and every tension re-adjusts at once; cut a support while a spring holds a block and, at that first instant, the spring is still stretched by the same $x$ and still exerts exactly $kx$ . Half of JEE's "suddenly cut" problems are this single sentence. Where does the linearity come from? Zoom into the metal: a solid is atoms in a lattice, each bond behaving — for small displacements — like a tiny spring. Coil the wire into a helix and the geometry converts small twists of the wire into large motions of the ends, but the proportionality survives: double the stretch, double every bond's strain, double the force. The linearity holds only while the deformation is small. On a load–extension graph the truth is drawn plainly:

Figure

The load–extension graph — where the law lives and where it dies

The straight region is Hooke's kingdom: slope $= k$ . The limit of proportionality is where the graph first bends; the elastic limit (at or just beyond it) is the point of no return — unload from beyond it and the spring stays permanently stretched. IGCSE and A-Level test the vocabulary and the graph-reading; JEE assumes you stay inside the linear region unless told otherwise.

The math: the law, the balance, the combinations

State the law with its bookkeeping:

F_{\text{spring}} = -k\,x \qquad (1)

where: $x$ — displacement of the spring's end from natural length (stretch positive, compression negative); $k$ — the spring constant in $\mathrm{N/m}$ , the force per unit stretch; the minus sign — the force opposes the deformation. A stiff car-suspension spring runs $k \sim 10^5\ \mathrm{N/m}$ ; a lab demonstration spring $\sim 20\ \mathrm{N/m}$ ; the effective spring of an AFM cantilever can be below $1\ \mathrm{N/m}$ . In problem-solving you usually work with magnitudes — $|F| = kx$ — and supply the direction ("toward natural length") by inspection on the FBD. Hang a mass on a vertical spring and let it settle. Equilibrium demands the spring force balance the weight:

x_0 = \frac{mg}{k} \qquad (2)

where: $x_0$ — the equilibrium stretch. This two-sided equation is a factory of exam questions: given $m$ and $x_0$ , it measures $k$ ; given $k$ and $x_0$ , it weighs the mass — which is precisely what a spring balance does. Note what it implies on other worlds: a spring balance reads $mg$ , so it would read one-sixth as much on the Moon, while a beam balance (comparing masses) would not change at all.

Derivation AThe spring balance — calibrating force with a ruler

We derive the working equation of the spring balance and use it to justify the claim that a spring converts force measurement into length measurement. The setup: a block of mass $m$ hangs at rest from a vertical spring of stiffness $k$ and natural length $L_0$ . Strategy: FBD of the block, equilibrium, then read the result as an instrument principle. Watch where $x$ is measured from — natural length, always.

Step 1. Forces on the block: weight $mg$ down; spring force $kx_0$ up (the spring is stretched, so it pulls its end — and the block — back up toward natural length).

Step 2. Equilibrium (block at rest):

k x_0 - mg = 0 \qquad\Rightarrow\qquad x_0 = \frac{mg}{k}

Step 3. Invert the reading. A scale marked in newtons beside the spring reports $F = k x_0$ for any steady load — the instrument: hang, wait for rest, read length, multiply by $k$ . Why: "wait for rest" is load-bearing. While the block bounces, the spring reads $kx$ at each instant — the instantaneous spring force — not the weight. A spring balance in a moving lift misreports weight for exactly this reason: it faithfully reports the spring force, and equilibrium is what ties that to $mg$ .

Step 4. Stack the loads: two identical 100 g masses stretch the spring twice as far as one — linearity is what makes the scale's markings evenly spaced. A scale with crowded markings at one end is confessing its spring has left the proportional region. Verified: $m = 0.50$ kg on $k = 98\ \mathrm{N/m}$ at $g = 9.8$ : $x_0 = 4.9/98 = 5.0$ cm — and 1.0 kg gives 10.0 cm, double, as linearity demands. Units: $\mathrm{N/m \times m = N}$ ✓.

Series and parallel — the combination preview

Exams love wiring springs together, and the two results are worth owning early (their full workout — oscillation periods, cut springs — belongs to the Cut & Combined Springs archetype in the oscillations chapter).

\text{series:}\quad \frac{1}{k_s} = \frac{1}{k_1} + \frac{1}{k_2} \qquad\qquad \text{parallel:}\quad k_p = k_1 + k_2 \qquad (3)

where: series (end-to-end) springs carry the same force and add their stretches — the combination is softer than either spring; parallel (side-by-side) springs share the load at a common stretch and add their stiffnesses — stiffer than either. Memory anchor: the rules are opposite to resistors' (series springs combine like parallel resistors). Two equal springs: series halves $k$ ; parallel doubles it.

Go deeper · Derivation B · 1 minEffective stiffness in series and in parallel

Derivation BEffective stiffness in series and in parallel

We derive both combination rules from the definitions — no memorization needed if you can re-run this in 60 seconds. Strategy: apply a force $F$ , find the total stretch, and read off $k_{\text{eff}} = F/x_{\text{total}}$ . The series case turns on a subtle point: why must both springs carry the same force?

Step 1. Series. Pull the bottom of a chain of two springs with force $F$ . Consider the (massless) junction between them: spring 2 pulls it down with the force it carries, spring 1 pulls it up with the force it carries; zero mass ⇒ the two forces match, and both equal $F$ . Same-force is the massless-string argument reborn.

Step 2. Each spring stretches by its own law: $x_1 = F/k_1$ , $x_2 = F/k_2$ . Total:

x = \frac{F}{k_1} + \frac{F}{k_2} \;\Rightarrow\; \frac{1}{k_s} = \frac{x}{F} = \frac{1}{k_1} + \frac{1}{k_2}

Why: flexibilities (1/k) add because stretches add at common force. Soft dominates: a weak spring in the chain is the bottleneck — as $k_2 \to 0$ , $k_s \to 0$ .

Step 3. Parallel. Two springs side by side gripping the same block: one displacement $x$ for both, forces $k_1 x$ and $k_2 x$ add on the block:

F = (k_1 + k_2)\,x \;\Rightarrow\; k_p = k_1 + k_2

Step 4. Check the equal-spring specials: two identical $k$ : series $k/2$ , parallel $2k$ . And $n$ identical springs: series $k/n$ , parallel $nk$ . Verified: limits — series with one rigid spring ( $k_2 \to \infty$ ) returns $k_1$ (a rigid link adds nothing); parallel with $k_2 = 0$ returns $k_1$ (a missing spring adds nothing). Numbers: $k_1 = 300$ , $k_2 = 600\ \mathrm{N/m}$ : series $200\ \mathrm{N/m}$ , parallel $900\ \mathrm{N/m}$ — softer than the softest, stiffer than the stiffest ✓.

Which fact answers which question

question	tool
force of a spring stretched/compressed by $x$	$\|F\| = kx$ , toward natural length
stretch under a hanging mass	$x_0 = mg/k$
find $k$ from a load–extension experiment	slope of the straight part of the graph
springs end-to-end	$1/k_s = 1/k_1 + 1/k_2$ (softer, same force)
springs side-by-side	$k_p = k_1 + k_2$ (stiffer, same stretch)
"the support is suddenly cut"	spring force is UNCHANGED at that instant — $x$ hasn't moved yet

Limiting cases

case	result	meaning
$x = 0$	$F = 0$	at natural length the spring is silent
$k \to \infty$	rigid link	an inextensible rod/string is an infinitely stiff spring
$k \to 0$	limp thread	no restoring ability
$n$ equal springs in series	$k/n$	a long spring is soft: same coil, more of it
$n$ equal springs in parallel	$nk$	load sharing: each spring carries $1/n$ of the force
beyond elastic limit	permanent set	Hooke's law void; graph curves, spring "remembers" the abuse

Common trap"x in F = kx is the length of the spring"

The deadliest substitution error in the chapter: $x$ is the deformation — current length minus natural length — never the length itself. A spring of natural length 20 cm stretched to 26 cm has $x = 6$ cm, and if $k = 50\ \mathrm{N/m}$ the force is $50 \times 0.06 = 3$ N, not $50 \times 0.26 = 13$ N. The same confusion returns in vertical oscillation problems dressed differently: with a mass hanging, the spring already carries the equilibrium stretch $x_0 = mg/k$ , and "the block is pulled down 4 cm" means 4 cm below equilibrium, i.e. a total deformation of $x_0 + 4$ cm and a spring force of $k(x_0 + 0.04)$ . Discipline: the moment you meet a spring, write down its natural length and mark where " $x = 0$ " lives; every later line inherits that choice. And convert to metres — a $k$ in N/m fed with centimetres silently produces answers a factor 100 off.

Worked examples

1 · Board level — reading a spring experiment (IGCSE/CBSE pattern, 3 marks)

A spring of natural length $12.0$ cm hangs vertically. With a $200$ g mass attached it settles at $16.0$ cm. Take $g = 9.8\ \mathrm{m/s^2}$ . Find (a) the spring constant, (b) the length when a $350$ g mass hangs instead (the spring stays within its limit of proportionality).

Step 1. (a) Extension: $x_0 = 16.0 - 12.0 = 4.0\ \mathrm{cm} = 0.040$ m. Load: $mg = 0.200 \times 9.8 = 1.96$ N.

Step 2. $k = \dfrac{mg}{x_0} = \dfrac{1.96}{0.040} = \boxed{49\ \mathrm{N/m}}$ .

Step 3. (b) New load: $0.350 \times 9.8 = 3.43$ N. New extension: $x = 3.43/49 = 0.070$ m $= 7.0$ cm.

Step 4. Length $= 12.0 + 7.0 = \boxed{19.0\ \mathrm{cm}}$ . Sanity: $1.75\times$ the load ⇒ $1.75\times$ the extension ( $4.0 \to 7.0$ cm) ✓ — linearity in action, and exactly what "within the limit of proportionality" licenses.

2 · Board level — spring pressed between hand and wall (push, not pull)

A block of mass $1.5$ kg on a smooth floor is pressed against a wall by a horizontal spring of $k = 400\ \mathrm{N/m}$ compressed by $6.0$ cm. Find the force the block exerts on the wall. The spring is then released and the block leaves it. What is the block's acceleration at the instant the compression has dropped to $2.0$ cm?

Step 1. Compressed spring pushes: on the block, $F = kx = 400 \times 0.060 = 24$ N toward the wall.

Step 2. Block in equilibrium (before release): wall pushes back on block with $24$ N; by Newton's third law the block pushes the wall with $\boxed{24\ \mathrm{N}}$ .

Step 3. After release, at $x = 2.0$ cm: spring force $= 400 \times 0.020 = 8.0$ N — the only horizontal force on the block now.

Step 4. $a = F/m = 8.0/1.5 = \boxed{5.3\ \mathrm{m/s^2}}$ , directed away from the wall. Note the force shrinks as the spring unloads — spring launches are not constant-force pushes, which is why energy methods will soon take over these problems (Block Meets Spring, in the energy chapter).

3 · Competitive level — series and parallel, same two springs (JEE-Main pattern, ~2.5 minutes)

Springs of $k_1 = 300\ \mathrm{N/m}$ and $k_2 = 600\ \mathrm{N/m}$ support a $6.0$ kg block, $g = 10\ \mathrm{m/s^2}$ . Find the total stretch when they hold the block (a) in series, (b) in parallel (both stretched equally). (c) In the series case, which spring stretches more?

Step 1. Load: $mg = 60$ N.

Step 2. (a) Series: $\dfrac{1}{k_s} = \dfrac{1}{300} + \dfrac{1}{600} = \dfrac{3}{600} \Rightarrow k_s = 200\ \mathrm{N/m}$ . Stretch: $x = 60/200 = \boxed{30\ \mathrm{cm}}$ .

Step 3. (b) Parallel: $k_p = 900\ \mathrm{N/m}$ . Stretch: $x = 60/900 = 6.7\ \mathrm{cm} \approx \boxed{6.7\ \mathrm{cm}}$ .

Step 4. (c) Series springs carry the same 60 N, so the softer stretches more: $x_1 = 60/300 = 20$ cm versus $x_2 = 60/600 = 10$ cm (their sum, 30 cm, checks Step 2 ✓). The soft spring is the bottleneck — it takes twice the deformation to deliver the same force. Timing note: series/parallel identification is the whole question — end-to-end shares force, side-by-side shares stretch. With that settled this is 2 minutes of arithmetic. MCQ distractor to expect: $k_s = 450$ (the average) — the series combination must be SOFTER than both, so 450 dies on sight.

4 · Competitive level — the spring balance in a lift (JEE-Main/NEET stretch, ~3 minutes)

A $2.0$ kg block hangs from a spring balance ( $k = 500\ \mathrm{N/m}$ ) fixed to the ceiling of a lift, $g = 10\ \mathrm{m/s^2}$ . Find the balance reading and the spring's extension when the lift (a) is at rest, (b) accelerates upward at $2.0\ \mathrm{m/s^2}$ , (c) is in free fall. (d) Would a beam balance comparing this block against standard masses change its verdict in case (b)?

Step 1. (a) At rest: reading $= mg = 20$ N; extension $= 20/500 = \boxed{4.0\ \mathrm{cm}}$ .

Step 2. (b) Newton's second law on the block, up positive: $F_{\text{spring}} - mg = ma \Rightarrow F = m(g + a) = 2.0 \times 12 = \boxed{24\ \mathrm{N}}$ ; extension $= 24/500 = 4.8$ cm. The balance reads the spring force — the apparent weight — not $mg$ .

Step 3. (c) Free fall: $a = g$ downward ⇒ $F = m(g - g) = \boxed{0}$ ; the spring returns to natural length. Balance reads zero — weightlessness is a spring balance's honest report of zero spring force.

Step 4. (d) No. A beam balance compares $m g_{\text{eff}}$ on both pans; whatever effective gravity does, it does to both sides equally, so the balance still matches 2.0 kg against 2.0 kg. Spring balances measure force; beam balances compare mass — the classic Moon question runs on the same distinction (spring reading drops to $g_{\text{moon}}/g \approx 1/6$ ; beam verdict unchanged). Timing note: (b) is the apparent-weight substitution $g \to g + a$ — one line if you've internalized the Lifts & Apparent Weight archetype, which owns the full family. The (d) concept pair "spring = force-meter, beam = mass-comparator" is a recurring one-mark MCQ; bank it. 3 minutes for all four parts.

Exam traps

$x$ from the wrong zero. Extension is measured from natural length — see the pitfall box. In vertical problems, decide explicitly whether a given displacement is from natural length or from the loaded equilibrium and write which; the difference is exactly $x_0 = mg/k$ .

Centimetres into a $\mathrm{N/m}$ formula. $k = 49\ \mathrm{N/m}$ with $x = 4$ must read $x = 0.04$ m. The give-away for a units slip is a spring force a hundred times too large — sanity-check against the weights in the problem.

"The spring's force adjusts like a string's." It cannot. A string's tension jumps instantly to whatever the constraint needs; a spring's force is locked to its current stretch. In any "support suddenly removed" problem, the spring force the instant after equals the spring force the instant before — then evolves as the spring's ends actually move.

Series/parallel swapped. End-to-end = series = softer ( $1/k$ 's add); side-by-side = parallel = stiffer ( $k$ 's add). Anchor: two equal springs in series must give $k/2$ — if your formula gives $2k$ , you swapped the rules. (Opposite to resistors.)

Applying $F = kx$ beyond the straight part. If the question's graph curves or mentions exceeding the elastic limit, the law is off-duty: read forces from the graph, not the formula, and expect permanent extension on unloading.

Assuming the balance reads $mg$ always. A spring balance reads the spring force. In equilibrium that equals the weight; in an accelerating lift it equals $m(g \pm a)$ ; in free fall, zero.

Which $g$ . IGCSE/A-Level typically use $9.8$ (CIE sometimes $9.81$ ); CBSE/ISC $9.8$ ; JEE/NEET declare $g = 10$ in the paper. Use what is stated — and note spring answers scale directly with it.

Origins · 1 minRobert Hooke and how this was discovered

Robert Hooke

1635–1703 · Royal Society, London

Robert Hooke may be the most under-credited genius of the scientific revolution. As the Royal Society's Curator of Experiments he had to demonstrate something new every week — and out of that furnace came the law of elasticity (hidden first in the anagram ceiiinosssttuv, unscrambled in 1678 as ut tensio, sic vis), the coining of the biological word "cell" from his microscope drawings in Micrographia, the balance-spring watch escapement he claimed Huygens took from him, and an inverse-square hunch about gravity he insisted Newton owed him. Newton, who feuded with Hooke for decades, wrote "if I have seen further it is by standing on the shoulders of giants" in a letter to Hooke — read by some as gracious, by others as a jab at Hooke's famously bent posture. No verified portrait of Hooke survives; the man who taught physics to measure force with a ruler left no measured likeness of himself.

History

1660

— Hooke discovers the proportionality of spring force and extension; hides it in an anagram (published 1676).

1678

— De Potentia Restitutiva unscrambles it: ut tensio, sic vis — as the extension, so the force.

1675

— Hooke and Huygens dispute priority on the balance-spring watch — elasticity becomes timekeeping.

1798

— Cavendish weighs the Earth with a torsion balance: Hooke's law in twist form measures gravity itself.

1986

— Binnig, Quate and Gerber's atomic force microscope reads atom-scale forces from a cantilever's deflection —

F = kx

at the nanoscale.

Going further

Texts. NCERT Physics Class 11, Ch. 4–5 · HC Verma, Concepts of Physics Vol. 1, Ch. 5 & 12 · CIE IGCSE 0625 topic 1.5.3 (load–extension graphs, limit of proportionality).

Provenance. All results are standard. A verifier should recompute: Example 1's $k = 49\ \mathrm{N/m}$ and 19.0 cm; Example 2's 24 N and $5.3\ \mathrm{m/s^2}$ ; Example 3's $k_s = 200$ , $k_p = 900\ \mathrm{N/m}$ , stretches 30 cm / 6.7 cm / 20+10 cm split; Example 4's readings 20 / 24 / 0 N and extensions 4.0 / 4.8 / 0 cm.

Prerequisites. Newton's Second Law · Weight & Normal Force · Free-Body Diagrams. Leads to. Spring Balances & Weighing Machines · Block Meets Spring (energy) · The Spring–Block Oscillator · Cut & Combined Springs (oscillations). Deep dive (UG). Spring Force & Hooke's Law (UG treatment, with the energy and SHM derivations).

What does the minus sign in F = −kx actually do in problems?

It encodes direction: the force opposes the displacement from natural length. In practice you rarely carry the sign through algebra at this level — you work with the magnitude $kx$ and point the arrow "toward natural length" on your FBD (stretched spring pulls its ends inward; compressed spring pushes them outward). The sign becomes load-bearing later, in oscillations: $F = -kx$ is precisely the restoring-force shape that produces SHM, so the minus sign is the reason springs vibrate rather than run away.

Is the spring constant a property of the material, like density?

No — it belongs to the object, not the material. Same steel, different geometry, different $k$ : a longer spring is softer (it is more coils in series — $k/n$ ), a thicker wire is stiffer, a wider coil is softer. That is why physics gives $k$ per spring but engineering speaks of Young's modulus — the material-level stiffness from which any particular spring's $k$ can be computed. You'll meet Young's modulus in the elasticity chapter; the mapping "wire = very stiff spring, $k = YA/L$ " is a lovely preview of it.

Why does cutting a spring in half make each piece stiffer?

Because the whole spring is the two halves in series: $1/k_{\text{whole}} = 1/k_{\text{half}} + 1/k_{\text{half}}$ , so $k_{\text{half}} = 2k$ . Physically: under the same force, each coil stretches the same small amount; half the coils means half the total stretch for the same force — stiffer by exactly the factor cut away. In general a piece with $1/n$ of the coils has $k' = nk$ . The full exam workout of this (with oscillation periods) lives in the Cut & Combined Springs archetype.

Does Hooke's law work for rubber bands?

Poorly, and examiner questions exploit it. A rubber band is roughly linear only for small stretches, stiffens dramatically as its polymer chains straighten, only pulls (it folds rather than push), and — most interestingly — does not retrace its loading curve on unloading. That loop (hysteresis) means energy is lost to heat each cycle, which is why repeatedly stretched rubber warms up and why IGCSE/A-Level force–extension questions about rubber show a loop instead of a line. Springs of metal, inside their elastic limit, retrace faithfully — that honesty is why instruments use metal.

Two spring balances are hooked in a line and I pull the free end with 10 N. What does each read?

Each reads 10 N — not 5 N each. They are in series: the same force threads through the whole chain (the massless-junction argument), so every link and every balance carries the full 10 N. The "shared out" intuition belongs to parallel arrangements, where side-by-side springs split the load. This is also why the series combination stretches more: each spring deforms as if it alone carried the entire force, and the stretches add.

What happens to k if a spring has non-negligible mass?

The static law survives — but the spring's own weight now stretches it, non-uniformly: coils at the top of a hanging heavy spring support all the coils below, so the spacing opens more near the support (the tension-profile logic from the strings article, applied to a spring). A hanging heavy spring extends by $Mg/2k$ under its own weight — as if half its mass hung at the end. In dynamics, a massive spring also carries kinetic energy; the standard correction adds one-third of the spring's mass to the load. Exam phrase "light spring" is the contract that lets you ignore all of this.

The problem says "the string holding the spring-block system is cut." Why does everyone insist the spring force doesn't change at that instant?

Because a spring's force is geometry: $F = kx$ , and $x$ is where the spring's ends are. Cutting a string removes the string's tension instantly (a constraint force can vanish in zero time), but the spring's ends haven't moved yet in the first instant — so its stretch, and hence its force, is exactly what it was. The accelerations right after the cut are computed with the OLD spring force and the string simply erased. One line of bookkeeping — "spring force cannot jump, string tension can" — is the key to a whole genre of JEE-Advanced instant-after problems.