The Potential Energy of a Spring (NCERT 5.9)
1. The Spring: A Rechargeable Energy Store
Compress a spring, let go, and it throws back everything you fed it — no losses, no leaks. The spring is mechanics' rechargeable battery, and NCERT 5.9 charges it with an exact number. The physics rests on Hooke's law from the force catalogue: within its elastic limit a spring pushes or pulls back toward its natural length with a force proportional to the deformation,
F = −k x (restoring form), k in N m⁻¹
where x is measured from the natural length and the minus sign says the force always opposes the displacement. Because the push grows as you push — 0 at natural length, kx at deformation x — the spring is the flagship variable-force problem of the chapter, and the area-under-the-graph method derives its energy in one triangle.
2. Complete Theory: Deriving U = ½kx² and Working the Chains
The derivation (a guaranteed derivation question — learn both routes). Compress the spring slowly through x. The applied force must match the spring's growing resistance, rising linearly from 0 to kx. Route one — average force: Favg = (0 + kx)/2 = kx/2, so W = Favg·x = ½kx². Route two — graph: the F–x plot is a triangle of base x and height kx, area ½·x·kx = ½kx². Choosing U = 0 at the natural length (the spring's minimum-energy state), this work is stored as
Us = ½ k x²
Two properties fall out of the square. First, symmetry: compressing by x and stretching by x store the same energy — x² erases the sign. Second, non-negativity: Us ≥ 0 always, minimal (zero) at natural length; the spring never owes energy, it only holds deposits.
Work done by the spring between two states. Generalising the triangle: the spring's work going from extension xi to xf is Wspring = ½k(xi² − xf²). Releasing from a compression x to natural length: xf = 0 → the spring pays out +½kx² (launch problems). Compressing it further from natural length: the spring does −½kx² (it resisted you the whole way). The sign is automatic from the formula — no case analysis needed.
The two energy chains NTA recycles yearly. Chain one — launch: spring energy ½kx² → kinetic energy ½mv², so v = x√(k/m). Chain two — max compression: a moving block meets a spring and stops momentarily; ½mv² = ½kxmax², so xmax = v√(m/k). Both are the conservation law of the previous page with the spring as the only U. On a vertical spring hanging at rest, gravity joins the books and equilibrium sits at extension mg/k — Hooke's law plus ΣF = 0, no energy needed for that particular question.
3. Visualising: Stretch, Compress — Same Deposit
Figure 5.6 — The square makes ±x equal
Left: natural length, U = 0. Middle: compressed x = 10 cm, coils squeezed (red), U = ½ × 1000 × 0.1² = 5 J. Right: stretched x = 10 cm, coils pulled, U = 5 J again.
Exam read-out: with k = 1000 N/m and x = 10 cm, both deformations bank exactly 5 J — and that 5 J launches a 0.1 kg ball at v = √(2 × 5/0.1) = 10 m/s. Every NTA spring numerical is one of these two panels feeding the conservation chain.
4. Solved Examples
Example 1 — Spring launches a ball
A spring of constant 1000 N m⁻¹ is compressed 10 cm and placed against a 0.1 kg ball on a frictionless floor. Find the energy stored and the ball's launch speed when the spring is released.
Solution (step by step): Stored: U = ½kx² = ½ × 1000 × (0.10)² = 5 J. Release: all of it converts (frictionless) → ½ × 0.1 × v² = 5 → v² = 100 → v = 10 m/s. Chain in one line: v = x√(k/m) = 0.1 × √(1000/0.1) = 0.1 × 100. Check the direction habit: the compressed spring pushes the ball away from the wall — the launch direction is decided by geometry, never by energy.
Example 2 — Maximum compression
A 2 kg block sliding at 6 m s⁻¹ on a frictionless floor hits a free-ended spring of constant 800 N m⁻¹. Find the maximum compression.
Solution: At maximum compression the block is momentarily at rest: Ki = ½ × 2 × 36 = 36 J has fully become Us = ½ × 800 × x². So 400x² = 36 → x² = 0.09 → x = 0.30 m. Reverse chain check: a block launched back by this spring from 0.3 m compression regains exactly 6 m/s — energy chains are reversible when friction is absent. Note what the answer ignores: the block's deceleration history; endpoints only, as always.
5. Practice Questions
Q1. A spring of constant 200 N m⁻¹ is stretched by 15 cm. Find the energy stored and the force required to hold it there.
ANSWER: U = ½kx² = ½ × 200 × (0.15)² = ½ × 200 × 0.0225 = 2.25 J. Holding force: F = kx = 200 × 0.15 = 30 N. Keep the two quantities separate in your head — the force is linear in x, the energy is quadratic; mixing them is the standard slip.
Q2 (MCQ). A spring is compressed by x, storing energy U. If the compression is reduced to x/2, the stored energy becomes: (a) U/2 (b) U/4 (c) 2U (d) 4U
ANSWER: (b) — U ∝ x², so halving x quarters the energy: U/4. The same ratio logic makes "doubling the compression gives 4U" and "one-third compression gives U/9" — three exam variants from one square.
Q3. A 1 kg block hangs at rest from a vertical spring of constant 500 N m⁻¹. Find the extension, and the energy stored in the stretched spring (g = 10 m s⁻²).
ANSWER: Equilibrium: kx = mg → x = 10/500 = 0.02 m = 2 cm (a force balance, no energy argument needed). Stored energy: U = ½kx² = ½ × 500 × 0.0004 = 0.1 J. Careful trap: gravity meanwhile lost mgh = 0.2 J of potential energy over the 2 cm descent — the difference 0.1 J is exactly the spring's gain, a three-line ledger worth reproducing if asked "where did the energy come from?".
6. Key Formulas & Takeaways
| Relation | Condition / remark |
|---|---|
| F = −kx | Hooke's law within elastic limit; x from natural length; k in N m⁻¹ |
| Us = ½kx² | Zero at natural length; same for stretch and compression; U ≥ 0 |
| Wspring = ½k(xi² − xf²) | Work by the spring; releasing → positive, compressing → negative |
| ½kx² = ½mv² (launch) | v = x√(k/m); frictionless floor |
| ½mv² = ½kx²max (stop) | xmax = v√(m/k); block momentarily at rest at full compression |
| Parallel: k = k₁ + k₂ · Series: k = k₁k₂/(k₁ + k₂) | Combos: parallel stiffer, series softer (ratio questions only) |
| Vertical equilibrium: x = mg/k | Force balance at rest; energy ledger still balances through ΔU_gravity |
Every spring question is one of three jobs: store (½kx²), pay out (launch chain), or absorb (max-compression chain). Identify the job, declare the zero at natural length, and let the conservation line finish.
7. Frequently Asked Questions
Derive the potential energy formula U = ½kx² for a spring.
Compressing a spring slowly through x means applying a force that grows linearly with Hooke's law from 0 to kx. The work done is the area under the F–x triangle: W = ½ · base · height = ½ · x · kx = ½kx². Equivalently, the average force over the stroke is (0 + kx)/2 = kx/2, and W = (kx/2)·x = ½kx². With U = 0 chosen at natural length, this work is stored as spring potential energy. The same ½kx² holds for stretch and compression alike because x enters squared.
What is the spring constant and what are its units?
The spring constant k measures stiffness: it is the force needed per unit extension, F = kx, so k = F/x with SI units newton per metre (N m⁻¹). A stiff spring (large k) gains a lot of energy for little stretch; a soft spring needs a long stretch for the same energy. Note from U = ½kx² that doubling k doubles the stored energy at the same x, while doubling x quadruples it — displacement is the more powerful lever, and NTA's ratio questions play on exactly that.
Why is the spring's potential energy never negative?
Because x enters as a square: U = ½kx² is zero at natural length and positive for compression (x negative) and stretch (x positive) alike. The spring's minimum energy state is its natural length, where it exerts no force. The energy is therefore always a positive deposit that the spring can pay back as work. The trap to avoid: while U itself is never negative, the work done BY the spring between two states, ½k(x_i² − x_f²), can be negative — compressing a spring further makes it do negative work on you.
How do springs combine in series and in parallel?
For two springs in parallel (side by side sharing the load) both stretch equally and their forces add: k_eq = k₁ + k₂. For two springs in series (end to end) both carry the same force and their extensions add: 1/k_eq = 1/k₁ + 1/k₂, i.e. k_eq = k₁k₂/(k₁ + k₂). These are the capacitor and resistor rules of electricity with the roles swapped — a parallel combination is stiffer than either spring, a series one is softer. Most Class 11 numericals stop at single springs; combinations appear as ratio questions.
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