For an ideal spring with force constant k stretched or compressed by amount x from its natural length, the restoring force is F = -k x and the elastic PE stored is U = ½ k x². Independent of sign of x.
-- NCERT Class 11 Physics, Ch. 5, p. 80Potential Energy Spring
8 MCQs2 revision cards9-step worked example
Source: NCERT Work, Energy and PowerPYQ coverage: NEET 2023Official key: NTA-verifiedLast updated: 21 Sep 2026
Potential Energy Spring, explained for NEET
The trap on this topic is almost always the same one: treating spring potential energy as if it scaled with the stretch. A spring stretched 2 cm stores some energy U; stretch it to 4 cm and the instinct says 2U. It is 4U. The square in U = ½kx² is the entire content of this topic, and a question that gives you U at one displacement and asks for U at another is testing only whether you squared the ratio.
NCERT Class 11 Physics Chapter 5 gives the result on page 80: the elastic potential energy of a spring displaced by x from its natural length is U = ½kx², with k the spring constant in N/m. Two features of this expression are worth fixing.
First, x is measured from the natural (unstretched) length, not from any other position. Second, U depends on x², so compression and extension of the same magnitude store the same energy — a spring squeezed 3 cm holds exactly what it holds when pulled 3 cm. The sign of x never reaches the answer.
The ratio form is the one to carry into the exam hall:
U₂/U₁ = (x₂/x₁)²
This is where NEET lives on this topic. Given "PE at 2 cm is U, find PE at 8 cm," the stretch ratio is 4, so the energy ratio is 16. Distractors are built to catch the two predictable slips: 4U (ratio not squared) and U/16 (ratio inverted). Both appear in the options; only one of them is your habit.
Two conditions attach to the formula. It holds while the spring obeys Hooke's law — past the elastic limit the expression fails. And it assumes an ideal spring with no internal damping; a real spring diverts some energy to internal heating.
Watch-out: when a question changes the spring rather than the stretch, k is linear — double k at fixed x and U doubles. The square belongs to x alone.
Can you answer these Potential Energy Spring MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In the expression for the elastic potential energy of a spring given in NCERT Class 11 Physics Chapter 5, the displacement x is measured from:
Show answer and why every option is right or wrong
Answer: B. B is correct. Page 80 of NCERT Class 11 Physics Chapter 5 defines x as the displacement from the spring's natural length, which is why U = 0 when the spring is unstretched.
Why A is wrong: A is wrong because the support's position is fixed and arbitrary; the energy depends on how far the spring is deformed, not on where it happens to be anchored.
Why C is wrong: C is wrong because the lowest point of a mass's motion is a feature of that particular problem, not of the spring; x is defined by the spring's own unstretched configuration.
Why D is wrong: D is wrong because it imports the gravitational-PE reference-level convention into the spring formula, where the zero is fixed by the natural length rather than chosen.
MCQ 2Easy RecallPractice
The SI unit of the spring constant k in U = ½kx² is:
Show answer and why every option is right or wrong
Answer: B. B is correct. k is force per unit displacement, so its unit is newton per metre, as listed with the formula on page 80 of NCERT Class 11 Physics Chapter 5.
Why A is wrong: A is wrong, although J/m is dimensionally equal to N — it is the unit of force, missing one further division by metre.
Why C is wrong: C is wrong because N·m is the joule, the unit of the energy U itself rather than of k.
Why D is wrong: D is wrong because J·m has dimensions of energy × length, which matches neither k nor any quantity in this expression.
MCQ 3Easy RecallPractice
An ideal spring is first extended by 3.0 cm and then, separately, compressed by 3.0 cm from its natural length. The elastic potential energy stored is:
Show answer and why every option is right or wrong
Answer: C. C is correct. U = ½kx² depends on x², so the sign of the displacement does not reach the answer — equal magnitudes of compression and extension store equal energy (NCERT Class 11 Physics Chapter 5, page 80).
Why A is wrong: A is wrong because it assumes the formula distinguishes the two directions; squaring x removes any such distinction for an ideal Hooke's-law spring.
Why B is wrong: B is wrong for the same reason as A — the asymmetry it assumes belongs to real springs with structural limits, not to the ideal spring the formula describes.
Why D is wrong: D is wrong because a negative x squares to a positive x²; energy is stored, and it is never negative in this expression.
MCQ 4Direct ApplicationPractice
A spring of force constant 2.0 × 10² N/m is stretched by 0.10 m from its natural length. The elastic potential energy stored is:
Show answer and why every option is right or wrong
Answer: A. A is correct: U = ½ × 2.0 × 10² × (0.10)² = ½ × 2.0 × 10² × 1.0 × 10⁻² = 1.0 J. The ½ is an exact factor from the formula on page 80 of NCERT Class 11 Physics Chapter 5 and does not affect the significant-figure count.
Why B is wrong: B is wrong because it uses kx rather than ½kx² — the displacement was not squared and the ½ was dropped.
Why C is wrong: C is wrong because it computes kx² without the factor of ½, doubling the stored energy.
Why D is wrong: D is wrong because it halves the correct result a second time, applying the ½ factor twice.
MCQ 5Direct ApplicationPractice
A spring stores potential energy U when stretched by 2.0 cm from its natural length. When the same spring is stretched by 8.0 cm, the energy stored is:
Show answer and why every option is right or wrong
Answer: C. C is correct. U scales as x², so U₂/U₁ = (8.0/2.0)² = 4² = 16, giving 16U (NCERT Class 11 Physics Chapter 5, page 80).
Why A is wrong: A is wrong because it inverts the ratio, treating energy as falling with increasing stretch; U rises with x.
Why B is wrong: B is wrong because it applies the stretch ratio of 4 directly without squaring it — the linear-scaling error this pattern is built to catch (trap: spring PE linear vs quadratic).
Why D is wrong: D is wrong because it cubes the stretch ratio (4³ = 64) rather than squaring it.
MCQ 6Direct ApplicationPractice
An ideal spring stores 8.0 J when stretched by a certain amount. To store 2.0 J, the same spring must be stretched by what fraction of that original displacement?
Show answer and why every option is right or wrong
Answer: A. A is correct. The energy ratio is 2.0/8.0 = ¼, and since U ∝ x², the displacement ratio is √(¼) = ½ (NCERT Class 11 Physics Chapter 5, page 80).
Why B is wrong: B is wrong because it reads the energy ratio of ¼ straight off as the displacement ratio, skipping the square root that the quadratic relation requires (trap: spring PE linear vs quadratic, run backwards).
Why C is wrong: C is wrong because it divides by the energy values rather than taking the square root of their ratio.
Why D is wrong: D is wrong because it squares the energy ratio instead of taking its square root, moving in the wrong direction along the quadratic relation.
MCQ 7Concept TrapPractice
Two ideal springs, of force constants k and 3k, are each stretched by the same displacement x from their natural lengths. The ratio of the energy stored in the stiffer spring to that in the softer one is:
Show answer and why every option is right or wrong
Answer: D. D is correct. In U = ½kx² the dependence on k is linear, so at equal x the energies are in the ratio 3k : k = 3 : 1 (NCERT Class 11 Physics Chapter 5, page 80).
Why A is wrong: A is wrong because it squares the spring-constant ratio; only x is squared in the expression, while k enters to the first power.
Why B is wrong: B is wrong because it inverts the ratio — the stiffer spring stores more energy for the same stretch, not less.
Why C is wrong: C is wrong because it takes a square root of the k-ratio, applying the inverse of the x-relation to a quantity that scales linearly.
MCQ 8CalculationPractice
An ideal spring is stretched from its natural length to a displacement x₀, and then stretched further to 3x₀. The additional energy that must be supplied during the second stage, expressed as a multiple of the energy already stored at x₀, is:
Show answer and why every option is right or wrong
Answer: D. D is correct. U(3x₀) = ½k(3x₀)² = 9 × ½kx₀² = 9U₀, and the energy already present is U₀, so the extra energy supplied is 9U₀ − U₀ = 8U₀ (NCERT Class 11 Physics Chapter 5, page 80).
Why A is wrong: A is wrong because it scales the extra energy with the extra stretch (from x₀ to 3x₀ adds 2x₀), which is the linear-scaling error applied to the increment (trap: spring PE linear vs quadratic).
Why B is wrong: B is wrong because 9U₀ is the total energy at 3x₀, not the additional amount — the U₀ already stored at the start of the second stage was not subtracted.
Why C is wrong: C is wrong because it reports the displacement ratio itself, leaving the quadratic relation unused.
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Potential Energy Spring: quick recall before you leave
How do you solve a Potential Energy Spring question? A worked example
Pattern: spring PE scaling with displacement (the highest-relevance topic-specific PYQ pattern in this dossier; anchored to a 2023 paper).
- 1
Given
An ideal spring stores elastic potential energy U₁ = 5.0 J when stretched by x₁ = 2.0 × 10⁻² m from its natural length. It is then stretched to x₂ = 6.0 × 10⁻² m.
- 2
Required
The elastic potential energy U₂ stored at the new displacement.
- 3
Concept
The elastic potential energy of a spring depends on the square of the displacement from the natural length. A ratio question therefore never needs k: the spring constant cancels between the two states, provided the spring is the same one and stays within its Hooke's-law range.
- 4
Formula
U = ½kx², so for one spring at two displacements:
U₂/U₁ = (x₂/x₁)² - 5
Substitution
U₂ = U₁ × (x₂/x₁)² = 5.0 × (6.0 × 10⁻² / 2.0 × 10⁻²)²
- 6
Calculation
The stretch ratio is (6.0 × 10⁻²)/(2.0 × 10⁻²) = 3.0. Squaring: 3.0² = 9.0.
U₂ = 5.0 × 9.0 = 45 J
The ½ in U = ½kx² is an exact factor from the formula and the exponent 2 is a counting integer; neither contributes to the significant-figure count. The two given quantities carry two significant figures each, so the answer carries two. - 7
Final answer
U₂ = 45 J, i.e. 4.5 × 10¹ J.
- 8
Common trap
The answer 15 J is waiting in the options. It comes from multiplying 5.0 J by the stretch ratio of 3.0 without squaring it — spring PE treated as linear in x. Check the exponent before you multiply: the phrase "three times the stretch" should immediately read as "nine times the energy." The mirror-image slip, 5.0/9.0 ≈ 0.56 J, comes from inverting the ratio and appears when the question is worded as a reduction in stretch.
- 9
Similar NEET-style question
An ideal spring stores 2.0 J of elastic potential energy when compressed by 1.0 × 10⁻² m. The compression is increased to 5.0 × 10⁻² m. Find the energy now stored, and the additional energy that had to be supplied. *(Answers: 50 J stored; 48 J supplied.)*
What to remember before solving Potential Energy Spring questions
Which Potential Energy Spring formulas do you need for NEET?
1 formula — click to collapse
Spring potential energy (Hooke's law regime)
Elastic potential energy stored in an ideal spring of force constant k that has been displaced by x from its natural length. Independent of the sign of x (compression or extension).
| Symbol | Quantity | SI Unit |
|---|---|---|
| U | Elastic PE | J |
| k | Spring constant (force per unit displacement) | N/m |
| x | Displacement from natural length | m |
Valid when
- Spring obeys Hooke's law (F = -k*x) over the displacement range
- No internal damping / hysteresis assumed
Do NOT use when
- Beyond elastic limit (Hooke's law fails)
- Real springs with finite damping (some elastic PE goes to internal energy)
Where do students lose marks on Potential Energy Spring?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
1 item — click to collapse
Category: Overthinking
Student treats spring PE as proportional to displacement (linear) instead of displacement-squared (quadratic). Common error: 'doubling the stretch doubles the PE'. Actual: doubling the stretch gives 4× the PE.
When it triggers
Question gives U at one stretch and asks for U at another. Distractors include linear-scaling answer (×2 instead of ×4 for double stretch).
How to avoid
U = ½ k x² is QUADRATIC. The PE-to-stretch ratio is the SQUARE of the stretch ratio: if stretch goes from x₁ to x₂, U_new / U_old = (x₂/x₁)².
More in Work, Energy and Power: 11 exam traps and mistakes · 9 formulas · 6 question patterns from its other lessons.
Potential Energy Spring questions from past NEET papers
1 question from NEET 2023. Answers verified against NTA official keys. — click to collapse
NEET 2023
14 U
28 U
316 U
42 U
NTA Answer: Option 3(final)
How does NEET ask about Potential Energy Spring?
1 recurring pattern from past papers — click to collapse
Spring PE = 1/2 k x^2; given U at one displacement, find U at another. Quadratic scaling: U(x2)/U(x1) = (x2/x1)^2. Common shape: U at 2 cm = U; find U at 8 cm -> 16 U. Distractors test linear (4U) and inverse (U/16) scaling errors.
Direct ApplicationEasy
Common distractors
linear scaling
Thinking PE scales as x not x^2
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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