The Elastic Potential Energy Formula gives the energy stored in a stretched or compressed elastic object, expressed as \ ( U = \frac{1}{2}kx^2 \), where k is the spring constant and x is the displacement from equilibrium. This formula is a core concept in Class 11 Physics (NCERT Chapter 6 — Work, Energy and Power) and appears regularly in CBSE board exams as well as JEE Main, JEE Advanced, and NEET. In this article, we cover the complete formula, its derivation, a full physics formula sheet, three progressive solved examples, CBSE exam tips for 2025-26, common mistakes, and JEE/NEET application patterns.
Key Elastic Potential Energy Formulas at a Glance
Quick reference for the most important elastic potential energy formulas.
- Elastic PE: \( U = \frac{1}{2}kx^2 \)
- Spring Force (Hooke's Law): \( F = -kx \)
- Spring Constant: \( k = \frac{F}{x} \)
- Work done by spring: \( W = -\frac{1}{2}kx^2 \)
- Total Mechanical Energy (SHM): \( E = \frac{1}{2}kA^2 \)
- Energy at position x (SHM): \( KE = \frac{1}{2}k(A^2 – x^2) \)
- Series springs: \( \frac{1}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2} \)
What is the Elastic Potential Energy Formula?
The Elastic Potential Energy Formula quantifies the energy stored in any elastic body — most commonly a spring — when it is deformed from its natural length. When you stretch or compress a spring by a distance x, work is done against the restoring force. This work is stored as potential energy inside the spring. The moment the spring is released, this stored energy converts into kinetic energy.
According to NCERT Class 11 Physics, Chapter 6 (Work, Energy and Power), elastic potential energy is a form of mechanical potential energy. It obeys the principle of conservation of energy. The formula is derived directly from Hooke's Law, which states that the restoring force in a spring is proportional to its displacement. The Elastic Potential Energy Formula is fundamental not only for springs but also for understanding molecular bonds, rubber bands, and any elastic deformation in engineering and biology.
The standard expression is:
\[ U = \frac{1}{2}kx^2 \]
Here, U is the elastic potential energy in Joules, k is the spring constant in N/m, and x is the displacement in metres.
Elastic Potential Energy Formula — Expression and Variables
The complete expression for elastic potential energy is:
\[ U = \frac{1}{2}kx^2 \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( U \) | Elastic Potential Energy | Joule (J) |
| \( k \) | Spring Constant (stiffness) | Newton per metre (N/m) |
| \( x \) | Displacement from equilibrium | Metre (m) |
| \( F \) | Restoring Force (Hooke's Law) | Newton (N) |
| \( A \) | Amplitude (in SHM context) | Metre (m) |
| \( W \) | Work done on/by spring | Joule (J) |
Derivation of the Elastic Potential Energy Formula
The derivation starts with Hooke's Law. The restoring force of a spring is \( F = -kx \). The negative sign indicates the force acts opposite to displacement. To stretch the spring by a small element \( dx \), the external applied force must equal \( +kx \).
The work done to stretch the spring from 0 to x is found by integration:
\[ W = \int_0^x kx\, dx = \frac{1}{2}kx^2 \]
This work is stored entirely as elastic potential energy. Therefore:
\[ U = \frac{1}{2}kx^2 \]
The same result applies for compression. The displacement x is always squared, so U is always positive regardless of the direction of deformation. This derivation is directly aligned with NCERT Class 11, Chapter 6, and is a standard derivation asked in CBSE board exams.
Complete Physics Formula Sheet — Springs and Elastic Energy
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Elastic Potential Energy | \( U = \frac{1}{2}kx^2 \) | k = spring constant, x = displacement | J | Class 11, Ch 6 |
| Hooke's Law | \( F = -kx \) | F = restoring force, k = spring constant, x = displacement | N | Class 11, Ch 6 |
| Spring Constant | \( k = \frac{F}{x} \) | F = applied force, x = extension/compression | N/m | Class 11, Ch 6 |
| Work Done by Spring | \( W = -\frac{1}{2}kx^2 \) | k = spring constant, x = displacement | J | Class 11, Ch 6 |
| Total Mechanical Energy (SHM) | \( E = \frac{1}{2}kA^2 \) | k = spring constant, A = amplitude | J | Class 11, Ch 14 |
| KE in SHM at position x | \( KE = \frac{1}{2}k(A^2 – x^2) \) | A = amplitude, x = displacement | J | Class 11, Ch 14 |
| Springs in Series | \( \frac{1}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2} \) | k1, k2 = individual spring constants | N/m | Class 11, Ch 6 |
| Springs in Parallel | \( k_{eff} = k_1 + k_2 \) | k1, k2 = individual spring constants | N/m | Class 11, Ch 6 |
| Gravitational Potential Energy | \( U_g = mgh \) | m = mass, g = 9.8 m/s², h = height | J | Class 11, Ch 6 |
| Kinetic Energy | \( KE = \frac{1}{2}mv^2 \) | m = mass, v = velocity | J | Class 11, Ch 6 |
| Conservation of Energy (spring-mass) | \( \frac{1}{2}kx^2 = \frac{1}{2}mv^2 \) | k = spring constant, x = compression, m = mass, v = velocity | J | Class 11, Ch 6 |
Elastic Potential Energy Formula — Solved Examples
Example 1 (Class 9-10 Level — Direct Application)
Problem: A spring with a spring constant of 200 N/m is compressed by 0.1 m from its natural length. Calculate the elastic potential energy stored in the spring.
Given:
- Spring constant, \( k = 200 \) N/m
- Compression, \( x = 0.1 \) m
Step 1: Write the Elastic Potential Energy Formula: \( U = \frac{1}{2}kx^2 \)
Step 2: Substitute the given values:
\[ U = \frac{1}{2} \times 200 \times (0.1)^2 \]
Step 3: Calculate step by step:
\[ U = \frac{1}{2} \times 200 \times 0.01 = 100 \times 0.01 = 1 \text{ J} \]
Answer
The elastic potential energy stored in the spring is 1 Joule.
Example 2 (Class 11-12 Level — Conservation of Energy)
Problem: A block of mass 0.5 kg is attached to a spring (k = 800 N/m) on a frictionless surface. The spring is compressed by 0.2 m and then released. Find the maximum velocity of the block when it passes through the equilibrium position.
Given:
- Mass, \( m = 0.5 \) kg
- Spring constant, \( k = 800 \) N/m
- Compression, \( x = 0.2 \) m
Step 1: At equilibrium, all elastic PE converts to kinetic energy:
\[ \frac{1}{2}kx^2 = \frac{1}{2}mv^2 \]
Step 2: Cancel \( \frac{1}{2} \) from both sides and solve for \( v \):
\[ v = x\sqrt{\frac{k}{m}} \]
Step 3: Substitute values:
\[ v = 0.2 \times \sqrt{\frac{800}{0.5}} = 0.2 \times \sqrt{1600} = 0.2 \times 40 = 8 \text{ m/s} \]
Answer
The maximum velocity of the block at the equilibrium position is 8 m/s.
Example 3 (JEE/NEET Level — Two Springs in Series)
Problem: Two springs with spring constants \( k_1 = 300 \) N/m and \( k_2 = 600 \) N/m are connected in series. A force of 30 N is applied to stretch the combination. Calculate (a) the effective spring constant, (b) the total extension, and (c) the total elastic potential energy stored.
Given:
- \( k_1 = 300 \) N/m, \( k_2 = 600 \) N/m
- Applied force, \( F = 30 \) N
Step 1: Find the effective spring constant for series combination:
\[ \frac{1}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2} = \frac{1}{300} + \frac{1}{600} = \frac{2}{600} + \frac{1}{600} = \frac{3}{600} \]
\[ k_{eff} = \frac{600}{3} = 200 \text{ N/m} \]
Step 2: Find the total extension using Hooke's Law:
\[ x = \frac{F}{k_{eff}} = \frac{30}{200} = 0.15 \text{ m} \]
Step 3: Apply the Elastic Potential Energy Formula:
\[ U = \frac{1}{2}k_{eff}x^2 = \frac{1}{2} \times 200 \times (0.15)^2 \]
\[ U = 100 \times 0.0225 = 2.25 \text{ J} \]
Answer
(a) Effective spring constant = 200 N/m
(b) Total extension = 0.15 m
(c) Total elastic potential energy stored = 2.25 J
CBSE Exam Tips 2025-26
- Memorise the formula with its derivation. CBSE 2025-26 board papers frequently ask for the derivation of \( U = \frac{1}{2}kx^2 \) as a 3-mark question. Write each integration step clearly.
- Always state units. We recommend writing “k in N/m, x in m, U in J” alongside every substitution. Forgetting units costs half marks in CBSE.
- Link elastic PE to conservation of energy. Many 5-mark problems combine elastic PE with kinetic energy. Set up \( \frac{1}{2}kx^2 = \frac{1}{2}mv^2 \) confidently.
- Know series vs. parallel spring combinations. The 2025-26 CBSE syllabus includes application-based questions on equivalent spring constants. Practice both configurations.
- Sketch the energy graph. The graph of \( U \) versus \( x \) is a parabola. Drawing it earns presentation marks and helps you verify your answer's sign and shape.
- Use dimensional analysis. Verify that \( [k][x^2] = \text{N/m} \times \text{m}^2 = \text{N} \cdot \text{m} = \text{J} \). This quick check prevents formula errors in exams.
Common Mistakes to Avoid
| Mistake | Why It Happens | Correct Approach |
|---|---|---|
| Writing \( U = kx^2 \) without the \( \frac{1}{2} \) | Confusing the force formula \( F = kx \) with the energy formula | Always include the factor of \( \frac{1}{2} \). Energy is the integral of force, not force itself. |
| Using displacement in cm instead of metres | Reading problem data carelessly | Convert all lengths to metres before substituting. 1 cm = 0.01 m. |
| Treating elastic PE as negative for compression | Confusing with the negative sign in Hooke's Law \( F = -kx \) | Elastic PE is always positive. \( x^2 \) is always positive regardless of compression or extension. |
| Adding spring constants directly for series combinations | Mixing up series and parallel rules | For series: use the reciprocal formula. For parallel: add directly. Remember — series gives a softer (smaller k) spring. |
| Ignoring the natural length when finding x | Using total length instead of extension/compression | Displacement \( x \) is always measured from the equilibrium (natural) position, not from a fixed wall or reference. |
JEE/NEET Application of the Elastic Potential Energy Formula
In our experience, JEE aspirants encounter the Elastic Potential Energy Formula in at least 2-3 questions per paper, spanning mechanics, SHM, and collision problems. Here are the most common application patterns:
Pattern 1 — Spring-Mass System and Maximum Velocity
JEE Main frequently tests the conversion of elastic PE to kinetic energy. A block compresses a spring, and you must find the block's velocity at the natural length. The key equation is \( \frac{1}{2}kx^2 = \frac{1}{2}mv^2 \). This gives \( v_{max} = x\sqrt{k/m} \), which is also the amplitude times angular frequency in SHM. Recognising this dual interpretation saves time in JEE Advanced multi-correct questions.
Pattern 2 — Elastic Collision via Spring Compression
JEE Advanced often models elastic collisions using a spring between two masses. At maximum compression, both masses move with the same velocity. You apply both conservation of momentum and conservation of energy (including \( \frac{1}{2}kx^2 \)) simultaneously. NEET also tests this in the context of molecular interactions modelled as spring forces.
Pattern 3 — SHM Energy Distribution
In Simple Harmonic Motion questions (Class 11, Chapter 14), the total energy is \( E = \frac{1}{2}kA^2 \). At position \( x \), elastic PE is \( \frac{1}{2}kx^2 \) and kinetic energy is \( \frac{1}{2}k(A^2 – x^2) \). JEE questions ask for the position where KE equals PE. Setting \( \frac{1}{2}kx^2 = \frac{1}{2}k(A^2 – x^2) \) gives \( x = \frac{A}{\sqrt{2}} \). Our experts suggest memorising this result directly to save 90 seconds in the exam hall.
Pattern 4 — NEET Biological Spring Analogy
NEET occasionally applies Hooke's Law to biological structures such as tendons and cartilage. The Elastic Potential Energy Formula is used to calculate energy stored in deformed biological tissues. The approach is identical — identify k (tissue stiffness) and x (deformation), then apply \( U = \frac{1}{2}kx^2 \).
FAQs on Elastic Potential Energy Formula
We hope this comprehensive guide on the Elastic Potential Energy Formula has strengthened your understanding for both CBSE board exams and competitive entrance tests. For further reading, explore our detailed articles on the Electric Field Formula, the Uniform Circular Motion Formula, and the Capacitance Formula. You can also browse our complete Physics Formulas hub for Class 11 and Class 12. For official NCERT textbook references, visit ncert.nic.in.