This MCQ module is based on: Potential Energy Conservation
TOPIC 21 OF 33
Potential Energy Conservation
🎓 Class 11
Physics
CBSE
Theory
Ch 5 – Work, Energy and Power
⏱ ~14 min
🌐 Language: [gtranslate]
🧠 AI-Powered MCQ Assessment
▲
📝 Worksheet / Assessment
▲
This assessment will be based on: Potential Energy Conservation
Upload images, PDFs, or Word documents to include their content in assessment generation.
Potential Energy Conservation
5.9 The Concept of Potential Energy
The word potential means stored, capacity to do work. A stretched bow stores energy ready to launch an arrow; a raised stone stores energy that becomes kinetic on falling; a wound spring drives a clock. The energy stored by virtue of position or configuration is called potential energy.
5.9.1 Gravitational Potential Energy (near Earth)
If we lift a body of mass \(m\) through a height \(h\) at constant velocity, the external force \(mg\) (upward) does work \(mgh\). This work is "stored" — let go and gravity will return it as kinetic energy. We define:
\[V(h) = mgh \quad \text{(measuring height from a reference level where } V = 0)\]
The reference level is arbitrary — only differences in PE have physical meaning. Setting \(V = 0\) at the floor, ceiling, or sea-level merely shifts numerical values without affecting forces or motion.
5.10 Conservative & Non-Conservative Forces
A force is called conservative if the work it does on a particle moving between two fixed endpoints does not depend on the path taken. Equivalently, the work in any closed loop is zero. For a conservative force, we can define a potential-energy function \(V(x)\) such that:
\[F = -\frac{dV}{dx} \quad \text{(in 1D)}\]
| Force | Conservative? | PE function (1D) |
|---|---|---|
| Gravity (uniform, near Earth) | Yes | V = mgh |
| Spring force F = −kx | Yes | V = ½kx² |
| Electrostatic (Coulomb) | Yes | V = kq₁q₂/r |
| Friction | No | None |
| Air drag | No | None |
| Tension (often) | Constraint force; W = 0 | — |
5.11 Conservation of Mechanical Energy
If only conservative forces act on a body, and we define mechanical energy \(E = K + V\), then:
Principle of Conservation of Mechanical Energy:
\[\boxed{\;K_i + V_i = K_f + V_f \;\Leftrightarrow\; E = \tfrac{1}{2}m v^2 + V(x) = \text{constant}\;}\]
Proof outline: Work-energy theorem says \(W_{\text{net}} = \Delta K\). For conservative force, \(W = -\Delta V\). Combining: \(\Delta K = -\Delta V\) ⇒ \(\Delta(K+V) = 0\) ⇒ \(K+V\) constant.
5.12 The Potential Energy of a Spring
For an ideal spring obeying Hooke's law, \(F = -k x\) where \(x\) is displacement from natural length. The work done by the spring as the block moves from 0 to x is:
\[W_{\text{spring}} = \int_0^x (-k x')\,dx' = -\tfrac{1}{2}k x^2\]
Hence the spring potential energy stored at displacement \(x\) is:
\[V(x) = \tfrac{1}{2}k x^2\]
This is symmetric about \(x = 0\) (natural length), and grows quadratically with stretch or compression.
🎯 Interactive Simulation: Conservation of Mechanical Energy
A 1 kg ball is dropped from height H. Drag the slider to choose its current height and watch how KE and PE share the constant total. (g = 10 m/s²)
PE = 50 J | KE = 50 J | Total = 100 J
Example 5.7: Pendulum Energy
A simple pendulum of length 1 m is pulled aside until it makes 30° with the vertical, then released. Find its speed at the lowest point. (g = 10 m/s²)
Height risen above lowest point: \(h = L(1 - \cos 30°) = 1 \times (1 - 0.866) = 0.134\) m.
Conservation of mechanical energy: \[mgh = \tfrac{1}{2}m v^2 \;\Rightarrow\; v = \sqrt{2gh} = \sqrt{2 \times 10 \times 0.134} \approx \boxed{1.64\text{ m/s}}\] Mass cancels — speed is independent of pendulum mass.
Conservation of mechanical energy: \[mgh = \tfrac{1}{2}m v^2 \;\Rightarrow\; v = \sqrt{2gh} = \sqrt{2 \times 10 \times 0.134} \approx \boxed{1.64\text{ m/s}}\] Mass cancels — speed is independent of pendulum mass.
Example 5.8: Spring Compression
A 0.5 kg block sliding on frictionless ice at 2 m/s strikes a spring of force constant 200 N/m. Find the maximum compression of the spring.
At maximum compression, block is momentarily at rest — all KE has become spring PE.
\[\tfrac{1}{2}m v^2 = \tfrac{1}{2}k x_{\max}^2\]
\[x_{\max} = v\sqrt{m/k} = 2 \times \sqrt{0.5/200} = 2 \times 0.05 = \boxed{0.10\text{ m}}\]
Example 5.9: Friction Reduces Mechanical Energy
A 2 kg block slides down a 30° incline of length 5 m. The coefficient of kinetic friction is 0.2. Find the speed at the bottom. (g = 10 m/s²)
Height drop: h = 5·sin 30° = 2.5 m. PE released = mgh = 2 × 10 × 2.5 = 50 J.
Friction force = μmg cos θ = 0.2 × 2 × 10 × cos 30° = 3.464 N.
Work by friction = −3.464 × 5 = −17.32 J.
By work-energy theorem (or energy balance): \[\tfrac{1}{2}m v^2 = 50 - 17.32 = 32.68\text{ J}\] \[v = \sqrt{2 \times 32.68/2} = \sqrt{32.68} \approx \boxed{5.72\text{ m/s}}\] With friction, mechanical energy is NOT conserved — some converts to heat.
Friction force = μmg cos θ = 0.2 × 2 × 10 × cos 30° = 3.464 N.
Work by friction = −3.464 × 5 = −17.32 J.
By work-energy theorem (or energy balance): \[\tfrac{1}{2}m v^2 = 50 - 17.32 = 32.68\text{ J}\] \[v = \sqrt{2 \times 32.68/2} = \sqrt{32.68} \approx \boxed{5.72\text{ m/s}}\] With friction, mechanical energy is NOT conserved — some converts to heat.
🎢 Activity 5.3 — Roller-Coaster Energy
L4 Analyse
Materials: Flexible plastic track, marble, ruler, stopwatch (or video), bookstand to set heights.
Procedure:
- Set up the track as a roller-coaster — first hill height H₁, second hill H₂ < H₁.
- Release a marble from H₁; record whether it makes it over H₂.
- Now make H₂ > H₁; the marble will not make it. Why?
- Try with a damp track (more friction); even an H₂ < H₁ will eventually fail.
Predict: What is the maximum height the marble can reach starting from H₁?
Observation: In the frictionless ideal case, the marble can reach a maximum height equal to H₁. Any second hill higher than H₁ requires more PE than the marble has — impossible.
Conclusion: Energy conservation predicts the maximum reachable height directly. Friction lowers the effective reachable height because it converts mechanical energy into heat. Real roller-coasters are designed so each successive peak is LOWER than the previous to allow for friction losses.
🎯 Competency-Based Questions
A child of mass 30 kg slides from rest down a smooth slide of vertical drop 4 m. At the bottom, she encounters a 2 m horizontal stretch of rough surface (μ = 0.3) before hitting a spring of force constant 1500 N/m. (g = 10 m/s²)
Q1. Find the child's speed at the bottom of the (smooth) slide.L3 Apply
Answer: (c). v = √(2gh) = √(2 × 10 × 4) = √80 ≈ 8.94 m/s.
Q2. Calculate the speed of the child just before the spring is touched. L4 Analyse
Answer: Friction work = −μmg·d = −0.3 × 30 × 10 × 2 = −180 J. KE at start of rough patch = ½(30)(8.94)² = 1200 J. KE just before spring = 1200 − 180 = 1020 J. v = √(2 × 1020/30) = √68 ≈ 8.25 m/s.
Q3. Find maximum spring compression. (Assume rough patch ends right at the spring.) L4 Analyse
Answer: KE at spring = 1020 J. Max compression: ½kx² = 1020 ⇒ x² = 2040/1500 = 1.36 ⇒ x ≈ 1.17 m.
Q4. State whether TRUE or FALSE: "The slide must lose mechanical energy because the surface is curved." Justify. L5 Evaluate
Answer: FALSE. Curvature alone does NOT cause energy loss. The normal reaction is always perpendicular to velocity, so it does zero work. Only the FRICTION on the slide loses energy. A perfectly smooth curved slide preserves mechanical energy.
Q5. HOT: Design a height-from-acceleration measurement using only conservation of mechanical energy and a fixed spring stiffness sensor at the bottom. Explain. L6 Create
Sample design: Drop a known-mass ball from unknown height onto a calibrated spring at the bottom. Measure maximum compression x. Then mgh = ½kx² gives h = kx²/(2mg). Spring stiffness k is known; m, g are known; x is measured — h is determined. Used in arrest barriers and impact testers.
🧠 Assertion–Reason Questions
Choose: (A) Both true, R explains A. (B) Both true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
Assertion (A): The mechanical energy of a swinging pendulum is conserved (assuming no air drag).
Reason (R): Tension does no work because it is perpendicular to velocity.
Answer: (A). Both true and R explains A. Only gravity (conservative) does work, so K+V is constant.
Assertion (A): Potential energy can be negative.
Reason (R): The reference level for potential energy is arbitrary.
Answer: (A). Both true and R explains A. Choosing the reference level above the body makes V negative, but only differences ΔV are physically meaningful.
Assertion (A): Friction is a conservative force.
Reason (R): Friction always converts mechanical energy into heat.
Answer: (D). Assertion FALSE — friction is non-conservative (work depends on path length). Reason TRUE — friction always dissipates energy as heat.
Frequently Asked Questions - Potential Energy Conservation
What is the main concept covered in Potential Energy Conservation?
In NCERT Class 11 Physics Chapter 5 (Work, Energy and Power), "Potential Energy Conservation" covers core principles and equations needed for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Potential Energy Conservation useful in real-life applications?
Real-life applications of Potential Energy Conservation from NCERT Class 11 Physics Chapter 5 include engineering design, satellite mechanics, sports biomechanics, transportation safety, and electrical/electronic devices. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Potential Energy Conservation?
Key formulas in Potential Energy Conservation (NCERT Class 11 Physics Chapter 5 Work, Energy and Power) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 5?
NCERT Class 11 Physics Chapter 5 (Work, Energy and Power) is structured so each part builds on the previous one. Potential Energy Conservation connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Potential Energy Conservation?
CBSE board questions from Potential Energy Conservation typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the Potential Energy Conservation lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
AI Tutor
Physics Class 11 Part I – NCERT (2025-26)
Ready