This MCQ module is based on: NCERT Exercises and Solutions: Work, Energy and Power
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NCERT Exercises and Solutions: Work, Energy and Power
🎓 Class 11
Physics
CBSE
Theory
Ch 5 – Work, Energy and Power
⏱ ~8 min
🌐 Language: [gtranslate]
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NCERT Exercises and Solutions: Work, Energy and Power
Chapter 5 — Summary
Key Equations & Concepts
- Scalar product: \(\vec{A}\cdot\vec{B} = AB\cos\theta = A_xB_x + A_yB_y + A_zB_z\)
- Work (constant force): \(W = \vec{F}\cdot\vec{d} = Fd\cos\theta\)
- Work (variable force): \(W = \int F(x)\,dx\) — area under F-x curve
- Kinetic energy: \(K = \tfrac{1}{2}mv^2\)
- Work-Energy Theorem: \(W_{\text{net}} = \Delta K\)
- Gravitational PE (near Earth): \(V = mgh\)
- Spring PE: \(V = \tfrac{1}{2}kx^2\)
- Conservation of mechanical energy: \(E = K + V = \text{const}\) (only conservative forces)
- Power (avg): \(P_{\text{avg}} = W/t\); (inst): \(P = \vec{F}\cdot\vec{v}\)
- Elastic 1D (m₂ at rest): \(v_1 = \frac{m_1-m_2}{m_1+m_2}u_1\), \(v_2 = \frac{2m_1}{m_1+m_2}u_1\)
- Perfectly inelastic: \(v = \frac{m_1u_1+m_2u_2}{m_1+m_2}\); KE always lost
| Quantity | SI Unit | Dimensions |
|---|---|---|
| Work / Energy | J (joule) | [ML²T⁻²] |
| Kinetic Energy | J | [ML²T⁻²] |
| Potential Energy | J | [ML²T⁻²] |
| Power | W (watt) = J/s | [ML²T⁻³] |
| Force | N | [MLT⁻²] |
| Spring constant k | N/m | [MT⁻²] |
🎯 Quick Self-Check Calculator
Try a quick KE calculation. Enter mass and speed; the calculator returns kinetic energy and equivalent work needed to bring the body to rest.
K = ½mv² = 250 J
Equivalent: lifting 5.10 m against gravity (g=9.8)
NCERT Exercises — Worked Solutions
Exercise 5.1: Sign of Work
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive, negative or zero:
(a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
(b) work done by gravitational force in the above case.
(c) work done by friction on a body sliding down an inclined plane.
(d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity.
(e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
(a) Positive: Force (upward) and displacement (upward) are in same direction.
(b) Negative: Gravity acts downward; bucket moves upward — opposite directions.
(c) Negative: Friction always opposes motion (180° to displacement).
(d) Positive: Applied force is in direction of motion (otherwise the body could not maintain uniform velocity against friction).
(e) Negative: Air resistance always opposes velocity.
(b) Negative: Gravity acts downward; bucket moves upward — opposite directions.
(c) Negative: Friction always opposes motion (180° to displacement).
(d) Positive: Applied force is in direction of motion (otherwise the body could not maintain uniform velocity against friction).
(e) Negative: Air resistance always opposes velocity.
Exercise 5.2: Work and Friction
A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute (a) work done by the applied force in 10 s, (b) work done by friction in 10 s, (c) work done by the net force on the body in 10 s, (d) change in kinetic energy of the body in 10 s, and interpret your results.
Friction force = μmg = 0.1 × 2 × 9.8 = 1.96 N.
Net force = 7 − 1.96 = 5.04 N. Acceleration a = 5.04/2 = 2.52 m/s².
Distance in 10 s: s = ½at² = ½(2.52)(100) = 126 m. Final speed v = at = 25.2 m/s.
(a) W_applied = 7 × 126 = 882 J.
(b) W_friction = −1.96 × 126 = −246.96 J (≈ −247 J).
(c) W_net = 5.04 × 126 = 635.04 J (≈ 635 J).
(d) ΔK = ½(2)(25.2)² − 0 = 635.04 J. Matches W_net — confirms work-energy theorem.
Net force = 7 − 1.96 = 5.04 N. Acceleration a = 5.04/2 = 2.52 m/s².
Distance in 10 s: s = ½at² = ½(2.52)(100) = 126 m. Final speed v = at = 25.2 m/s.
(a) W_applied = 7 × 126 = 882 J.
(b) W_friction = −1.96 × 126 = −246.96 J (≈ −247 J).
(c) W_net = 5.04 × 126 = 635.04 J (≈ 635 J).
(d) ΔK = ½(2)(25.2)² − 0 = 635.04 J. Matches W_net — confirms work-energy theorem.
Exercise 5.3: Potential Energy from F(x)
Given F(x) for 0 ≤ x ≤ a as F = +F₀ and for a ≤ x ≤ 2a as F = −F₀ (a step force), sketch V(x) given V(0) = 0.
Since F = −dV/dx, V = −∫F dx + const.
For 0 ≤ x ≤ a: V = −F₀x (decreases linearly to V = −F₀a at x = a).
For a ≤ x ≤ 2a: V = −F₀a + F₀(x − a) = F₀(x − 2a) (rises linearly back to V = 0 at x = 2a).
Shape: V is V-shaped with minimum at x = a, value −F₀a; symmetric about x = a.
For 0 ≤ x ≤ a: V = −F₀x (decreases linearly to V = −F₀a at x = a).
For a ≤ x ≤ 2a: V = −F₀a + F₀(x − a) = F₀(x − 2a) (rises linearly back to V = 0 at x = 2a).
Shape: V is V-shaped with minimum at x = a, value −F₀a; symmetric about x = a.
Exercise 5.4: Power & Lifting
An electric motor of power 5 kW is used to lift water out of a 50 m deep well. How much water in litres can it lift in one hour? (g = 9.8 m/s², density of water = 1000 kg/m³)
Energy in 1 hr = 5000 W × 3600 s = 1.8 × 10⁷ J.
For mass m of water, mgh = energy ⇒ m = E/(gh) = 1.8 × 10⁷/(9.8 × 50) ≈ 36 735 kg.
Volume = m/ρ = 36 735/1000 = 36.735 m³ = 36 735 L.
(Realistically, motors are not 100% efficient — actual lift would be ~70% of this.)
For mass m of water, mgh = energy ⇒ m = E/(gh) = 1.8 × 10⁷/(9.8 × 50) ≈ 36 735 kg.
Volume = m/ρ = 36 735/1000 = 36.735 m³ = 36 735 L.
(Realistically, motors are not 100% efficient — actual lift would be ~70% of this.)
Exercise 5.5: Spring & Energy
A spring of force constant 1200 N/m is mounted on a horizontal table. A 3 kg block is pushed against the spring, compressing it by 2.0 cm. Then released. Find the speed of the block when the spring reaches its natural length. (frictionless surface)
Spring PE released = ½kx² = ½(1200)(0.02)² = 0.24 J.
At natural length all of this is KE: ½(3)v² = 0.24 ⇒ v² = 0.16 ⇒ v = 0.40 m/s.
At natural length all of this is KE: ½(3)v² = 0.24 ⇒ v² = 0.16 ⇒ v = 0.40 m/s.
Exercise 5.6: Conservative Field
State if each of the following statements is true or false. Give reasons.
(a) The total energy of a system is always conserved if all forces are conservative.
(b) Work done by a conservative force on a particle in a closed loop is always zero.
(c) Total mechanical energy of a system is conserved if external and conservative forces act on it.
(d) In an inelastic collision, the final KE is always less than initial KE.
(a) True. Conservative forces preserve K + V.
(b) True. By definition of conservative force.
(c) Partially true. Conservative forces do; external forces may add or remove energy. Strictly the statement as given would require external forces to do net zero work.
(d) True. Inelastic by definition has KE loss.
(b) True. By definition of conservative force.
(c) Partially true. Conservative forces do; external forces may add or remove energy. Strictly the statement as given would require external forces to do net zero work.
(d) True. Inelastic by definition has KE loss.
Exercise 5.7: Work in Various Cases
State if each is true or false:
(a) Work done by static friction is always zero.
(b) The kinetic energy of a body is always non-negative.
(c) An object can have zero velocity and still have non-zero acceleration.
(d) Two satellites of equal mass orbiting at different altitudes have the same KE.
(a) False. Static friction can do work (e.g., on a person stepping forward — the friction force on the foot is in the direction of motion of the body, so positive work is transferred to the body).
(b) True. K = ½mv² ≥ 0.
(c) True. A ball at the top of its toss has v = 0 but a = g.
(d) False. KE depends on orbital speed, which decreases with altitude (for circular orbits, v² ∝ 1/r).
(b) True. K = ½mv² ≥ 0.
(c) True. A ball at the top of its toss has v = 0 but a = g.
(d) False. KE depends on orbital speed, which decreases with altitude (for circular orbits, v² ∝ 1/r).
Exercise 5.8: Variable-Force Work
A particle of mass 0.5 kg is acted on by a force F(x) = (3 + 2x) N along the +x axis, where x is in metres. Find the work done from x = 0 to x = 4 m, and the speed at x = 4 m if it started from rest.
\[W = \int_0^4 (3 + 2x)\,dx = \left[3x + x^2\right]_0^4 = 12 + 16 = 28\text{ J}\]
By work-energy theorem: ½(0.5)v² = 28 ⇒ v² = 112 ⇒ v ≈ 10.58 m/s.
Exercise 5.9: Elastic Collision Mass Ratio
A 4 kg ball moving at 10 m/s collides elastically head-on with a 6 kg ball at rest. Find post-collision velocities and KE of each ball.
\[v_1 = \frac{4-6}{10}(10) = -2\text{ m/s}\]
\[v_2 = \frac{2 \times 4}{10}(10) = 8\text{ m/s}\]
KE of m₁ after = ½(4)(4) = 8 J. KE of m₂ after = ½(6)(64) = 192 J. Total = 200 J. Check: initial KE = ½(4)(100) = 200 J. ✓
🔄 Concept Check — Energy Pathways
L4 Analyse
Trace the energy path in each scenario: identify whether energy is converted between K, V, or lost as heat/light.
- A pendulum swinging in air gradually loses amplitude.
- A coin spinning on a table eventually stops.
- A bouncing ball rebounds to lower height each time.
- An orbiting satellite has constant total mechanical energy.
Predict: What energy form is the final destination of all dissipation?
- Pendulum: K↔V cycles, but air drag converts mechanical → thermal energy of air.
- Coin: Rotational K → heat (friction with table) + sound.
- Ball: Each bounce converts some KE to internal heat/sound; height decreases as energy budget shrinks.
- Satellite: Constant E because gravity is conservative and air drag is negligible at orbital altitudes.
The "graveyard" of all dissipation is thermal energy (random molecular motion) — the second law of thermodynamics in action.
🎯 Competency-Based Questions
A 2.0 kg block on a horizontal surface (μ = 0.2) is given an initial speed of 5.0 m/s. It slides freely until coming to rest. (g = 10 m/s²)
Q1. What kind of force is friction here?L1 Remember
Answer: (b). Friction is non-conservative — work depends on path length, mechanical energy is lost as heat.
Q2. Distance travelled before stopping?L3 Apply
Answer: KE = friction work ⇒ ½(2)(5)² = μmg·d ⇒ 25 = 0.2 × 2 × 10 × d ⇒ d = 6.25 m.
Q3. Heat generated by friction over this distance?L3 Apply
Answer: All initial KE → heat = 25 J.
Q4. State whether TRUE or FALSE: "If the surface were frictionless and there were no air drag, the block would slide forever at 5 m/s." L5 Evaluate
Answer: TRUE. By Newton's first law and energy conservation, with no horizontal force the block would maintain its velocity indefinitely. This is why friction-free experiments (e.g., on air tables) are important pedagogical tools.
Q5. HOT: The same block is fired at 5 m/s up a 30° incline (same μ). Will it travel further or shorter than 6.25 m? Justify quantitatively. L6 Create
Answer: On incline both gravity component (mg·sin θ) AND friction (μmg·cos θ) decelerate the block. Total decel-force/mass = g·sin30° + μg·cos30° = 5 + 0.2 × 10 × 0.866 = 6.732 m/s². Distance = v²/(2a) = 25/13.46 ≈ 1.86 m — much shorter than 6.25 m on flat ground because gravity now joins friction in opposing motion.
🧠 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): Energy can never be created or destroyed.
Reason (R): The total energy of an isolated system is conserved (first law of thermodynamics).
Answer: (A). Both true and R explains A.
Assertion (A): A heavy car has more KE than a light bicycle even at the same speed.
Reason (R): KE is directly proportional to mass.
Answer: (A). Both true; R explains A. K = ½mv² is linear in m at fixed v.
Assertion (A): When a body is in equilibrium, no work is done on it.
Reason (R): Net force on a body in equilibrium is zero.
Answer: (D). Assertion is FALSE — even in equilibrium INDIVIDUAL forces may do work (e.g., a body moving with constant velocity: applied force does +W, friction does −W, net 0). Reason TRUE — net force is zero ⇒ NET work zero. Subtle distinction.
Frequently Asked Questions - NCERT Exercises and Solutions: Work, Energy and Power
What are the key NCERT exercise types in Chapter 5 Work, Energy and Power?
NCERT Class 11 Physics Chapter 5 Work, Energy and Power exercises cover conceptual MCQs, short numerical problems (2-3 marks), long numerical derivations (5 marks), and assertion-reason questions. The MyAiSchool solution set provides step-by-step worked solutions for every NCERT exercise question in the chapter, aligned with the CBSE board exam pattern. Students should focus on dimensional analysis, formula application, and unit consistency to score full marks.
How should students approach numerical problems in Work, Energy and Power?
For numerical problems in NCERT Class 11 Physics Chapter 5 Work, Energy and Power: (1) list all given data with units, (2) write the relevant formula(e), (3) substitute values carefully, (4) calculate with proper significant figures, (5) state the final answer with the correct unit. Always draw a free-body or schematic diagram where relevant. The MyAiSchool solutions follow this 5-step CBSE-aligned format consistently.
What are the most-asked CBSE board questions from Chapter 5?
From NCERT Class 11 Physics Chapter 5 (Work, Energy and Power), the most-asked CBSE board questions test conceptual understanding of fundamental principles, application of derived formulas to standard scenarios, and proof-style derivations. 5-mark questions usually combine derivation + application. The MyAiSchool exercise set tags each question by board frequency so students can prioritize high-yield problems before exams.
How do I check the dimensional correctness of my answer?
Dimensional correctness in NCERT Class 11 Physics is verified by ensuring the LHS and RHS of any equation have the same dimensional formula (M^a L^b T^c). For Chapter 5 Work, Energy and Power problems, write the dimensions of each quantity, substitute, and simplify. If dimensions match the equation is dimensionally valid (necessary but not sufficient). The MyAiSchool solutions include dimensional checks in every numerical answer.
What are common mistakes students make in Chapter 5 exercises?
Common mistakes in NCERT Class 11 Physics Chapter 5 Work, Energy and Power exercises include: (1) unit conversion errors (CGS vs SI), (2) sign convention mistakes for vectors, (3) forgetting to consider all forces in free-body diagrams, (4) algebra errors during derivations, and (5) misreading the problem statement. The MyAiSchool solutions highlight these traps with red-flag annotations so students learn to avoid them.
How does the MyAiSchool solution differ from other NCERT solution sets?
MyAiSchool Class 11 Physics Chapter 5 Work, Energy and Power solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with reasoning, dimensional checks, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Each solution is verified against NCERT and CBSE marking schemes for accuracy.
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