This MCQ module is based on: Introduction Work
TOPIC 19 OF 33
Introduction Work
🎓 Class 11
Physics
CBSE
Theory
Ch 5 – Work, Energy and Power
⏱ ~14 min
🌐 Language: [gtranslate]
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Introduction Work
5.1 Introduction
The terms work, energy, and power are commonplace in everyday life, but their meanings in physics are precise and sometimes surprising. A man pushing a wall but failing to move it does NO physical work, yet feels tired. A weight-lifter holding a barbell motionless overhead does no work either, despite exhaustion. We must therefore reserve these words for tightly defined quantities.
This chapter develops the scalar product of two vectors (since work is a scalar quantity even though force and displacement are vectors), then defines work, kinetic energy, and the work–energy theorem — a cornerstone result that links the integrated effect of force on motion (work) to the change in motion (kinetic energy).
5.2 Notions of Work and Kinetic Energy: The Work-Energy Theorem
For one-dimensional motion under a constant acceleration \(a\) and starting velocity \(u\), the kinematic equation is:
\[v^2 - u^2 = 2as\]
Multiplying both sides by \(m/2\):
\[\tfrac{1}{2}mv^2 - \tfrac{1}{2}mu^2 = mas = Fs\]
where \(F = ma\) is the net force. We recognise the left side as the change in kinetic energy \(K\), and the right side as work \(W = Fs\) done by the net force. Hence \(K_f - K_i = W_{\text{net}}\) — the work-energy theorem.
5.3 Notions of Work
If a constant force \(\vec{F}\) acts on an object that undergoes a displacement \(\vec{d}\), the work done is the component of force along the displacement multiplied by the magnitude of displacement. Hence the natural mathematical tool is the scalar (dot) product of vectors.
5.4 Scalar Product
The scalar product of two vectors \(\vec{A}\) and \(\vec{B}\) is defined as:
\[\vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}|\cos\theta\]
where \(\theta\) is the angle between \(\vec{A}\) and \(\vec{B}\) when their tails coincide.
Key Properties of the Scalar Product:
- Commutative: \(\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}\)
- Distributive: \(\vec{A} \cdot (\vec{B} + \vec{C}) = \vec{A} \cdot \vec{B} + \vec{A} \cdot \vec{C}\)
- \(\vec{A} \cdot \vec{A} = |\vec{A}|^2\)
- If \(\vec{A} \perp \vec{B}\), then \(\vec{A} \cdot \vec{B} = 0\) (angle 90°, cos 90° = 0)
- Unit vectors: \(\hat{i}\cdot\hat{i}=\hat{j}\cdot\hat{j}=\hat{k}\cdot\hat{k}=1\); \(\hat{i}\cdot\hat{j}=\hat{j}\cdot\hat{k}=\hat{k}\cdot\hat{i}=0\)
If \(\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}\) and \(\vec{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k}\), then expanding using distributivity:
\[\vec{A} \cdot \vec{B} = A_xB_x + A_yB_y + A_zB_z\]
5.5 Work Done by a Constant Force
If a constant force \(\vec{F}\) produces a displacement \(\vec{d}\), the work done is:
\[W = \vec{F} \cdot \vec{d} = F\,d\,\cos\theta\]
where \(\theta\) is the angle between force and displacement.
Three cases of work:
- Positive work (0° ≤ θ < 90°): force has a component along the motion. Example: gravity on a falling stone.
- Zero work (θ = 90°): force perpendicular to motion. Example: gravity on horizontal motion; centripetal force on a circling body.
- Negative work (90° < θ ≤ 180°): force opposes motion. Example: friction on a sliding block; gravity on a stone thrown upward.
| Situation | θ (force vs displacement) | Sign of W |
|---|---|---|
| Pulling a sled with a rope angled upward | Acute (0° < θ < 90°) | Positive |
| Carrying a bag horizontally (gravity vs motion) | 90° | Zero |
| Friction on a sliding box | 180° | Negative |
| Earth's pull on an Earth-orbiting satellite | 90° | Zero |
| Engine pulling a train forward | 0° | Positive (max) |
5.6 Work Done by a Variable Force
When the force changes with position, we cannot simply multiply force by displacement. Instead, we divide the path into infinitesimal pieces \(dx\) so small that \(F(x)\) is essentially constant over each piece, then sum (integrate):
\[W = \int_{x_i}^{x_f} F(x)\,dx\]
Geometrically, the work done is the area under the F–x curve between \(x = x_i\) and \(x = x_f\).
🎯 Interactive Simulation: Work Calculator (Constant Force)
Adjust force magnitude, displacement, and the angle between them. Watch how the sign and value of work change.
W = F·d·cos θ
43.30 J
Positive work — force aids motion
Example 5.1: Work by Constant Force
A box is pushed along a horizontal floor by a force of 100 N at 60° above the horizontal. The box moves 5 m. Find the work done by the applied force and by gravity.
Given: F = 100 N, d = 5 m, θ = 60°, m·g acts vertically downward, motion is horizontal.
Work by applied force: \[W_F = F\,d\cos 60° = 100 \times 5 \times 0.5 = \boxed{250\text{ J}}\] Work by gravity: Since gravity is perpendicular to horizontal displacement, θ = 90°, cos 90° = 0: \[W_g = 0\] The normal reaction also does zero work (perpendicular to motion).
Work by applied force: \[W_F = F\,d\cos 60° = 100 \times 5 \times 0.5 = \boxed{250\text{ J}}\] Work by gravity: Since gravity is perpendicular to horizontal displacement, θ = 90°, cos 90° = 0: \[W_g = 0\] The normal reaction also does zero work (perpendicular to motion).
Example 5.2: Scalar Product in Components
Find \(\vec{A}\cdot\vec{B}\) for \(\vec{A} = 3\hat{i} + 4\hat{j} - 2\hat{k}\) and \(\vec{B} = 2\hat{i} - \hat{j} + \hat{k}\). What is the angle between them?
Dot product:
\[\vec{A}\cdot\vec{B} = (3)(2)+(4)(-1)+(-2)(1) = 6 - 4 - 2 = 0\]
Conclusion: Since \(\vec{A}\cdot\vec{B} = 0\), the vectors are perpendicular (θ = 90°).
Verification of magnitudes (optional): \(|\vec{A}| = \sqrt{9+16+4}=\sqrt{29}\), \(|\vec{B}| = \sqrt{4+1+1}=\sqrt{6}\). Since dot product is 0, cos θ = 0, so θ = 90°.
Verification of magnitudes (optional): \(|\vec{A}| = \sqrt{9+16+4}=\sqrt{29}\), \(|\vec{B}| = \sqrt{4+1+1}=\sqrt{6}\). Since dot product is 0, cos θ = 0, so θ = 90°.
Example 5.3: Work by Variable Force
A spring force is given by \(F = -kx\) with \(k = 200\) N/m. Compute the work done by the spring as the block moves from \(x = 0\) to \(x = 0.10\) m.
\[W = \int_0^{0.10} (-kx)\,dx = -\frac{1}{2}k x^2\Big|_0^{0.10}\]
\[= -\tfrac{1}{2}(200)(0.10)^2 = -1.0\text{ J}\]
The work done by the spring is −1.0 J. The negative sign indicates that the spring opposes the displacement.
🧪 Activity 5.1 — When Is Work Zero?
L3 Apply
Materials: A book, a tray, a piece of string, a small box.
Procedure:
- Hold a heavy book stationary in your outstretched hand for 30 seconds. Note your sensation.
- Walk steadily across the room while carrying the book at constant height.
- Tie a string to a small box and twirl it in a horizontal circle by holding the string.
Predict before answering: In which case(s) does the muscular effort produce zero physical work on the book/box?
Observation:
- Step 1: No displacement of book → W = 0 (despite muscle exhaustion).
- Step 2: Displacement is horizontal but the force you apply (vertical, against gravity) is perpendicular → W = 0.
- Step 3: The string force on the box (centripetal) is always perpendicular to its velocity → W = 0.
Conclusion: Three distinct mechanisms produce zero work: (i) no displacement, (ii) perpendicular force, (iii) perpendicular force at every instant of curved motion.
🎯 Competency-Based Questions
A porter at a railway station carries a 40 kg suitcase on his head while walking 100 m horizontally on the platform, then climbs a staircase 3 m vertical, and finally pulls the suitcase along the floor through 5 m using a rope angled 30° below horizontal with 80 N tension. (g = 10 m/s²)
Q1. Calculate the work done by gravity on the suitcase during the horizontal walk of 100 m.L3 Apply
Answer: (d) 0 J. Gravity acts vertically, displacement is horizontal — they are perpendicular. cos 90° = 0, so W_grav = 0.
Q2. Compute the work done against gravity in climbing the 3 m staircase. L3 Apply
Answer: W = mgh = 40 × 10 × 3 = 1200 J against gravity. Equivalently, gravity does −1200 J on the suitcase.
Q3. The rope is angled 30° BELOW horizontal. Find work done by the rope tension over 5 m. L4 Analyse
Answer: W = F·d·cos θ = 80 × 5 × cos 30° = 400 × 0.866 = 346.4 J. The rope does positive work because its horizontal component aids the motion.
Q4. State whether TRUE or FALSE: "If a force does negative work on a body, the body must be slowing down." Justify with reasoning. L5 Evaluate
Answer: FALSE in general. A single force may do negative work while OTHER forces do enough positive work to keep the body accelerating. Only the NET work determines kinetic-energy change. Example: gravity does −mgh on a rocket lifting off, but engine thrust still accelerates it.
Q5. HOT: Devise an experiment to verify that the work done by the variable force \(F = -kx\) equals the area under the F–x curve. Specify apparatus, measurements, and analysis. L6 Create
Sample design: Attach a spring to a smooth cart; record force using a force-sensor while displacement is logged via a motion-sensor. Plot F vs x — should be a straight line through origin (slope k). The integral (area of triangle) = ½kx² should match the kinetic-energy gain of the cart upon release. Compare graphical area to the measured ΔKE; agreement within experimental error confirms the work-as-area principle.
🧠 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): Work done by the centripetal force in uniform circular motion is zero.
Reason (R): Centripetal force is always perpendicular to velocity.
Answer: (A). Both A and R are true. Since F⊥v, F·v = 0 at every instant, so cumulative work is zero. R correctly explains A.
Assertion (A): A man pushing a wall does positive work as long as he applies force.
Reason (R): Work done equals force times displacement, and displacement of the wall is zero.
Answer: (D). Assertion is FALSE — no displacement, no work. Reason is TRUE and shows why A is wrong.
Assertion (A): The dot product \(\hat{i}\cdot\hat{j} = 0\).
Reason (R): The unit vectors along orthogonal axes are perpendicular and the dot product of perpendicular vectors is zero.
Answer: (A). Both true and R explains A.
Frequently Asked Questions - Introduction Work
What is the main concept covered in Introduction Work?
In NCERT Class 11 Physics Chapter 5 (Work, Energy and Power), "Introduction Work" 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 Introduction Work useful in real-life applications?
Real-life applications of Introduction Work 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 Introduction Work?
Key formulas in Introduction Work (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. Introduction Work 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 Introduction Work?
CBSE board questions from Introduction Work 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 Introduction Work 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.
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