This MCQ module is based on: Work, Energy, and Simple Machines — NCERT Exercises
TOPIC 26 OF 50
Work, Energy, and Simple Machines — NCERT Exercises
🎓 Class 9
Science
CBSE
Theory
Ch 7 — Work, Energy, and Simple Machines
⏱ ~13 min
🌐 Language: [gtranslate]
🧠 AI-Powered MCQ Assessment
▲
📝 Worksheet / Assessment
▲
This assessment will be based on: Work, Energy, and Simple Machines — NCERT Exercises
Upload images, PDFs, or Word documents to include their content in assessment generation.
Chapter 7 — Summary at a Glance
Work
W = F s cos θ. SI unit: joule (J). Work can be positive (θ<90°), zero (θ=90°), or negative (θ>90°).
Energy
The capacity of a body to do work. SI unit: joule (J). Forms: mechanical, heat, light, sound, electrical, chemical, nuclear.
Kinetic Energy
Energy of motion: KE = ½ m v². Always non-negative; doubling v makes KE four times larger.
Potential Energy
Energy of position: PE = m g h above the chosen reference level.
Conservation
Energy can neither be created nor destroyed, only transformed. KE + PE = constant in free fall (no friction).
Power
Rate of doing work: P = W/t = F v. SI unit: watt (W). 1 hp ≈ 746 W.
Commercial Unit
1 kWh = 1 kW × 1 hour = 3.6 × 10⁶ J. The "unit" on your electricity bill.
Simple Machines
Lever, pulley, inclined plane, wheel and axle, screw, wedge — they redirect or multiply effort.
MA, VR, η
MA = Load/Effort; VR = effort distance / load distance; η = (MA/VR) × 100%. Real machines have η < 100%.
Keyword Grid
Work
F × s × cos θ; transferred energy.
Joule
Work done by 1 N over 1 m.
Energy
Capacity to do work; unit J.
Kinetic Energy
½ m v² — energy of motion.
Potential Energy
mgh — energy of position.
Conservation
Total energy stays constant.
Power
Work per unit time; unit W.
Watt
1 J per second.
Horsepower
≈ 746 W (non-SI).
Kilowatt-hour
3.6 × 10⁶ J — commercial unit.
Lever
Bar pivoted at fulcrum.
Fulcrum
Pivot point of a lever.
Pulley
Grooved wheel carrying a rope.
Inclined Plane
Sloping surface; MA = L/h.
Wheel and Axle
Big wheel + small axle on shaft.
Screw
Inclined plane wrapped on a cylinder.
Wedge
Movable inclined plane that splits.
MA
Load ÷ Effort.
VR
Effort distance ÷ load distance.
Efficiency
Useful output ÷ total input × 100%.
NCERT Exercises (with Step-by-Step Solutions)
Q1. A force of 7 N acts on an object. The displacement is, say, 8 m, in the direction of the force. Calculate the work done by the force. L3
F = 7 N, s = 8 m, θ = 0°.
W = F s cos θ = 7 × 8 × 1 = 56 J.
Q2. When a body falls freely towards the earth, then what is the work done by the force of gravity on it? L2
Gravitational force acts vertically downward. The displacement during free fall is also vertically downward, so the angle between them is 0°.
Therefore work done is positive: W = F s cos 0° = F s. Energy is being added to the body — its kinetic energy increases as it falls.
Q3. A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15 m long. How much work is done in ploughing the length of the field? L3
F = 140 N, s = 15 m, θ = 0° (force along ploughing direction).
W = 140 × 15 = 2100 J.
Q4. What is the kinetic energy of an object of mass 10 kg moving with a uniform velocity of 5 m s⁻¹? L3
m = 10 kg, v = 5 m s⁻¹.
KE = ½ m v² = ½ × 10 × 25 = 125 J.
Q5. The mass of a cyclist together with the bicycle is 90 kg. Calculate the work done by the cyclist if the speed increases from 0 to 6 m s⁻¹. L4
By the work–energy theorem, work done = change in KE = ½ m (v² − u²).
= ½ × 90 × (36 − 0) = ½ × 90 × 36 = 1620 J.
Q6. Is it possible that a force is acting on a body but still the work done is zero? Explain with the help of an example. L2
Yes. Whenever the force is perpendicular to the displacement (θ = 90°), work done is zero because cos 90° = 0.
Examples: (i) The Moon orbiting the Earth — gravity acts toward the Earth, motion is tangent to the orbit. (ii) A coolie carrying a bag horizontally on his head — gravity is vertical, displacement horizontal.
Q7. An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half-way down. (g = 10 m s⁻²) L4
Initial PE = mgh = 40 × 10 × 5 = 2000 J.
Half way down (height 2.5 m): PE = 40 × 10 × 2.5 = 1000 J.
By conservation: KE = 2000 − 1000 = 1000 J.
Q8. What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer. L3
For a circular orbit, the gravitational pull on the satellite acts toward the centre of the Earth, while the satellite's instantaneous displacement is tangent to its orbit — perpendicular to the gravitational force.
θ = 90° → W = F s cos 90° = 0 J. Hence the satellite's kinetic energy (and speed) does not change in a circular orbit.
Q9. Can there be displacement of an object in the absence of any force acting on it? Think and explain. L4
Yes, by Newton's first law a body in motion continues to move in a straight line at constant velocity if no net force acts on it.
A spacecraft far from any star or planet, with engines off, continues to drift through deep space — there is displacement without any external force.
Q10. A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer. L3
In the scientific sense, no external work is done on the bundle because there is no displacement (s = 0 ⇒ W = 0).
However, the person's muscles continuously contract to support the bundle. Internal physiological processes consume chemical energy (ATP) and produce heat, which is why he gets tired. The fatigue is real, but the formula W = F s gives zero on the bundle.
Q11. An electric heater is rated 1500 W. How much energy does it use in 10 hours? Give your answer in kilowatt-hour and in joules. L3
Energy = P × t = 1500 W × 10 h = 15 000 Wh = 15 kWh.
In joules: 15 × 3.6 × 10⁶ = 5.4 × 10⁷ J.
Q12. Illustrate, by an example, the law of conservation of energy with reference to a freely falling body. L4
Drop a 2 kg ball from a height of 10 m (g = 10 m s⁻²).
At top: PE = mgh = 200 J, KE = 0 → Total = 200 J.
After falling 5 m: v² = 2g(5) = 100, KE = ½(2)(100) = 100 J; PE = 2 × 10 × 5 = 100 J → Total = 200 J.
Just before impact: PE = 0; v² = 2g(10) = 200; KE = ½(2)(200) = 200 J → Total = 200 J.
Total mechanical energy stays at 200 J throughout — confirming conservation.
Q13. What is the work done by the force of gravity in the following cases: (a) a man pushes his car horizontally for 50 m, (b) a man lifts a 10 kg bag through 1.5 m. Take g = 10 m s⁻². L3
(a) Gravity acts vertically downward; displacement is horizontal → θ = 90°.
W = F s cos 90° = 0 J.
W = F s cos 90° = 0 J.
(b) Gravity acts downward (10 × 10 = 100 N); displacement upward (1.5 m) → θ = 180°.
W = 100 × 1.5 × cos 180° = −150 J (work done by gravity is negative; the man does +150 J against gravity).
W = 100 × 1.5 × cos 180° = −150 J (work done by gravity is negative; the man does +150 J against gravity).
Q14. A water pump rated 800 W lifts 200 kg of water through a vertical height of 6 m in 1 minute. Find (i) the useful output power, (ii) the efficiency of the pump. (g = 10 m s⁻²) L5
Useful work output = mgh = 200 × 10 × 6 = 12 000 J.
Time = 60 s → Useful power output = 12 000 / 60 = 200 W.
Efficiency η = (output / input) × 100% = (200 / 800) × 100% = 25%.
Q15. An effort of 250 N raises a 1000 N load using a system of pulleys, when the effort moves through 2 m and the load rises by 0.4 m. Calculate (i) Mechanical Advantage, (ii) Velocity Ratio, (iii) Efficiency of the system. L5
(i) MA = Load / Effort = 1000 / 250 = 4.
(ii) VR = effort distance / load distance = 2 / 0.4 = 5.
(iii) η = (MA / VR) × 100% = (4 / 5) × 100% = 80%.
20% of the input is lost — mostly to friction in the pulley axles and rope.
Well done! You have completed Chapter 7 — Work, Energy and Simple Machines. Be sure you can write the formulas W = F s cos θ, KE = ½ m v², PE = mgh, P = W/t, and the MA expressions for each simple machine without looking — they are frequently asked in Class 9 board-pattern tests.
Frequently Asked Questions — NCERT Exercises & Intext Questions
How do I solve NCERT Class 9 Science Chapter 7 (Work, Energy, and Simple Machines) exercise questions for the CBSE board exam?
Solve NCERT Chapter 7 — Work, Energy, and Simple Machines — exercise questions by first reading the question carefully, writing down the given data, recalling the relevant concepts like work, kinetic energy, potential energy, and applying them step by step. This Part 4 covers every intext and end-of-chapter exercise from the NCERT textbook. Write balanced equations, label diagrams clearly and show each step — CBSE Class 9 examiners award step marks even if the final answer has a small slip. Practising these solutions strengthens conceptual clarity and builds speed for both the school exam and the upcoming Class 10 board exam.
Are the NCERT intext questions from Work, Energy, and Simple Machines important for the Class 9 Science exam?
Yes, NCERT intext questions for Chapter 7 Work, Energy, and Simple Machines are highly important for the CBSE Class 9 Science exam. Many questions in school and competitive papers are directly lifted or only slightly modified from these intext questions, and they test the foundational concepts — work, kinetic energy, potential energy — that chapter-end questions and the Class 10 board build on. Attempt every intext question first, then move on to the exercises. This practice ensures complete NCERT coverage, which is the CBSE syllabus's primary source.
What types of questions from Work, Energy, and Simple Machines are asked in the Class 9 Science exam?
The Class 9 Science paper (CBSE pattern) asks a mix of question types from Work, Energy, and Simple Machines: 1-mark MCQ and assertion-reason, 2-mark short answers, 3-mark explanations, 5-mark long answers with diagrams or derivations, and 4-mark competency-based / case-study questions. These test understanding of work, kinetic energy, potential energy, conservation of energy. Practising every NCERT exercise and intext question prepares you to answer all of these formats with confidence.
How many marks does Chapter 7 — Work, Energy, and Simple Machines — typically carry in the Class 9 Science paper?
Chapter 7 — Work, Energy, and Simple Machines — is part of the CBSE Class 9 Science syllabus and typically contributes 5–9 marks in the annual paper, depending on the year's weightage. Questions are drawn from definitions, reasoning, numerical/descriptive problems and diagrams on topics like work, kinetic energy, potential energy. Solving the NCERT exercises in this part is essential because CBSE directly references the NCERT Exploration textbook for question design.
Where can I find step-by-step NCERT solutions for Chapter 7 Work, Energy, and Simple Machines Class 9 Science?
You can find complete, step-by-step NCERT solutions for Chapter 7 Work, Energy, and Simple Machines Class 9 Science on MyAiSchool. Every intext and end-of-chapter exercise question is solved with full working, labelled diagrams and CBSE-aligned mark distribution. Solutions highlight key points about work, kinetic energy, potential energy that examiners look for. This makes revision quick and exam-focused for Class 9 CBSE students.
What is the best way to revise Work, Energy, and Simple Machines for the Class 9 Science exam?
The best way to revise Work, Energy, and Simple Machines for the CBSE Class 9 Science exam is a three-pass approach. First pass: skim the chapter and note down key terms like work, kinetic energy, potential energy in a one-page mind map. Second pass: solve every NCERT intext and exercise question without looking at the solution, then self-check. Third pass: attempt sample papers and competency-based questions under timed conditions. This structured revision secures full marks for this chapter.
AI Tutor
Science Class 9 — Exploration
Ready