Chapter 11 — Work and Energy
Welcome to HSLC Guru! This page presents complete English-medium notes, solved textbook questions, additional practice questions, glossary and a formula reference for Class 9 Science Chapter 11 — Work and Energy as per the ASSEB (Assam State School Education Board) syllabus. Every concept is explained in clear language with worked numerical examples so that students can prepare confidently for school tests, half-yearly and final examinations. The work-energy chapter forms the foundation of mechanics and is essential for higher classes, so go through each section carefully and revise the formulas given at the end.
Summary of the Chapter
In ordinary language the word “work” is used very loosely. A student reading a book, a person sitting and thinking, or a labourer carrying a load — all are said to be doing work. However, in physics, work has a precise meaning. Work is said to be done only when a force acts on a body and the body undergoes a displacement in the direction of the force (or having a component along the direction of the force). Mathematically, if a constant force F acts on a body and produces a displacement s making an angle θ with the direction of force, then work done is given by W = F · s · cos θ. The SI unit of work is the joule (J), where 1 J = 1 N × 1 m. Work is a scalar quantity. When the force and displacement are in the same direction (θ = 0°), cos θ = 1, and the work done is positive. When force and displacement are in opposite directions (θ = 180°), cos θ = −1, and work done is negative (e.g. work done by friction). When force is perpendicular to displacement (θ = 90°), cos θ = 0, and work done is zero (e.g. a coolie carrying a load on his head walking on a horizontal road, or the gravitational force on a satellite moving in a circular orbit).
The capacity of a body to do work is called its energy. Energy exists in many forms — mechanical, heat, light, sound, chemical, electrical, nuclear, etc. The SI unit of energy is also the joule (J). Mechanical energy is of two types — kinetic and potential. Kinetic energy (KE) is the energy possessed by a body by virtue of its motion. For a body of mass m moving with velocity v, the kinetic energy is given by KE = ½ m v². This relation is derived from the equation of motion v² = u² + 2as. The work-energy theorem states that the net work done by the resultant force on a body is equal to the change in its kinetic energy: W = ½ m v² − ½ m u². Potential energy (PE) is the energy possessed by a body due to its position or configuration. The gravitational potential energy of a body of mass m raised through a height h above the ground is PE = m g h, where g is the acceleration due to gravity (9.8 m/s²).
The total mechanical energy of a body is the sum of its kinetic and potential energies, i.e. E = KE + PE. According to the law of conservation of energy, energy can neither be created nor destroyed; it can only be transformed from one form to another. The total energy of an isolated system always remains constant. For example, when a body of mass m falls freely from a height h, its total mechanical energy (mgh + 0) at the top becomes (0 + ½mv²) at the bottom, and at any intermediate point KE + PE = mgh = constant. Energy transformations occur in everyday devices — an electric bulb converts electrical energy into light and heat; a fan converts electrical energy into mechanical energy; a battery converts chemical energy into electrical energy.
Power is defined as the rate of doing work or the rate of transfer of energy. If work W is done in time t, then power P = W / t. The SI unit of power is the watt (W), where 1 W = 1 J/s. Larger units include 1 kilowatt (kW) = 1000 W and 1 horsepower (hp) ≈ 746 W. Since the joule is a small unit, the commercial unit of energy is the kilowatt-hour (kWh), often called “1 unit” by electricity boards. 1 kWh = 1 kW × 1 h = 1000 W × 3600 s = 3.6 × 10⁶ J. The electricity bill we pay every month is calculated in kWh. Numerical examples like calculating the work done in lifting a 10 kg box to a 2 m high shelf (W = mgh = 10 × 9.8 × 2 = 196 J), or the kinetic energy of a 2 kg ball moving at 5 m/s (KE = ½ × 2 × 25 = 25 J), help students apply these concepts confidently.
Textbook Questions and Answers
A. Very Short Answer Questions (1 mark)
Q1. Define work in physics.
Answer: Work is said to be done when a force acting on a body produces a displacement of the body in the direction of the force. Mathematically, W = F · s · cos θ.
Q2. What is the SI unit of work?
Answer: The SI unit of work is the joule (J). 1 joule = 1 newton × 1 metre.
Q3. Define 1 joule of work.
Answer: 1 joule is the work done when a force of 1 newton displaces a body through a distance of 1 metre in the direction of the force.
Q4. What is energy?
Answer: The capacity of a body to do work is called energy. Its SI unit is the joule (J).
Q5. What is kinetic energy?
Answer: The energy possessed by a body by virtue of its motion is called kinetic energy. KE = ½ m v².
Q6. What is potential energy?
Answer: The energy possessed by a body by virtue of its position or configuration is called potential energy. Gravitational PE = m g h.
Q7. Define power. State its SI unit.
Answer: Power is the rate of doing work, P = W / t. SI unit is the watt (W); 1 W = 1 J/s.
Q8. What is the commercial unit of energy?
Answer: The commercial unit of energy is the kilowatt-hour (kWh). 1 kWh = 3.6 × 10⁶ J.
Q9. When is work done by a force zero?
Answer: Work done is zero when (i) displacement is zero, or (ii) force is perpendicular to displacement (θ = 90°, cos 90° = 0).
Q10. Name two forms of mechanical energy.
Answer: Kinetic energy and potential energy.
B. Short Answer Questions (2-3 marks)
Q1. A body of mass 2 kg is lifted to a height of 5 m. Find the work done against gravity. (g = 9.8 m/s²)
Answer: Work done = m g h = 2 × 9.8 × 5 = 98 J.
Q2. Distinguish between positive and negative work with one example each.
Answer: Positive work — when the force and displacement are in the same direction (θ = 0°). Example: a falling stone — gravity does positive work. Negative work — when force and displacement are in opposite directions (θ = 180°). Example: work done by frictional force on a moving body.
Q3. A ball of mass 0.5 kg is moving with a velocity of 10 m/s. Calculate its kinetic energy.
Answer: KE = ½ m v² = ½ × 0.5 × (10)² = ½ × 0.5 × 100 = 25 J.
Q4. State and explain the law of conservation of energy.
Answer: The law states that energy can neither be created nor destroyed; it can only be transformed from one form to another. The total energy of an isolated system remains constant. For example, in a freely falling body the loss in PE is exactly equal to the gain in KE, so total mechanical energy is conserved.
Q5. A machine does 1500 J of work in 30 seconds. Calculate its power.
Answer: P = W / t = 1500 / 30 = 50 W.
Q6. Convert 1 kWh into joules.
Answer: 1 kWh = 1000 W × 3600 s = 3,600,000 J = 3.6 × 10⁶ J.
Q7. Why is the work done by a coolie carrying a load on his head while walking on a horizontal road said to be zero?
Answer: The force applied by the coolie acts vertically upward (to support the load), but his displacement is horizontal. The angle between force and displacement is 90°, so cos 90° = 0, and W = F·s·cos θ = 0.
C. Long Answer Questions (5-6 marks)
Q1. Derive the expression for kinetic energy of a body of mass m moving with a velocity v.
Answer: Consider a body of mass m initially at rest (u = 0). Let a constant force F be applied on it, producing an acceleration a. Suppose it travels a distance s and attains velocity v.
Using v² = u² + 2as → v² = 0 + 2as → s = v² / (2a).
Work done by the force: W = F × s = (m a) × (v² / 2a) = ½ m v².
This work is stored in the body as kinetic energy. Hence KE = ½ m v².
Q2. State and prove the work-energy theorem.
Answer: Statement: The work done by the net force acting on a body is equal to the change in its kinetic energy.
Proof: Let a body of mass m moving with initial velocity u be acted upon by a constant force F producing acceleration a. After covering a distance s, its velocity becomes v.
From v² = u² + 2as → 2as = v² − u² → s = (v² − u²) / 2a.
Work done: W = F·s = (ma) × (v² − u²)/(2a) = ½ m v² − ½ m u² = KE_final − KE_initial = ΔKE. Hence proved.
Q3. A body of mass 5 kg is raised vertically through a height of 4 m. Calculate (i) the work done against gravity, (ii) the potential energy gained. Take g = 10 m/s².
Answer:
(i) Work done = F × s = m g × h = 5 × 10 × 4 = 200 J.
(ii) Potential energy gained = m g h = 5 × 10 × 4 = 200 J.
The work done against gravity is stored as gravitational potential energy in the body.
Q4. Show that the total mechanical energy of a freely falling body is conserved.
Answer: Consider a body of mass m falling freely from a height H. Let it have fallen a distance x at any instant, so it is at height (H − x) above the ground.
At point A (top, height H): KE = 0; PE = mgH; Total E = mgH.
At point B (after falling x): velocity v² = 2gx; KE = ½ m v² = ½ m (2gx) = mgx; PE = mg(H − x); Total E = mgx + mg(H − x) = mgH.
At point C (just before hitting ground): v² = 2gH; KE = ½ m × 2gH = mgH; PE = 0; Total E = mgH.
At all positions total mechanical energy = mgH = constant. Hence the law of conservation of mechanical energy is verified.
Q5. An electric bulb of 60 W is used for 6 hours daily. Calculate the energy consumed in 30 days in kWh and in joules.
Answer:
Power P = 60 W = 0.06 kW.
Time t = 6 h × 30 days = 180 h.
Energy = P × t = 0.06 × 180 = 10.8 kWh.
In joules: 10.8 × 3.6 × 10⁶ = 3.888 × 10⁷ J.
Additional Practice — Multiple Choice Questions
Q1. The SI unit of work is —
(a) newton (b) joule (c) watt (d) erg
Answer: (b) joule.
Q2. Kinetic energy of a body is given by —
(a) m g h (b) ½ m v² (c) m v (d) F × s
Answer: (b) ½ m v².
Q3. The commercial unit of energy is —
(a) joule (b) watt (c) kilowatt-hour (d) calorie
Answer: (c) kilowatt-hour.
Q4. 1 kWh equals —
(a) 3.6 × 10³ J (b) 3.6 × 10⁶ J (c) 3.6 × 10⁵ J (d) 36 J
Answer: (b) 3.6 × 10⁶ J.
Q5. When force and displacement are perpendicular, work done is —
(a) maximum (b) zero (c) negative (d) infinite
Answer: (b) zero.
Q6. Power is defined as —
(a) work × time (b) work / time (c) energy × time (d) force / time
Answer: (b) work / time.
Q7. The energy possessed by a body due to its position is —
(a) kinetic energy (b) potential energy (c) heat energy (d) sound energy
Answer: (b) potential energy.
Q8. 1 horsepower is approximately equal to —
(a) 100 W (b) 500 W (c) 746 W (d) 1000 W
Answer: (c) 746 W.
Q9. If the velocity of a body becomes double, its kinetic energy becomes —
(a) double (b) half (c) four times (d) same
Answer: (c) four times.
Q10. Work done by friction is generally —
(a) positive (b) zero (c) negative (d) infinite
Answer: (c) negative.
Fill in the Blanks
Q1. The SI unit of energy is __________.
Answer: joule.
Q2. Work done = force × __________ × cos θ.
Answer: displacement.
Q3. The energy of a moving body is called __________ energy.
Answer: kinetic.
Q4. 1 watt = 1 __________ per second.
Answer: joule.
Q5. Energy can neither be __________ nor destroyed.
Answer: created.
True or False
Q1. Work is a vector quantity.
Answer: False. Work is a scalar quantity.
Q2. Kinetic energy can never be negative.
Answer: True.
Q3. 1 kWh = 3.6 × 10⁵ J.
Answer: False. 1 kWh = 3.6 × 10⁶ J.
Q4. A body at rest has zero potential energy.
Answer: False. A body at rest can still have potential energy due to its position (e.g. a stone on a hill).
Q5. The total mechanical energy of a freely falling body remains constant in the absence of air resistance.
Answer: True.
Glossary of Important Terms
| Term | Meaning |
|---|---|
| Work | Product of force and displacement in the direction of force; W = F·s·cos θ. |
| Joule | SI unit of work and energy; 1 J = 1 N × 1 m. |
| Energy | The capacity of a body to do work. |
| Kinetic Energy | Energy possessed by a body due to its motion; KE = ½ m v². |
| Potential Energy | Energy due to position or configuration; PE = m g h. |
| Mechanical Energy | Sum of kinetic and potential energy; E = KE + PE. |
| Work-Energy Theorem | Net work done on a body equals the change in its KE. |
| Conservation of Energy | Energy cannot be created or destroyed, only transformed. |
| Power | Rate of doing work; P = W / t. SI unit watt. |
| Watt | 1 W = 1 J/s; SI unit of power. |
| Kilowatt-hour | Commercial unit of energy; 1 kWh = 3.6 × 10⁶ J. |
| Horsepower | Practical unit of power; 1 hp ≈ 746 W. |
Formula Reference Table
| Quantity | Formula | SI Unit |
|---|---|---|
| Work | W = F · s · cos θ | joule (J) |
| Kinetic Energy | KE = ½ m v² | joule (J) |
| Potential Energy | PE = m g h | joule (J) |
| Work-Energy Theorem | W = ½ m v² − ½ m u² | joule (J) |
| Mechanical Energy | E = KE + PE | joule (J) |
| Power | P = W / t | watt (W) |
| Power (with velocity) | P = F · v | watt (W) |
| Commercial Energy | 1 kWh = 3.6 × 10⁶ J | kilowatt-hour |
| Horsepower | 1 hp ≈ 746 W | watt (W) |
| Free fall energy | mgh = ½ m v² (top to bottom) | joule (J) |
Practise the numerical problems and revise the derivations of KE = ½ m v² and the work-energy theorem thoroughly. Make sure you understand when work done is positive, negative or zero, and how energy transforms in everyday devices. For more ASSEB Class 9 Science chapter notes, MCQs and previous-year questions, keep visiting HSLC Guru. All the best for your examinations!