Navigating through the newly revised CBSE Class 9 Science curriculum (Exploration) requires a thorough physical and mathematical understanding of mechanical work (positive, negative, and zero work), forms of energy, the Work-Energy Theorem, kinetic energy (KE = 1/2 × m × v²), gravitational potential energy (PE = m × g × h), the Law of Conservation of Mechanical Energy in freely falling bodies, power ratings (Watts and Kilowatts), commercial electrical units (kilowatt-hour), and the newly integrated physics of simple machines (levers, inclined planes, pulleys, mechanical advantage, and the impossibility of perpetual motion machines). Chapter 7 of Class 9 Physics, “Work, Energy, and Simple Machines”, bridges fundamental mechanics with real-world technological applications. It explores why holding a heavy barbell steady performs zero mechanical work despite muscular exertion; proves why the speed of an object sliding down frictionless ramps of equal height is independent of the path shape; analyzes the energy losses caused by friction in roller-coaster loops; explains why winding roads and inclined ladders multiply human effort; and derives step-by-step energy conservation proofs. To help students master every aspect of this high-weightage chapter, this comprehensive solutions guide offers textbook-accurate, highly structured, and pedagogically sound responses strictly aligned with the latest CBSE Class 9 evaluation standards.
Every question presented in the official NCERT textbook—ranging from all introductory “Think It Over” sections and in-text “Pause and Ponder” prompts (Pages 119, 121, 123, 126, 129, and 132) to the complete end-of-chapter “Revise, Reflect, Refine” exercises (Questions 1 to 12 on Pages 136–138)—has been solved with exhaustive detail. Numerical problems follow a step-by-step box format with explicit variable legends, standard SI unit conversions, and algebraic substitutions using clean plain-text symbols without raw LaTeX tags. Key scoring terms, official CBSE exam tags, and dynamic summary tables have been highlighted to ensure students secure maximum marks in their examinations.
Master Concept & Comparative Summary Tables
1. Master Formula Sheet for Work, Energy, Power, and Machines
| Physical Quantity / Concept | Standard Formula | SI Unit & Symbol | Key Mathematical Notes |
|---|---|---|---|
| Mechanical Work (W) | Work = Force × Displacement = F × s | Joule (J) | 1 Joule = 1 Newton × 1 metre (1 N · m). Zero work if displacement is zero or angle is 90°. |
| Kinetic Energy (KE) | KE = 1/2 × mass × (velocity)² = 1/2 × m × v² | Joule (J) | Proportional to mass (m) and square of velocity (v²). Doubling speed quadruples KE. |
| Potential Energy (PE) | PE = mass × gravity × height = m × g × h | Joule (J) | Gravitational energy relative to ground; depends only on vertical height (h), not path. |
| Work-Energy Theorem | Work Done = Change in Kinetic Energy = 1/2·m·v² - 1/2·m·u² | Joule (J) | Net work done on a body equals the net increase in its kinetic energy. |
| Mechanical Power (P) | Power = Work Done / Time = W / t | Watt (W) | 1 Watt = 1 Joule/second (1 J/s). 1 kW = 1000 W. 1 Horsepower (hp) = 746 W. |
| Commercial Energy (E) | Energy = Power (in kW) × Time (in hours) | Kilowatt-hour (kWh / Unit) | 1 kWh = 1 Board of Trade Unit = 3.6 × 10⁶ Joules (3.6 MJ). |
| Mechanical Advantage (MA) | MA = Load / Effort = Effort Arm / Load Arm | Dimensionless (No unit) | If MA > 1, the machine acts as a force multiplier (reduces effort required). |
2. Nature of Work Done Under Different Physical Conditions
| Nature of Work Done | Angle Between Force & Motion | Mathematical Sign | Real-World Physical Example |
|---|---|---|---|
| Positive Work (W > 0) | Angle = 0° (Force is in the same direction as displacement). | Positive (+) | A horse pulling a cart forward; gravity acting on a falling apple. |
| Negative Work (W < 0) | Angle = 180° (Force is in the opposite direction to displacement). | Negative (-) | Kinetic friction opposing a sliding coin; gravity acting on an upward-thrown ball. |
| Zero Work (W = 0) | Angle = 90° (Force is perpendicular to displacement) OR Displacement = 0. | Zero (0) | • A coolie carrying luggage on his head walking horizontally. • Gravitational force on a satellite in a circular orbit. • Pushing against a stationary concrete wall. |
NCERT In-Text Questions: “Think It Over”
Page No. 116: Think It Over (Questions 1 to 3)
Question 1 What will be the magnitude of the velocity of the child at the bottom of the blue slide? [Exam Favorite]
Answer: The magnitude of the velocity of the child at the bottom of the slide is given by: Velocity = Square root of (2 × g × h).
- Scientific Reason: According to the Law of Conservation of Energy, in the absence of friction, the child’s initial gravitational potential energy at the top of the slide (PE = m × g × h) converts entirely into kinetic energy at the bottom (KE = 1/2 × m × v²): 1/2 × m × v² = m × g × h v² = 2 × g × h v = Square root of (2 × g × h).
Question 2 Will two children of different masses reach the bottom of the same slide with the same velocity? [Exam Favorite]
Answer: Yes, both children will reach the bottom of the slide with the exact same velocity (assuming friction is negligible).
- Explanation: In the energy conservation equation (1/2 × m × v² = m × g × h), the mass term (m) cancels out from both sides, yielding v = Square root of (2 × g × h). The final speed depends solely on the vertical height (h) and gravitational acceleration (g), making it completely independent of the mass of the child.
Question 3 Which of the slides (steep, gentle, or curved, all having the exact same vertical height) will result in the largest magnitude of velocity for the child at its bottom? [Exam Favorite]
Answer: All slides of the same vertical height will result in the EXACT SAME magnitude of velocity at the bottom (assuming frictionless surfaces).
- Reason: Gravitational potential energy depends strictly on the vertical height (h) of the slide, not on the length, shape, or slope angle of the sliding path. Since all slides start from the same vertical height, the potential energy converted into kinetic energy is identical, producing the same final velocity at the bottom. (However, the child on the steeper slide will reach the bottom in a shorter time due to higher initial acceleration).
NCERT In-Text Questions: “Pause and Ponder”
Page No. 119: Pause and Ponder (Questions 1 & 2)
Question 1 In the previous chapter, a weightlifter is shown holding a barbell steady in her hands (Fig. 6.8). Is she doing any work on the barbell while holding it steady? [Exam Favorite]
Answer: No, the weightlifter performs ZERO mechanical work on the barbell while holding it steady.
- Physical Reason: In physics, mechanical work is defined as the product of force and displacement in the direction of the force (Work = Force × Displacement).
- Although the weightlifter exerts a massive upward muscular force to support the barbell against gravity, the displacement of the barbell is zero (s = 0). Therefore: Work Done = Force × 0 = 0 Joules. (The muscular tiredness experienced is due to internal physiological contractions of muscle fibers, not external mechanical work).
Question 2 Is the work done by friction on a stack of coins that travels on a rough surface positive, negative, or zero? [Exam Favorite]
Answer: The work done by friction is Negative.
- Scientific Reason: Friction is a resistive force that always acts in the direction opposite to the displacement of the moving coins (Angle = 180°). When force and displacement are in opposite directions, the work done is negative: Work = – (Frictional Force × Displacement).
Page No. 121: Pause and Ponder (Question 3)
Question 3 When a person pedals a bicycle, in what forms does this muscular energy appear as you ride? [Exam Favorite]
Answer: When a person pedals a bicycle, the chemical energy stored in their muscles undergoes three continuous energy transformations:
- Kinetic Energy: The major portion converts into the mechanical kinetic energy of the moving bicycle, rider, and spinning wheels.
- Thermal (Heat) Energy: A portion is dissipated as heat in the chain, gears, bearings, and tires due to frictional resistance.
- Sound Energy: A small fraction is converted into sound vibrations generated by the mechanical drivetrain.
Page No. 123: Pause and Ponder (Question 4)
Question 4 Does the kinetic energy of an object which moves with constant velocity change with its position? [Exam Favorite]
Answer: No, the kinetic energy does not change with position.
- Explanation: Kinetic energy depends exclusively on two physical parameters: mass (m) and velocity (v) according to the formula KE = 1/2 × m × v². If an object moves with a constant velocity, its speed and mass remain unchanged at every spatial position; hence, its kinetic energy remains constant throughout its trajectory.
Page No. 126: Pause and Ponder (Question 5)
Question 5 Does the gravitational potential energy of an object near the surface of the Earth change if it moves with constant velocity in the horizontal direction? What if the object is gradually raised in the vertical direction? [Exam Favorite]
Answer:
- Moving in Horizontal Direction: No, the potential energy does not change. Gravitational potential energy depends on vertical height above the reference level (PE = m × g × h). In horizontal motion, the height (h) remains constant, so PE remains unchanged.
- Raised in Vertical Direction: Yes, the potential energy increases. As the object is lifted to a greater vertical height (h), positive work is done against gravity, which is stored as increased gravitational potential energy in direct proportion to height (\text{PE} \propto h).
Page No. 129: Pause and Ponder (Question 6)
Question 6 You may have seen an exhibit in a science park where a ball is released from the highest point of a track with successive humps (A, B, C, D, E). Describe how kinetic and potential energy change. Why do subsequent peaks have lower heights? [Exam Favorite]
Answer:
================================================================================ ENERGY TRANSFORMATION IN TRACK ROLLER-COASTER: -------------------------------------------------------------------------------- 1. At Peak A (Highest Point): Ball is at rest; Potential Energy is MAXIMUM, Kinetic Energy is ZERO. 2. Descending from A to Valley B: As the ball rolls down, height decreases and speed increases. Potential energy transforms into Kinetic Energy (KE is maximum at valley B). 3. Ascending from B to Peak C: As the ball climbs, speed decreases and height increases; Kinetic Energy transforms back into Potential Energy. WHY SUBSEQUENT PEAKS ARE LOWER: • Real-World Energy Losses: In any real physical system, the rolling ball experiences frictional resistance against the track and aerodynamic air drag. • Dissipation: A portion of total mechanical energy is continuously dissipated as non-recoverable Heat and Sound energy. • Reduced Total Mechanical Energy: Because total usable mechanical energy decreases over time, the ball does not possess sufficient energy to climb back to its original release height. Hence, subsequent humps must be engineered lower. ================================================================================
Page No. 132: Pause and Ponder (Questions 7 to 10)
Question 7 Explain why roads on hills are built to wind around in gentle slopes rather than going straight up. [Exam Favorite]
Answer: Winding hill roads function as natural Inclined Planes (Simple Machines):
- A straight road directly up a steep mountain slope requires vehicles to exert a massive vertical lifting force against gravity over a short distance, which can stall engines or cause wheels to slip.
- By building winding roads with gentle gradients, the total path distance is increased.
- According to the principle of machines (Work = Force × Distance), increasing the distance allows the vehicle to climb to the same vertical mountain height with a much smaller engine driving force (effort), providing a high Mechanical Advantage.
Question 8 To reach a higher floor, we find climbing an inclined ladder easier in comparison to climbing a vertical ladder. Explain why. [Exam Favorite]
Answer:
- Vertical Ladder: The climber must exert a muscular force equal to their entire body weight (F = m \times g) directly upward against gravity.
- Inclined Ladder: The inclined ladder acts as a ramp. The climber’s weight is partially supported by the ladder structure, so the required forward muscular effort along the incline is significantly less than their full body weight (F = m \times g \times \text{height} / \text{length}). The longer distance travelled reduces the instantaneous force required, making climbing easier.
Question 9 Why do you push an object closer to the scissors’ fulcrum when you want to cut an object which is hard? [Exam Favorite]
Answer: Scissors operate as a Class-1 Lever:
- Mechanical Advantage Formula:
MA = Effort Arm / Load Arm. - When a tough object (like thick cardboard or wire) is pushed closer to the central pivot screw (fulcrum), the Load Arm distance is minimized.
- A shorter load arm produces a higher Mechanical Advantage (MA > 1). Consequently, a normal hand squeezing effort applied on the handles is magnified into a much larger cutting force at the blades.
Question 10 Throughout history, many designs of perpetual motion machines have been proposed, but none actually work. Why do all real machines eventually slow down and stop? [Exam Favorite]
Answer: Perpetual motion machines are physically impossible due to the Law of Conservation of Energy and Friction:
- A perpetual machine claims to produce continuous mechanical work indefinitely without an external energy source.
- In every real mechanical system, moving contact parts experience friction and air drag, which continuously convert useful mechanical energy into thermal (heat) and acoustic (sound) energy.
- Because energy cannot be created from nothing, the total mechanical energy rapidly depletes to zero, causing the machine to slow down and stop.
NCERT Chapter-End Exercises: “Revise, Reflect, Refine” (Pages 136–138)
Question 1 State whether the following statements are True or False: [Exam Favorite] (i) Work done by gravity on a satellite revolving in a circular orbit around Earth is zero. (ii) A body can have energy without having momentum. (iii) A body can have momentum without having energy. (iv) Simple machines can multiply energy.
Answer:
- (i) True (The gravitational force acts radially inward toward the Earth’s center, perpendicular to the satellite’s tangential displacement at all points; Angle = 90°, so Work = 0).
- (ii) True (A stationary object elevated at a height has gravitational potential energy (PE = m \cdot g \cdot h), but its velocity is zero, so its momentum is zero).
- (iii) False (Momentum (p = m \cdot v) requires non-zero velocity; any body with velocity must possess kinetic energy (KE = p² / 2m > 0)).
- (iv) False (Simple machines multiply force or speed, but they can never multiply energy; Work Output is always less than or equal to Work Input due to conservation of energy).
Question 2 A force of 10 N acts on an object displacing it by 5 m in the direction of the force. Calculate the work done. [Exam Favorite]
Answer:
================================================================================
CALCULATION (WORK DONE):
--------------------------------------------------------------------------------
GIVEN DATA:
• Applied Force (F) = 10 N
• Displacement (s) = 5 m (in direction of force, Angle = 0°)
CALCULATION:
Formula: Work Done (W) = Force × Displacement
W = 10 N × 5 m = 50 Joules (50 J)
FINAL ANSWER:
The work done on the object is 50 Joules (50 J).
================================================================================
Question 3 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? [Exam Favorite]
Answer:
================================================================================
CALCULATION (WORK DONE BY BULLOCKS):
--------------------------------------------------------------------------------
GIVEN DATA:
• Pulling Force (F) = 140 N
• Distance/Length (s) = 15 m
CALCULATION:
Formula: Work Done (W) = Force × Displacement
W = 140 N × 15 m = 2100 Joules (2.1 kJ)
FINAL ANSWER:
The work done in ploughing the length of the field is 2100 Joules (2.1 kJ).
================================================================================
Question 4 The kinetic energy of an object of mass m moving with a velocity of 5 m/s is 25 J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times? [Exam Favorite]
Answer:
================================================================================
NUMERICAL SOLUTION (KINETIC ENERGY SCALING):
--------------------------------------------------------------------------------
GIVEN DATA:
• Initial Velocity (v₁) = 5 m/s
• Initial KE (KE₁) = 25 J
STEP 1: Finding Mass of Object (m)
Formula: KE = 1/2 × m × v²
25 = 1/2 × m × (5)²
25 = 1/2 × m × 25
m = 2 kg
STEP 2: When Velocity is Doubled (v₂ = 2 × 5 = 10 m/s)
KE₂ = 1/2 × m × (v₂)² = 1/2 × 2 kg × (10)² = 100 Joules
*(Since KE ∝ v², doubling velocity increases KE by 2² = 4 times: 4 × 25 J = 100 J)*.
STEP 3: When Velocity is Increased Three Times (v₃ = 3 × 5 = 15 m/s)
KE₃ = 1/2 × m × (v₃)² = 1/2 × 2 kg × (15)² = 225 Joules
*(Since KE ∝ v², tripling velocity increases KE by 3² = 9 times: 9 × 25 J = 225 J)*.
FINAL ANSWER:
• KE when velocity is doubled = 100 Joules (100 J)
• KE when velocity is tripled = 225 Joules (225 J)
================================================================================
Question 5 What is the work done by the force of gravity on a flying airplane moving horizontally at constant speed? [Exam Favorite]
Answer: The work done by gravity on the horizontally flying airplane is ZERO.
- Physical Justification: The gravitational force acts vertically downward toward the center of the Earth, while the displacement of the airplane is along the horizontal plane. The angle between the gravitational force vector and the displacement vector is 90° (perpendicular).
Work = Force × Displacement × cos(90°) = Force × Displacement × 0 = 0 Joules.
Question 6 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. (Take g = 9.8 m/s²). [Exam Favorite]
Answer:
================================================================================
NUMERICAL SOLUTION (POTENTIAL & KINETIC ENERGY IN FREE FALL):
--------------------------------------------------------------------------------
GIVEN DATA:
• Mass (m) = 40 kg
• Total Height (h) = 5 m
• Acceleration due to gravity (g) = 9.8 m/s²
STEP 1: Potential Energy at Top Height (5 m)
Formula: PE_initial = m × g × h
PE_initial = 40 kg × 9.8 m/s² × 5 m = 1960 Joules
STEP 2: Kinetic Energy Half-Way Down (h' = 2.5 m)
According to the Law of Conservation of Energy:
Total Mechanical Energy at top = 1960 J (Pure Potential Energy).
Half-way down, half of the potential energy is converted into Kinetic Energy:
• Loss in PE = m × g × (h / 2) = 40 × 9.8 × 2.5 = 980 J
• Gain in KE = Loss in PE = 980 Joules
FINAL ANSWER:
• Potential Energy at 5 m height = 1960 Joules
• Kinetic Energy half-way down = 980 Joules
================================================================================
Question 7 A ball of mass 2 kg is thrown vertically upward with an initial velocity of 20 m/s. It reaches a maximum height of 19.4 m. Calculate: (i) Initial kinetic energy, (ii) Potential energy at maximum height, and (iii) Work done against air resistance. (Take g = 10 m/s²). [Exam Favorite]
Answer:
================================================================================ NUMERICAL SOLUTION (ENERGY DISSIPATION & AIR DRAG): -------------------------------------------------------------------------------- GIVEN DATA: • Mass of ball (m) = 2 kg • Initial launch velocity (u) = 20 m/s • Maximum height reached (h) = 19.4 m • Gravity (g) = 10 m/s² STEP 1: Initial Kinetic Energy at Ground Level KE_initial = 1/2 × m × u² KE_initial = 1/2 × 2 kg × (20 m/s)² = 400 Joules STEP 2: Gravitational Potential Energy at Maximum Height (19.4 m) PE_top = m × g × h PE_top = 2 kg × 10 m/s² × 19.4 m = 388 Joules STEP 3: Work Done Against Air Resistance (Energy Lost) By Energy Conservation: Initial Energy = Final Energy + Energy Lost to Air Drag Energy Lost = KE_initial - PE_top Energy Lost = 400 J - 388 J = 12 Joules FINAL ANSWER: • Initial Kinetic Energy = 400 Joules • Potential Energy at Maximum Height = 388 Joules • Work Done Against Air Resistance = 12 Joules ================================================================================
Question 8 What are simple machines? Define Mechanical Advantage (MA). Give examples of levers, inclined planes, and pulleys. [Exam Favorite]
Answer:
- Definition of Simple Machines: A simple machine is a mechanical device that alters the magnitude, direction, or point of application of an applied force, enabling heavy physical tasks to be accomplished with less human effort.
- Mechanical Advantage (MA): The ratio of the load (resistance force overcome) to the effort (applied input force): Mechanical Advantage = Load / Effort.
- Examples of Simple Machines:
- Levers: Crowbar, scissors, see-saw, bottle opener (pivots about a fulcrum).
- Inclined Planes: Wheelchair ramps, winding mountain roads, sloping planks.
- Pulleys: Well water pulley, construction crane cable pulleys (changes direction of pulling force).
Question 9 Why does an inclined plane make it easier to lift heavy loads into a truck compared to lifting them vertically? [Exam Favorite]
Answer: An inclined plane acts as a force-multiplying simple machine:
- Lifting a 100 kg crate vertically into a truck requires an upward effort equal to its full gravitational weight (W = 1000 N).
- Rolling the crate up a smooth inclined ramp requires an effort only to overcome the component of gravity along the incline (F = \text{Weight} \times \text{Height} / \text{Length of ramp}).
- Because the ramp length is much greater than the vertical height, the required pushing effort is significantly reduced, making loading manageable for human workers.
Question 10 A woman pulls a bucket of water of mass 15 kg from a well of depth 10 m in 20 s. Calculate: (i) Work done, and (ii) Power consumed by the woman. (Take g = 10 m/s²). [Exam Favorite]
Answer:
================================================================================
NUMERICAL SOLUTION (WORK AND POWER CALCULATION):
--------------------------------------------------------------------------------
GIVEN DATA:
• Mass of bucket with water (m) = 15 kg
• Vertical Depth/Height (h) = 10 m
• Time taken (t) = 20 s
• Gravity (g) = 10 m/s²
STEP 1: Calculating Work Done (W)
Formula: Work Done = Force × Height = (m × g) × h
W = (15 kg × 10 m/s²) × 10 m
W = 150 N × 10 m = 1500 Joules
STEP 2: Calculating Power Consumed (P)
Formula: Power = Work Done / Time
P = 1500 J / 20 s = 75 Watts (75 W)
FINAL ANSWER:
• Work Done = 1500 Joules (1.5 kJ)
• Power Consumed = 75 Watts
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Question 11 An electric lamp of 60 W is used for 6 hours per day. Calculate the ‘units’ of electrical energy consumed in one day by the lamp. [Exam Favorite]
Answer:
================================================================================
NUMERICAL SOLUTION (COMMERCIAL ELECTRICAL ENERGY):
--------------------------------------------------------------------------------
GIVEN DATA:
• Power of lamp (P) = 60 W = 60 / 1000 kW = 0.06 kW
• Time per day (t) = 6 hours
CALCULATION:
Formula: Energy Consumed (in kWh / Units) = Power (in kW) × Time (in hours)
Energy = 0.06 kW × 6 h = 0.36 kWh (or 0.36 Units)
FINAL ANSWER:
The electrical energy consumed by the lamp in one day is 0.36 Units (0.36 kWh).
================================================================================
Question 12 State the Law of Conservation of Energy. Show that the total mechanical energy of a freely falling body remains constant at all points of its path. [Exam Favorite]
Answer:
- Law of Conservation of Energy: Energy can neither be created nor destroyed; it can only be transformed from one form into another. The total energy of an isolated system remains constant.
================================================================================ PROOF: CONSERVATION OF MECHANICAL ENERGY FOR A FREELY FALLING BODY -------------------------------------------------------------------------------- Let an object of mass m be dropped from rest (u = 0) at height h (Point A). 1. AT POINT A (Top Height h): • Velocity u = 0 ===> Kinetic Energy KE_A = 0 • Potential Energy PE_A = m × g × h • Total Energy E_A = KE_A + PE_A = 0 + m·g·h = m·g·h ----------------- (1) 2. AT POINT B (After falling distance x, height is h - x): • Velocity v_B: Using v² - u² = 2as ===> v_B² - 0 = 2·g·x ===> v_B² = 2·g·x • Kinetic Energy KE_B = 1/2 · m · v_B² = 1/2 · m · (2·g·x) = m·g·x • Potential Energy PE_B = m · g · (h - x) = m·g·h - m·g·x • Total Energy E_B = KE_B + PE_B = m·g·x + (m·g·h - m·g·x) = m·g·h ---- (2) 3. AT POINT C (Just touching the ground, height = 0): • Total distance fallen = h • Velocity v_C: v_C² - 0 = 2·g·h ===> v_C² = 2·g·h • Kinetic Energy KE_C = 1/2 · m · (2·g·h) = m·g·h • Potential Energy PE_C = m · g · 0 = 0 • Total Energy E_C = KE_C + PE_C = m·g·h + 0 = m·g·h ----------------- (3) CONCLUSION: From (1), (2), and (3): E_A = E_B = E_C = m·g·h = Constant. Total mechanical energy is strictly conserved at every instant of free fall. ================================================================================
Frequently Asked Questions (FAQs) – Class 9 Science Chapter 7
Question 1: Define 1 Joule of work. [Exam Favorite] Answer: One Joule (1 J) is defined as the amount of mechanical work done when a constant force of 1 Newton displaces an object through a distance of 1 metre in the direction of the force (1 J = 1 N · m).
Question 2: What is the work done on an object moving in a circular path? [Exam Favorite] Answer: The work done is Zero. The centripetal force is directed radially inward, while the instantaneous displacement is tangential, making the angle between force and motion 90° (Work = F × s × cos 90° = 0).
Question 3: Differentiate between Potential Energy and Kinetic Energy. [Exam Favorite] Answer:
- Kinetic Energy: The energy possessed by an object by virtue of its motion (KE = 1/2 × m × v²).
- Potential Energy: The energy possessed by an object by virtue of its position, height, or configuration (PE = m × g × h).
Question 4: What is the relationship between Kinetic Energy (KE) and Linear Momentum (p)? [Exam Favorite] Answer: Kinetic Energy is related to linear momentum by the equation: KE = (Momentum)² / (2 × mass) = p² / 2m.
Question 5: Define 1 Watt of power. [Exam Favorite] Answer: One Watt (1 W) is the rate of doing work when 1 Joule of energy is transferred or consumed in 1 second (1 W = 1 J / 1 s).
Question 6: How many Joules are present in 1 kilowatt-hour (1 kWh)? [Exam Favorite] Answer: 1 kWh = 1 kilowatt × 1 hour = 1000 Watts × 3600 seconds = 3,600,000 Joules = 3.6 × 10⁶ Joules (3.6 MJ).
Question 7: What is a Lever? Name its three critical points. [Exam Favorite] Answer: A lever is a rigid rod that turns freely about a fixed pivot. Its three points are: (i) Fulcrum (pivot point), (ii) Effort (point where force is applied), and (iii) Load (point where resistance is overcome).
Question 8: When is Mechanical Advantage (MA) greater than 1? Give an example. [Exam Favorite] Answer: MA is greater than 1 when the effort arm is longer than the load arm (Effort < Load), acting as a force multiplier (e.g., Crowbar, Wheelbarrow, Nutcracker).
Question 9: Can a simple machine have an efficiency of 100%? Why or why not? [Exam Favorite] Answer: No real machine can achieve 100% efficiency because a portion of input work is always wasted overcoming friction in moving joints and moving the machine’s own components, generating non-usable heat.
Question 10: What energy transformations occur in an electric hydroelectric dam? [Exam Favorite] Answer: Potential Energy of stored reservoir water → Kinetic Energy of falling water → Mechanical Rotational Energy of turbines → Electrical Energy via generators.
Question 11: A stretched bow possesses which form of energy? [Exam Favorite] Answer: Elastic Potential Energy, which converts into Kinetic Energy of the arrow when the bowstring is released.
Question 12: If the mass of a moving car is doubled while its velocity remains constant, what happens to its Kinetic Energy? [Exam Favorite] Answer: Because kinetic energy is directly proportional to mass (KE ∝ m), doubling the mass doubles the kinetic energy.
Question 13: If the speed of a moving car is doubled while its mass remains constant, what happens to its Kinetic Energy? [Exam Favorite] Answer: Because kinetic energy is proportional to the square of velocity (KE ∝ v²), doubling the speed increases the kinetic energy by four times (2² = 4).
Question 14: What is the work done by a student holding a 20 kg school bag on their shoulders for 10 minutes? [Exam Favorite] Answer: Zero Joules. Since there is no displacement of the school bag (s = 0), mechanical work done is zero.
Question 15: Why is a single fixed pulley used if its Mechanical Advantage is only 1? [Exam Favorite] Answer: Although a single fixed pulley does not multiply force (MA = 1), it changes the direction of the applied effort to a convenient downward direction, allowing the person to use their own body weight to lift the load easily.
Mastering the NCERT Solutions for Class 9 Science Chapter 7 (Exploration), “Work, Energy, and Simple Machines”, equips students with the computational tools, energy conversion theorems, and machine mechanics required for top performance in CBSE physics evaluations. Review the numerical box solutions, the master summary tables, and the 15 high-yield FAQs above to secure full marks in your examinations.