Here’s your guide to Class 9 Science Chapter 11: Work and Energy! Perfect for Class 9 students, parents, and teachers prepping for the 2025 CBSE exams, this post covers in-text questions and answers, exercise solutions, and detailed notes, explaining work, energy, and power with clarity. Let’s dive into the Class 9 Science Chapter 11 questions, answers, and notes!
Class 9 Science Chapter 11 Notes
1. Work
- Definition: Work = Force × Displacement × cosθ (W = F·s·cosθ), where θ is the angle between force and displacement.
- Unit: Joule (J), 1 J = 1 N·m.
- Conditions: No work if displacement is zero or force is perpendicular (θ = 90°).
2. Energy
- Definition: Capacity to do work.
- Forms:
- Kinetic Energy (KE): KE = ½mv² (motion energy).
- Potential Energy (PE): PE = mgh (stored energy due to position).
- Unit: Joule (J).
3. Law of Conservation of Energy
- Energy cannot be created or destroyed, only transformed (e.g., PE to KE in a falling object).
4. Power
- Definition: Rate of doing work (P = W/t).
- Unit: Watt (W), 1 W = 1 J/s.
5. Applications
- Lifting objects (work), falling water (energy conversion), engines (power).
These notes summarize the chapter—use them as a reference for the Q&As!
In-Text Questions and Answers: Page 148
Question 1: Explain in detail when work is considered to be done in scientific terms, including the specific conditions that must be met and examples where work is or isn’t done despite applied force.
Answer:
In scientific terms, work is considered done when a force causes an object to move through a displacement in the direction of the force, defined by the formula W = F·s·cosθ, where F is force, s is displacement, and θ is the angle between them. For work to occur, three conditions must be met: a force must be applied, the object must move (displacement ≠ 0), and the force must have a component along the displacement (cosθ ≠ 0). If θ = 0° (force and displacement aligned), work is maximum; if θ = 90°, work is zero.
Examples clarify this: pushing a box 5 m with 10 N force in the same direction does work (W = 10 × 5 × cos0° = 50 J). But holding a 10 kg bag stationary, despite a 98 N force against gravity, does no work because displacement is zero (W = 98 × 0 = 0 J). Similarly, a waiter carrying a tray horizontally applies upward force, but the tray moves forward (θ = 90°), so W = F·s·cos90° = 0 J. Contrast this with lifting the tray 2 m upward (W = 20 × 2 × cos0° = 40 J). These cases show work depends on movement and alignment, not just effort, distinguishing scientific work from everyday usage.
Question 2: Provide a comprehensive analysis of how lifting an object against gravity constitutes work, including the forces involved, the energy transfer, and a real-world example with calculations.
Answer:
Lifting an object against gravity constitutes work because a force is applied to cause displacement in the direction of the force, transferring energy to the object. The force involved is the upward force you exert, equal to the object’s weight (W = mg), opposing gravity’s downward pull (g = 9.8 m/s²). Work done is W = F·s·cosθ, where θ = 0° (force and displacement both upward), so W = mgh.
This work increases the object’s potential energy (PE = mgh), storing energy due to its higher position. For example, lifting a 2 kg book 1.5 m: weight = 2 × 9.8 = 19.6 N; work = 19.6 × 1.5 × cos0° = 29.4 J. This 29.4 J becomes the book’s PE at 1.5 m, convertible to kinetic energy if dropped. In real life, cranes lifting 500 kg steel beams 10 m do work (W = 500 × 9.8 × 10 = 49,000 J), transferring energy to position the beam for construction. The forces balance—your push counters gravity—and the energy shift from your muscles to the object’s PE exemplifies work’s role in altering an object’s state, critical in engineering and daily tasks.
Question 3: Discuss extensively why no work is done when a person pushes a wall that doesn’t move, even though effort is exerted, and relate this to the scientific definition and practical implications.
Answer:
No work is done when a person pushes a wall that doesn’t move because, scientifically, work requires displacement (W = F·s·cosθ), and if displacement (s) is zero, work is zero, regardless of the force applied. The definition hinges on movement: force alone isn’t enough; it must cause the object to shift in the force’s direction. When pushing a wall with, say, 200 N, the wall’s rigidity resists, producing an equal opposite force (Newton’s Third Law), so it stays put—s = 0, thus W = 200 × 0 = 0 J.
Despite muscle effort and energy expenditure (converted to heat), no mechanical work occurs because the wall’s position doesn’t change. This aligns with θ = 0° (force and intended motion aligned), but s = 0 overrides it. Practically, this explains why static efforts—like holding a heavy load without lifting—or pushing immovable objects (e.g., a car stuck in mud that doesn’t budge) don’t count as work in Physics, though they tire you out. It highlights a key distinction: biological energy use versus mechanical work, impacting how we analyze tasks in sports, construction, or machinery, where only successful motion counts.
In-Text Questions and Answers: Page 152
Question 1: Elaborate in detail on how energy enables an object to do work, including the types of energy involved and examples showing energy transformations in everyday activities.
Answer:
Energy enables an object to do work by providing the capacity to exert a force over a distance, with forms like kinetic (KE = ½mv²) and potential energy (PE = mgh) driving this process. Energy is the stored or active ability to cause change—work is its manifestation. Kinetic energy, from motion, lets a moving car (KE = ½ × 1000 × 20² = 200,000 J) push air aside or crash into a barrier, doing work. Potential energy, from position, allows a 5 kg rock at 10 m (PE = 5 × 9.8 × 10 = 490 J) to fall and crush something below, converting PE to KE then work.
Examples abound: stretching a bow stores PE in the string (e.g., 50 J), which becomes KE in the arrow, doing work to pierce a target. A child sliding down a slide transforms PE (mgh) at the top into KE, pushing against friction (work). In a windmill, wind’s KE turns blades, doing work to grind grain. These transformations—PE to KE, or chemical energy (food) to mechanical work (running)—show energy as the driver, enabling objects to alter their state or surroundings, a principle powering daily life and technology.
Question 2: Explain thoroughly how the kinetic energy of a moving object is calculated, including the formula, its derivation, and a practical example with step-by-step computation.
Answer:
The kinetic energy (KE) of a moving object is calculated using the formula KE = ½mv², where m is mass (kg) and v is velocity (m/s), measuring energy due to motion in joules (J). This formula derives from work-energy principles: work done to accelerate an object from rest to velocity v equals the force (F = ma) times distance (s). Using the equation v² = u² + 2as (u = 0), s = v²/2a, so work = F·s = ma × (v²/2a) = ½mv², which becomes KE. For a practical example, consider a 2 kg ball moving at 5 m/s:
- Step 1: Identify m = 2 kg, v = 5 m/s.
- Step 2: Square velocity: v² = 5 × 5 = 25.
- Step 3: Multiply by mass: m·v² = 2 × 25 = 50.
- Step 4: Take half: KE = ½ × 50 = 25 J.
This 25 J is the energy the ball can transfer—say, denting a wall. The derivation ties force to motion, and the example (a rolling ball, a speeding car) shows KE’s real-world relevance, like calculating crash energy or sports dynamics, emphasizing mass and speed’s squared impact.
Question 3: Analyze in depth why the potential energy of an object increases when it is raised to a higher position, including the role of gravity and a numerical example to illustrate the change.
Answer:
The potential energy (PE) of an object increases when raised to a higher position because it gains the capacity to do more work due to gravity, defined as PE = mgh, where m is mass, g is gravity (9.8 m/s²), and h is height. Gravity pulls downward, so lifting against it requires work (W = F·s = mg·h), which is stored as PE. The higher the object, the greater the distance it can fall, increasing its potential to convert PE into KE or other work. For a 3 kg box: at 2 m, PE = 3 × 9.8 × 2 = 58.8 J; at 5 m, PE = 3 × 9.8 × 5 = 147 J.
The increase (147 – 58.8 = 88.2 J) reflects the extra work done to lift it 3 m more (W = 3 × 9.8 × 3 = 88.2 J), matching the PE gain. This stored energy could break something if dropped. In reality, a water tank raised on a tower gains PE to supply homes, or a roller coaster at its peak holds more PE to thrill riders, showing how height amplifies gravitational potential, directly tied to Earth’s pull.
Exercise Questions and Answers: Page 166
Question 1: A force of 7 N acts on an object, causing a displacement of 8 m in the direction of the force. Calculate in detail the work done by the force, explaining the steps and the significance of the result.
Answer:
To calculate the work done by a 7 N force causing an 8 m displacement in its direction:
- Formula: W = F·s·cosθ.
- Given: F = 7 N, s = 8 m, θ = 0° (force and displacement aligned, cos0° = 1).
- Step 1: Multiply force and displacement: 7 × 8 = 56.
- Step 2: Apply angle: W = 56 × 1 = 56 J.
The work done is 56 joules. This means 56 J of energy is transferred to the object, perhaps increasing its KE (if moving) or PE (if lifted). For example, pushing a crate 8 m across a floor with 7 N adds 56 J to its motion, overcoming friction or accelerating it. The result’s significance lies in quantifying energy transfer, crucial for analyzing mechanical systems, like engines or human effort, showing how force and distance combine to produce tangible effects.
Question 2: An object of mass 15 kg is moving with a uniform velocity of 4 m/s. Compute in detail the kinetic energy possessed by the object, including all steps and an explanation of what this energy represents.
Answer:
To compute the kinetic energy (KE) of a 15 kg object moving at 4 m/s:
- Formula: KE = ½mv².
- Given: m = 15 kg, v = 4 m/s.
- Step 1: Square velocity: v² = 4 × 4 = 16.
- Step 2: Multiply by mass: 15 × 16 = 240.
- Step 3: Take half: KE = ½ × 240 = 120 J.
The kinetic energy is 120 joules. This represents the energy the object has due to its motion—120 J it could transfer, like stopping by friction or impacting something. For instance, a 15 kg cart at 4 m/s could push debris aside or dent a wall with this energy. KE’s dependence on v² (not just v) means speed has a big effect—doubling to 8 m/s would quadruple KE to 480 J—highlighting its role in dynamics, from vehicles to falling objects.
Class 9 Science Chapter 11: Work and Energy ties force to energy transformations, key for your 2025 CBSE exams. With these notes and detailed Q&As, you’re set to succeed. Check NCERT solutions for more practice, and comment your doubts—we’re here to help!
Class 9 Science Chapter 12: Sound – Questions, Answers, Exercise Solutions, and Notes
For all Class 9 Science Chapters: Notes and Question Answers. Click Here.