Chapter 6

Chapter 6: Work, Energy, and Power

Chapter Overview

Physics is a fundamental branch of science that deals with the study of matter, energy, and the fundamental forces of nature. In Class 11, we have been studying the basics of physics, and in this chapter, we will delve deeper into the concepts of work, energy, and power. This chapter is essential as it lays the foundation for understanding more complex topics in physics.

Learning Objectives

• Define work and its units.
• Explain the concept of energy and its types.
• Understand the concept of power and its relation to work and energy.
• Derive the formula for work done by a constant force.
• Apply the concept of work, energy, and power to real-life situations.
• 💡 Pro Tip: To better understand the concepts, try to relate them to everyday life situations, such as lifting a heavy object or riding a bicycle.

Important Concepts

Work Done by a Constant Force

Work done by a constant force is defined as the product of the force applied and the displacement of the object in the direction of the force. Mathematically, it is represented as:

W = F × d × cos(θ)

where W is the work done, F is the force applied, d is the displacement, and θ is the angle between the force and displacement.

• 🧠 Trick to Remember: Use the mnemonic "WFD" to remember the formula, where W is work, F is force, and D is displacement.

Deep-Dive Case Studies and Real-Life Applications

• Example 1: A person is lifting a heavy box from the ground to a height of 2 meters. If the force applied is 100 N and the angle between the force and displacement is 30°, calculate the work done.
• Solution: Using the formula W = F × d × cos(θ), we get W = 100 N × 2 m × cos(30°) = 173.2 J.
• Real-Life Application: This concept is essential in understanding the energy required to lift heavy objects in various fields such as construction, manufacturing, and logistics.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. It is defined as half the product of the mass and the square of the velocity of the object.

K.E. = 1/2 mv^2

• Example: A car moving at a speed of 60 km/h has a certain amount of kinetic energy due to its motion.

Real-Life Applications

• Example 2: A cyclist is moving at a speed of 20 km/h on a flat road. If the mass of the cyclist is 70 kg, calculate the kinetic energy of the cyclist.
• Solution: Using the formula K.E. = 1/2 mv^2, we get K.E. = 1/2 × 70 kg × (20 km/h)^2 = 1400 J.
• Real-Life Application: This concept is essential in understanding the energy required to move objects in various fields such as transportation, sports, and recreation.

Potential Energy

Potential energy is the energy an object possesses due to its position or configuration. It is defined as the product of the mass and the height of the object.

P.E. = mgh

• Example: A ball at the top of a hill has potential energy due to its height, which is converted to kinetic energy as it rolls down the hill.

Real-Life Applications

• Example 3: A water tank is filled with water to a height of 10 meters. If the mass of the water is 1000 kg, calculate the potential energy of the water.
• Solution: Using the formula P.E. = mgh, we get P.E. = 1000 kg × 9.8 m/s^2 × 10 m = 98000 J.
• Real-Life Application: This concept is essential in understanding the energy required to lift water in various fields such as water supply, irrigation, and hydroelectric power generation.

Power

Power is the rate at which work is done or energy is transferred. It is defined as the ratio of work done to the time taken to do the work.

P = W/t

• 💡 Pro Tip: Power is an important concept in understanding the efficiency of machines and devices.

Real-Life Applications

• Example 4: A machine is doing work at a rate of 100 W. If the time taken to do the work is 10 seconds, calculate the work done.
• Solution: Using the formula P = W/t, we get W = P × t = 100 W × 10 s = 1000 J.
• Real-Life Application: This concept is essential in understanding the efficiency of machines and devices in various fields such as engineering, manufacturing, and energy production.

Work-Energy Theorem

The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy.

W = ΔK.E.

• Example: When a car accelerates from rest to a certain speed, the work done by the engine is equal to the change in the car's kinetic energy.

Deep-Dive Case Studies and Real-Life Applications

• Example 5: A car is accelerating from rest to a speed of 60 km/h. If the mass of the car is 1500 kg, calculate the work done by the engine.
• Solution: Using the formula W = ΔK.E., we get W = 1/2 mv^2 - 0 = 1/2 × 1500 kg × (60 km/h)^2 = 270000 J.
• Real-Life Application: This concept is essential in understanding the energy required to accelerate objects in various fields such as transportation, sports, and recreation.

Key Definitions

• Work: The product of the force applied and the displacement of the object in the direction of the force.
• Energy: The capacity to do work.
• Power: The rate at which work is done or energy is transferred.
• Efficiency: The ratio of output work to input work, often expressed as a percentage.

Important Terms

Term	Meaning
Kinetic Energy	The energy an object possesses due to its motion.
Potential Energy	The energy an object possesses due to its position or configuration.
Work-Energy Theorem	The net work done on an object is equal to the change in its kinetic energy.
Efficiency	The ratio of output work to input work.

Important Formulas

• W = F × d × cos(θ)
• K.E. = 1/2 mv^2
• P.E. = mgh
• P = W/t
• W = ΔK.E.
• Efficiency = (Output Work / Input Work) × 100%

Step-by-Step Problem Solving Strategies & Detailed Proofs

Problem 1: Work Done by a Constant Force

A force of 100 N is applied to an object, displacing it by 2 meters in the direction of the force. If the angle between the force and displacement is 30°, calculate the work done.

Solution:

1. Identify the given values: F = 100 N, d = 2 m, θ = 30°
2. Use the formula W = F × d × cos(θ) to calculate the work done.
3. Substitute the values into the formula: W = 100 N × 2 m × cos(30°) = 173.2 J

Problem 2: Kinetic Energy

A car is moving at a speed of 60 km/h. If the mass of the car is 1500 kg, calculate the kinetic energy of the car.

Solution:

1. Identify the given values: m = 1500 kg, v = 60 km/h
2. Use the formula K.E. = 1/2 mv^2 to calculate the kinetic energy.
3. Substitute the values into the formula: K.E. = 1/2 × 1500 kg × (60 km/h)^2 = 270000 J

Problem 3: Potential Energy

A water tank is filled with water to a height of 10 meters. If the mass of the water is 1000 kg, calculate the potential energy of the water.

Solution:

1. Identify the given values: m = 1000 kg, h = 10 m
2. Use the formula P.E. = mgh to calculate the potential energy.
3. Substitute the values into the formula: P.E. = 1000 kg × 9.8 m/s^2 × 10 m = 98000 J

Higher-Order Thinking Skills (HOTS) Questions

1. A force of 200 N is applied to an object, displacing it by 3 meters in the direction of the force. If the angle between the force and displacement is 45°, calculate the work done.
2. A car is accelerating from rest to a speed of 80 km/h. If the mass of the car is 1800 kg, calculate the work done by the engine.
3. A water tank is filled with water to a height of 15 meters. If the mass of the water is 1200 kg, calculate the potential energy of the water.

Previous Year Questions (PYQs) with solutions

1. A force of 300 N is applied to an object, displacing it by 4 meters in the direction of the force. If the angle between the force and displacement is 60°, calculate the work done. (Solution: W = 300 N × 4 m × cos(60°) = 600 J)
2. A car is moving at a speed of 70 km/h. If the mass of the car is 1600 kg, calculate the kinetic energy of the car. (Solution: K.E. = 1/2 × 1600 kg × (70 km/h)^2 = 392000 J)
3. A water tank is filled with water to a height of 12 meters. If the mass of the water is 900 kg, calculate the potential energy of the water. (Solution: P.E. = 900 kg × 9.8 m/s^2 × 12 m = 105840 J)

NCERT Textbook Questions & Detailed Answers

Question 1: A force of 500 N is applied to an object, displacing it by 5 meters in the direction of the force. If the angle between the force and displacement is 30°, calculate the work done.

Solution: W = 500 N × 5 m × cos(30°) = 2083.3 J

Question 2: A car is moving at a speed of 50 km/h. If the mass of the car is 1200 kg, calculate the kinetic energy of the car.

Solution: K.E. = 1/2 × 1200 kg × (50 km/h)^2 = 150000 J

Question 3: A water tank is filled with water to a height of 8 meters. If the mass of the water is 600 kg, calculate the potential energy of the water.

Solution: P.E. = 600 kg × 9.8 m/s^2 × 8 m = 47040 J