Chapter 4

Chapter 4: Work and Energy

Chapter Overview

Physics is the study of the natural world around us. It involves the study of matter, energy, and the fundamental laws that govern the behavior of the physical universe. In this chapter, we will explore the fundamental principles of physics that govern the behavior of physical systems, with a focus on the concepts of work, energy, and power.

Learning Objectives

• Understand the concept of work and energy.
• Learn to calculate work done by a constant force.
• Understand the concept of kinetic energy and potential energy.
• Learn to calculate the work-energy theorem.
• Understand the concept of power.

Important Concepts

Work Done by a Constant Force

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

W = F × s × cos(θ)

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

Real-World Application: Consider a scenario where a person is pushing a block of mass 10 kg up a frictionless incline with a force of 20 N. If the block is displaced by 5 m in the direction of the force, the work done can be calculated using the formula:

W = F × s × cos(θ) W = 20 N × 5 m × cos(0°) W = 100 J

This means that the person has done 100 J of work in pushing the block up the incline.

Kinetic Energy

Kinetic energy is the energy of motion. It is the energy an object possesses due to its motion. The kinetic energy of an object is given by the formula:

K = (1/2) × m × v^2

where K is the kinetic energy, m is the mass of the object, and v is its velocity.

Real-World Application: Consider a scenario where a car of mass 1500 kg is traveling at a speed of 50 m/s. The kinetic energy of the car can be calculated using the formula:

K = (1/2) × m × v^2 K = (1/2) × 1500 kg × (50 m/s)^2 K = 375,000 J

This means that the car has a kinetic energy of 375,000 J due to its motion.

Potential Energy

Potential energy is the energy an object possesses due to its position or configuration. It is the energy an object has due to its height, compression, or tension. The potential energy of an object is given by the formula:

U = m × g × h

where U is the potential energy, m is the mass of the object, g is the acceleration due to gravity, and h is its height.

Real-World Application: Consider a scenario where a block of mass 5 kg is placed at a height of 10 m above the ground. The potential energy of the block can be calculated using the formula:

U = m × g × h U = 5 kg × 9.8 m/s^2 × 10 m U = 490 J

This means that the block has a potential energy of 490 J due to its position.

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. Mathematically, it is represented as:

W_net = ΔK

where W_net is the net work done and ΔK is the change in kinetic energy.

Real-World Application: Consider a scenario where a block of mass 10 kg is pushed up a frictionless incline with a force of 20 N. If the block is displaced by 5 m in the direction of the force, the work done can be calculated using the formula:

W = F × s × cos(θ) W = 20 N × 5 m × cos(0°) W = 100 J

If the block is then released and allowed to roll down the incline, its kinetic energy will increase due to the loss of potential energy. The work-energy theorem states that the net work done on the block is equal to the change in its kinetic energy.

Power

Power is the rate at which work is done or energy is transferred. It is measured in watts (W) and is given by the formula:

P = W / t

where P is the power, W is the work done, and t is the time taken.

Real-World Application: Consider a scenario where a person is lifting a block of mass 10 kg up a frictionless incline with a force of 20 N. If the block is displaced by 5 m in the direction of the force in 2 s, the power can be calculated using the formula:

P = W / t P = 100 J / 2 s P = 50 W

This means that the person is doing 50 W of power in lifting the block up the incline.

Advanced Sections

Deep-Dive Case Studies and Real-Life Applications

• Case Study 1: A car of mass 1500 kg is traveling at a speed of 50 m/s. The kinetic energy of the car can be calculated using the formula:

  K = (1/2) × m × v^2 K = (1/2) × 1500 kg × (50 m/s)^2 K = 375,000 J

• Case Study 2: A block of mass 5 kg is placed at a height of 10 m above the ground. The potential energy of the block can be calculated using the formula:

  U = m × g × h U = 5 kg × 9.8 m/s^2 × 10 m U = 490 J

Step-by-Step Problem Solving Strategies & Detailed Proofs

• Problem 1: A person is pushing a block of mass 10 kg up a frictionless incline with a force of 20 N. If the block is displaced by 5 m in the direction of the force, calculate the work done.

  1. Identify the given information: m = 10 kg, F = 20 N, s = 5 m.
  2. Use the formula W = F × s × cos(θ) to calculate the work done.
  3. W = F × s × cos(0°)
  4. W = 20 N × 5 m × cos(0°)
  5. W = 100 J

• Problem 2: A car of mass 1500 kg is traveling at a speed of 50 m/s. Calculate the kinetic energy of the car.

  1. Identify the given information: m = 1500 kg, v = 50 m/s.
  2. Use the formula K = (1/2) × m × v^2 to calculate the kinetic energy.
  3. K = (1/2) × 1500 kg × (50 m/s)^2
  4. K = 375,000 J

Higher-Order Thinking Skills (HOTS) Questions

• Question 1: A block of mass 5 kg is placed at a height of 10 m above the ground. If the block is allowed to roll down the incline, what is the change in its kinetic energy?

  1. Identify the given information: m = 5 kg, h = 10 m.
  2. Use the formula U = m × g × h to calculate the potential energy.
  3. U = 5 kg × 9.8 m/s^2 × 10 m
  4. U = 490 J
  5. The change in kinetic energy is equal to the loss of potential energy.
  6. ΔK = -U
  7. ΔK = -490 J

• Question 2: A person is lifting a block of mass 10 kg up a frictionless incline with a force of 20 N. If the block is displaced by 5 m in the direction of the force in 2 s, what is the power?

  1. Identify the given information: m = 10 kg, F = 20 N, s = 5 m, t = 2 s.
  2. Use the formula P = W / t to calculate the power.
  3. P = W / t
  4. P = (F × s × cos(θ)) / t
  5. P = (20 N × 5 m × cos(0°)) / 2 s
  6. P = 50 W

Previous Year Questions (PYQs) with solutions

• PYQ 1: A car of mass 1500 kg is traveling at a speed of 50 m/s. Calculate the kinetic energy of the car.

  1. Use the formula K = (1/2) × m × v^2 to calculate the kinetic energy.
  2. K = (1/2) × 1500 kg × (50 m/s)^2
  3. K = 375,000 J

• PYQ 2: A block of mass 5 kg is placed at a height of 10 m above the ground. Calculate the potential energy of the block.

  1. Use the formula U = m × g × h to calculate the potential energy.
  2. U = 5 kg × 9.8 m/s^2 × 10 m
  3. U = 490 J

NCERT Textbook Questions & Detailed Answers

Question 1: A person is pushing a block of mass 10 kg up a frictionless incline with a force of 20 N. If the block is displaced by 5 m in the direction of the force, calculate the work done.

1. Identify the given information: m = 10 kg, F = 20 N, s = 5 m.
2. Use the formula W = F × s × cos(θ) to calculate the work done.
3. W = F × s × cos(0°)
4. W = 20 N × 5 m × cos(0°)
5. W = 100 J

Question 2: A car of mass 1500 kg is traveling at a speed of 50 m/s. Calculate the kinetic energy of the car.

1. Identify the given information: m = 1500 kg, v = 50 m/s.
2. Use the formula K = (1/2) × m × v^2 to calculate the kinetic energy.
3. K = (1/2) × 1500 kg × (50 m/s)^2
4. K = 375,000 J

Question 3: A block of mass 5 kg is placed at a height of 10 m above the ground. Calculate the potential energy of the block.

1. Identify the given information: m = 5 kg, h = 10 m.
2. Use the formula U = m × g × h to calculate the potential energy.
3. U = 5 kg × 9.8 m/s^2 × 10 m
4. U = 490 J