Chapter 5: Work, Energy and Power

Physics Part-1 • Class 11

Download PDF

Loading PDF...

Chapter Analysis

Intermediate20 pages • English

Quick Summary

The chapter 'Work, Energy and Power' delves into the definitions and concepts of work, energy, and power in physics. It highlights the work-energy theorem and explores kinetic energy, potential energy, and their interrelations. Additionally, the chapter discusses the conservation of mechanical energy, the concept of power, and the dynamics of collisions, illustrating these ideas with examples and exercises.

Key Topics

  • Work and its calculation
  • Kinetic energy and its relation to work
  • Work done by variable forces
  • Conservation of mechanical energy
  • Potential energy
  • Power and its calculation
  • Collisions and energy conservation in collisions

Learning Objectives

  • Understand and calculate work done by a constant force
  • Describe and apply the work-energy theorem
  • Differentiate between kinetic and potential energy
  • Apply the concept of conservation of mechanical energy
  • Calculate power in physical processes
  • Analyze collisions in terms of momentum and energy conservation

Questions in Chapter

The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: (a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket. (b) work done by gravitational force in the above case, (c) work done by friction on a body sliding down an inclined plane, (d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity, (e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest.

Page 88

A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the (a) work done by the applied force in 10 s, (b) work done by friction in 10 s, (c) work done by the net force on the body in 10 s, (d) change in kinetic energy of the body in 10 s.

Page 89

Given in Fig. 5.11 are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case.

Page 89

The potential energy function for a particle executing linear simple harmonic motion is given by V(x) = kx^2/2, where k is the force constant of the oscillator. For k = 0.5 N m-1, the graph of V(x) versus x is shown in Fig. 5.12. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m.

Page 89

Answer the following: (a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere? (b) Comets move around the sun in highly elliptical orbits. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why? (c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth? (d) In Fig. 5.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. 5.13(ii), he walks the same distance pulling the rope behind him. In which case is the work done greater?

Page 89

Additional Practice Questions

Calculate the work done by a force of 10 N acting parallel to a displacement of 5 m.

easy

Answer: The work done W is given by the formula W = F.d = 10 N * 5 m = 50 J (Joules).

A car accelerates from rest under a constant force. If its mass is 1,500 kg and it reaches a speed of 20 m/s in 30 seconds, what is the work done on the car?

medium

Answer: The kinetic energy gained by the car is given by KE = 0.5 * m * v^2 = 0.5 * 1500 kg * (20 m/s)^2 = 300,000 J. So, the work done is 300,000 J.

Explain how energy conservation applies to a pendulum swinging without friction.

medium

Answer: In a frictionless pendulum, mechanical energy is conserved. At the highest point, energy is all potential; at the lowest point, it is all kinetic. The total energy remains constant, showing energy conservation.

Describe the concept of power and how it relates to work and energy.

easy

Answer: Power is defined as the rate at which work is done or energy is transferred. It can be calculated as P = W/t where W is work and t is time. Its unit is the watt, which is one joule per second.

If an object moves in a circle at a constant speed, what can you say about the work done by the centripetal force?

hard

Answer: No work is done by the centripetal force because it acts perpendicular to the velocity and thus does not cause displacement in the direction of the force.