NCERT Solution (Class 11) > Physics Chapter 6 Work, Energy And Power

NCERT Solution (Class 11) > Physics Chapter 6 Work, Energy And Power

NCERT Solution Class 11 Physics Chapter 6 Work, Energy And Power: National Council of Educational Research and Training Class 11 Physics Chapter 6 Solution – Work, Energy And Power. Free PDF Download facility available at our website.

Board

NCERT
Class

11

Subject

Physics
Chapter

6

Chapter Name

Work, Energy And Power
Topic

Exercise Solution

Work, Energy And Power Chapter all Questions and Numericals Solution

 

Chapter6 – Work energy and power

6.1 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.

ANSWER-

in general sign of work done depends upon direction of work done and displacements . if both are in same direction then work done is positive otherwise negative.

(a)  Positive

Here force by a man on bucket and displacement of bucket both are in the same direction which is upward. Hence, the sign of work done is positive. In this case, the work is done on the bucket.

(b) Negative

Here the direction of gravitational force is always vertically downward and displacement vertically upward as bucket is going up are opposite to each other. Hence, the sign of work done is negative.

(c) Negative

Since the direction of frictional force is always opposite to the direction of motion, the work done by frictional force is negative.

(d) Positive

Here the body is moving on a rough horizontal plane. To maintain the uniform velocity force must be applied opposite to frictional force and in the direction of motionhence the work done is positive.

(e) Negative

The resistive force of air is in the opposite direction to the direction of the pendulum. Hence, the work done is negative.

 

6.3 Given in Fig. 6.11 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. Think of simple physical contexts for which these potential energy shapes are relevant.

ANSWER-

Case 1

Total energy of a system is given by the relation:

E = P.E. + K. E.

∴K.E. = E – P.E.

Kinetic energy of a body is a positive quantity. It cannot be negative. Therefore, the

particle will not exist in a region where K.E. becomes negative. From graph x > a;

particle will not exists as the potential energy of the particle for x > a becomes greater than total

energy (E) and kinetic energy becomes negative.

Case 2

Total energy of a system is given by the relation:

E = P.E. + K. E.

∴K.E. = E – P.E.

here, the potential energy (V0) is greater than total energy (E) in all regions. Kinetic energy become negative.

Hence, the particle will not exist in this region.

Case 3

x > a and x < b; –V1

, the condition regarding the positivity of K.E. is satisfied only in the region between  a<x <b. hence for the region  x <a and b<x particle does not exist.

Minimum energy needed is –V1 .

Case4

From graph potential energy is greater than total energy in the region between –b/2 to b/2 .for remaining region kinetic energy will be negative and particle does not exist. Minimum energy needed is –V1 .

 

6.4) The potential energy function for a particle executing linear simple harmonic motion is given by

V (x) =kx2/2, 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. 6.12. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m.

ANSWER-

 

6.5) 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. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. 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. 6.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig.6.13(ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater?

ANSWER-

(a) Heat energy required for burning of rocket comes from rocket .as rocket goes up it counters the friction .it has to do work against the friction which results in decrease in kinetic energy of rocket. This work done against friction results into heat.

(b) We know that work done by conservative force in closed path is equal to zero. Gravitational force is also a conservative force hence its work done is independent on path followed and equal to zero in closed path.

(c) The total energy of the system remains constant. When an artificial satellite, orbiting around earth, moves closer to earth, its potential energy decreases due to decrease in height. This reduction in potential energy is responsible in an increase in K.E. to make energy conserved. Hence, due to increase in kinetic energy, the velocity of the satellite increases.

(d) From first scenario,

Mass, m = 15 kg

Displacement, s = 2 m

Work done is given by,

W = f x s x cos (θ)

W = mg x s x cos (θ)

 

6.6) Underline the correct alternative:

(a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered.

(b) Work done by a body against friction always results in a loss of its kinetic/potential energy.

(c) The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system.

(d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies.

ANSWER-

(a) When a conservative force does positive work on a body, the potential energy of the body decreases.

(b) Work done by a body against friction always results in a loss of its kinetic energy.

(c) The rate of change of total momentum of a many-particle system is proportional to the external force.

(d) In an inelastic collision of two bodies, the quantities which do not change after the collision is the total linear momentum.

 

6.7) State if each of the following statements is true or false. Give reasons for your answer.

(a) In an elastic collision of two bodies, the momentum and energy of each body is conserved.

(b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present.

(c) Work done in the motion of a body over a closed loop is zero for every force in nature.

(d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.

ANSWER-

(a) False. In an elastic collision, the total momentum and energy of system is conserved but momentum of individual need not be conserved.

(b) False.as the external forces that have the ability to do work. Hence, external forces are able to change the energy of a system.

(c) False. for conservation force only The work done in the motion of a body over a closed loop is zero.

(d) True. In an inelastic collision, there is always loss of energy hence the final kinetic energy is always less than the initial kinetic energy of the system.

 

6.8) Answer carefully, with reasons:

(a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact)?

(b) Is the total linear momentum conserved during the short time of an elastic collision of two balls?

(c) What are the answers to (a) and (b) for an inelastic collision?

(d) If the potential energy of two billiard balls depends only on the separation distance between their centers, is the collision elastic or inelastic? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).

ANSWER-

(a)In an elastic collision the total kinetic energy of the system is conserved. But at the time of collision when they are in contact, the kinetic energy of the balls will get converted into potential energy.

(b)The total linear momentum of the system always remains conserved for elastic collision.

(c)for inelastic collision, there is always a loss of kinetic energy and if we consider momentum then it will remain conserved no matter the collision is elastic or inelastic in nature.

(d)here in this case potential energy depend on the separation between the centres of the billiard balls hence the forces are conservative in nature Hence, the collision is elastic.

 

6.9) A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to

(i) t1/2 (ii) t (iii) t3/2   (iv) t2

 

 

6.10) A body is moving unidirectional under the influence of a source of constant power. Its displacement in time tis proportional to

(i) t1/2 (ii) t (iii) t3/2  (iv) t2

Answer –

 

6.11) 

 

6.13) 

 

 

6.14) 

 

6.16) Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V. If the collision is elastic, which of the following (Fig. 6.14) is a possible result after collision?

ANSWER-

For an elastic to be elastic collision, the total kinetic energy of a system should remain conserved before and after collision. Let’s check for conservation of kinetic energy.

For mass of each ball bearing m, Total kinetic energy of the system before collision

Total initial kinetic energy =  mv2 + 2m(0)2

                                          =  mv2.

For case i:

Final kinetic energy = m(0)2 +  2m (v/2)2

                                    =  m(v)2

As kinetic energy is not conserved hence it is not the case of elastic collision.

For case ii:

Final kinetic energy = 2m(0)2 +  m (v)2

                                    =  m (v)2

As kinetic energy is conserved hence it is the case of elastic collision.

For case iii:

Final kinetic energy = + 3m (v/3)2 =  m(v)2

As kinetic energy is not conserved hence it is not the case of elastic collision.

 

In case you are missed :- Previous Chapter Solution

 

6.17) 

 

6.19)

 

6.21)

 

6.22) 

 

6.23)

 

6.25)

 

6.26)

 

6.27)

 

6.28) 

 

6.29) Which of the following potential energy curves in Fig. 6.18 cannot possibly describe the elastic collision of two billiard balls? Here r is the distance between centres of the balls.

ANSWER-

We know that The potential energy of a system of two masses is inversely proportional to the separation between them(r). The potential energy of the system of the two balls will decrease as they come closer to each other and  will become zero when the two balls touch each other where  r = 2R, where R is the radius of each billiard ball.

The potential energy curves given in figures (i), (ii), (iii), (iv), and (vi) do not satisfy these two conditions.

 

6.30) Consider the decay of a free neutron at rest : n —> p + e

Show that the two-body decay of this type must necessarily give an electron of fixed energy and, therefore, cannot account for the observed continuous energy distribution in the b-decay of a neutron or a nucleus (Fig. 6.19).

[Note: The simple result of this exercise was one among the several arguments advanced by W.Pauli to predict the existence of a third particle in the decay products of b-decay. This particle is known as neutrino. We now know that it is a particle of intrinsic spin ½ (like e, p or n), but is neutral, and either massless or having an extremely small mass(compared to the mass of electron) and which interacts very weakly with matter. The correct decay process of neutron is : n –> p + e + n ] .

ANSWER-

The decay process of free neutron at rest is given by

n —> p + e

by law of conservation of momentum

momentum of neutron = momentum of proton + momentum of electron

From above expression it is proved that emitted electron is having constant velocity and hence the constant energy hence we can’t not account for the continuous energy distribution in b-decay.

In case you are missed :- Next Chapter Solution

Updated: April 17, 2023 — 11:29 am

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