NCERT Solutions for Class 11 Physics | Chapter 6 | Work, Energy and Power

1 Star2 Stars3 Stars4 Stars5 Stars (3 votes, average: 4.67 out of 5)

Loading...Here we are providing free NCERT Solutions for Class 11 Physics, Chapter 6, Work, Energy and Power.

Salient Features of Solutions for Class 11 NCERT Physics, Chapter 6 Work, Energy and Power provided by SciPhy.in.

  • SciPhy.in is managed by ex-IITians and experts from reputed institutes.
  • Questions are solved by experienced physics teachers, assistant professors and field experts.
  • We have shown proper free body diagram, whenever needed.
  • Solutions are explained in simple English for better understandings.
  • Solutions are accompanied with related links and additional thoughts.

 

Question 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:
(a) Positive [ In this case, force (vertically upward) and displacement (vertically upward) have the same directions. ]
(b) Negative [ In this case, the force (vertically downward) and displacement (vertically upward) are in opposite directions. ]
(c) Negative [ The direction of the frictional force is always opposite to the relative motion. ]
(d) Positive [ The displacement of the body is along the applied force. ]
(e) Negative [ The direction of the resistive force is opposite to the direction of motion of the pendulum. ]

Question 6. 2. 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 and interpret your results.

Answer:
Mass of the body, m = 2 kg
Applied force, F = 7 N
Coefficient of kinetic friction, \mu = 0.1
Initial velocity, u = 0
Time, t = 10 s
Gravitational acceleration, g = 10 m/s^2
Normal reaction N acts vertically upward.
As the object does not have any acceleration in the vertical direction, total force in the vertical direction must be zero. We conclude N-mg=0 or N=mg
The frictional force is given as:
Fk = \mu N= \mu mg= 2 N (acts opposite to the direction of motion)
Thus, the effective force, along the direction of motion, acting on the object is F-F_k= 7-2 =5 N
The acceleration produced in the body due to this effective force is
a = 5/2 = 2.5 ms^{-2}
The distance travelled by the body in 10 sec due to this acceleration can be found from
s = ut + (1/2)at^2 = 0 + (1/2) × 2.5 × (10)^2 = 125 m
(a) Work done by the applied force, W_a = Fs = 7×125 = 875 J
(b) Work done by the frictional force, W_F= F_K s = –250 J
(c) Work done by the net force, W_{net}= 5×125 = 625 J
(d) Final velocity of the object is given by
v = u + at = 0 + 2.5 × 10 = 25 m/s
Change in kinetic energy = (1/2) mv^2 – (1/2) mu^2 = (1/2) x 2(v^2 - u^2) = (25)^2 - 0^2 = 625 J

Conclusion: Change in kinetic energy = Work done by the net force

Question 6.3. Given figures 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 some physical contexts for which these potential energy shapes are relevant.Solutions for Class 11 NCERT Physics, Chapter 6 Work, Energy and Power provided by SciPhy.in fig 1

Answer:
Total energy of a system is given by the relation: E = P.E. + K. E.
Therefore K.E. = E – P.E.
Kinetic energy of a body is a positive quantity and can never be negative. Therefore, the particle will not exist in a region where K.E. becomes negative. Alternatively, the particle will not exist in a region where P.E becomes greater than the total energy.

(a) For x > a P.E. (V_0) > E Hence, the object cannot exist in this region.
(b) As the potential energy is always greater than the total energy the particle can not exist anywhere.
(c) Particle can be found only in the region x > a and x < b as P.E < E. Particle can not be found elsewhere.
(d) Particle can not be found in the region -b/2 <x<-a/2 and a/2< x < b/2

As the minimum value of the kinetic energy is zero, the total energy of the particle must be greater than -V_1. So, the minimum total energy the particle must be -V_1.
N.B. You will be more familiar with these potential diagrams, popularly known as potential well in the course of Quantum Mechanics. Do a Google search with “finite potential well in quantum mechanics”.

Question 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 Nm-1 , the graph of V (x) versus x is shown in Fig. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m.Solutions for Class 11 NCERT Physics, Chapter 6 Work, Energy and Power provided by SciPhy.in fig 2

Answer:
Total energy E = Potential energy + Kinetic energy.
At the extremum kinetic energy becomes zero and E= Potential energy
Given values are, force constant k = 0.5 Nm-1 and total energy of particle E = 1J.
Let the maximum distance be x_m, Thus \frac{1}{2}kx_m^2 = E or
x_m=\sqrt{\frac{2E}{k}}=\sqrt{4}=2
Thus the maximum distance will be ±2m before turning back.

Question 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.(i), the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. (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?Solutions for Class 11 NCERT Physics, Chapter 6 Work, Energy and Power provided by SciPhy.in fig 3

Answer:
(a) Heat energy required for burning of casing of rocket comes from the rocket itself.
(b) This is because gravitational force is a conservative force. For a conservative force the work done over a path is the decrease in potential energy. Therefore work done depends only on the final and initial position. For a close loop which is zero.
(c) An artificial satellite gradually loses its energy due to dissipation against atmospheric resistance. This leads to the decrease in potential energy. However the kinetic energy increases slightly leading to the increase in speed.
(d) In the first case, force applied on the mass by the man is in vertically upward direction but the distance travelled is along the horizontal direction. As the angle between the force and displacement is 90^o, work done is zero.
In the second case, the directions of force and displacement are in the horizontal direction and there will be finite work done.
W = 15 x 10 x 2 x 1 = 300 joule.
Thus, the work done in case (ii) is greater.

Hope this NCERT Solutions for Class 11 Physics, Chapter 6 Work, Energy and Power provided by SciPhy.in was helpful for you. Please share with your friends.

Question 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 of 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) Potential energy of the body decreases, as the work is done by losing potential energy.
(b) Work done by a body against friction always results in a loss of its kinetic energy, as the friction does work against the relative motion.
(c) According to Newton’s law if there is no external force the total momentum of a body can not change. A many particle system can be considered as a superposition of several single body where the law of motion is valid individually. Therefore to change the total momentum of many particle system we need an external force.
(d) Linear momentum is conserved in any type of collision. Therefore linear momentum is conserved here. Also the total energy can not be destroyed thus conserved.
N.B: Kinetic energy is conserved in elastic collision only.

Question 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, the total momentum, total energy and total kinetic energy are conserved individually. However, there is no such constraint to conserve momentum or energy of each body.
(b) False, If there is no external force on the system, only then the total energy is conserved.
(c) False, This is true only for a conservative force field like gravitational or electrostatic force. Friction is a non-conservative force and the above statement is not true. Can you name another non-conservative force?
(d) True. You can check the case where two particles stick together after collision.

Question 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) No. During the collision (i.e., when they are in contact) total kinetic energy converts into potential energy. [they become stationary for a moment]
(b) Yes, Total momentum is always conserved when there is no external force.
(c) Total momentum is conserved but kinetic energy is not conserved even after the collision.
(d) It will be elastic collision.

Question 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) t^\frac{1}{2} (ii) t (iii) t^\frac{3}{2} (iv) t^2

Answer:
The body was initially at rest and underwent one-dimensional motion with constant acceleration. Thus the final velocity v = u + at where a is the constant acceleration and u is zero. Thus
v = 0 + at = at
Power (P) = F\times v
Therefore, P = (ma) x at = ma^2t
Correct answer (ii) P\alpha t

Question 6. 10. A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to (i) t^\frac{1}{2} (ii) t (iii) t^\frac{3}{2} (iv) t^2

Answer:
In an alternative form, we can express Power P=\frac{1}{2}\frac{d}{dt}(mv^2)
Here, P=\frac{1}{2}\frac{d}{dt}(mv^2)= Const.
Now, integrating both side with respect to t mv^2=ct
Or, v=\frac{dx}{dt} \propto \sqrt{t}
Again integrating w.r.t t
x \propto t^{\frac{3}{2}}

Please share this NCERT Solutions for Class 11 Physics, Chapter 6, Work, Energy and Power by SciPhy.in with your friends.

Question 6. 11. A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by \vec{F}=-\hat{i}+2\hat{j}+3\hat{k} N where i, j, k, are unit vectors along the x- y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis?

Answer:
Work done = \vec{F}\cdot \vec{S}=(-\hat{i}+2\hat{j}+3\hat{k}) \cdot 4\hat{k}=12 J

Question 6. 12. An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton? Obtain the ratio of their speeds, (electron mass = 9.11 x 10-31 kg, proton mass = 1.67 x 10-27 kg, 1 eV= 1.60 x 1019J).

Answer:
K_e= 10 keV and K_p= 100 keV
m_e=9.11 \times 10^{-31}kg m_p=1.67 \times 10^{-27} kg
K=\frac{1}{2}mv^2 or v=\sqrt{\frac{2K}{m}}
Therefore, \frac{v_e}{v_p}= \sqrt{\left \frac{K_e}{K_p}}{\frac{m_p}{m_e}\right }=13.54

Question 6. 13. A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 ms^{-1} ?

Answer:
Radius of rain water r = 2 \times 10^{-3} m.
Density of water, \rho = 10^3 kg/m^3
Mass of raindrop =\frac{4}{3} \pi r^3 \times \rho = 3.35 \times 10^-5 kg
Work done by the gravitational force in the first half is
W_1=mgh/2= 0.08375 Joule [g=10 m/s^2]
Work done by the gravitational force in the second half is
W_2=mgh/2= 0.08375 Joule
Notice that the work done in this case id not depend on the velocity of the raindrop.

Total work done in absence of any resistive force
W=W_1+W_2= 0.1675 J

Actual work done in presence of resistive force is gain in kinetic energy
K.E. = \frac{1}{2}mv^2 = 1.675 \times 10^{-3}J

Thus, work done by the resistive forces is W – K.E = 0.164 - 1.675 \times 10^{-3} = 0.165825 joule.

Question 6. 14. A molecule in a gas container hits a horizontal wall with speed 200 m/s and angle 30^o with the normal, and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?

Answer:
Yes. Momentum of the composite system is always conserved and in this case the total momentum of the gas molecule an the container is also conserved.
Recoil velocity of the wall can be considered to be zero for any practical calculation. In that case the final K.E = initial K.E an the collision is elastic in nature.
N.B.: We do not need to perform explicit calculation to conclude the result.

Question 6. 15. A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m^3 in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump?

Answer:
Volume of water tank= 30 m^3 and the corresponding of water is m = 30 \times 10^3 kg
Time required to fill up the tank t = 15 min = 15 x 60 = 900s
Height of the tank from the ground h = 40 m
Efficiency of the pump n = 30%

Work done by the pump in 15 min W=mgh=12 \times 10^6 Joule [g=10 m/s^2]
Power required to fill the tank in 15 min is P= W/t = 13.33 KW

As the pump is working at 30% efficiency, Power consumed by the pump is 13.33 x 100/30= 44.43 KW

Note: Here we are assuming a tank with finite volume, however having negligible height. Can you solve the problem if we mention the cross sectional area of the cylindrical tank to be 10 m^2. [Hint: here you have to check the displacement of the center of mass of water cylinder]

Solutions for Question 6.16 to 6.30

Please share this Solutions for Class 11 NCERT Physics, Chapter 6 Work, Energy and Power provided by SciPhy.in with your friends.