NCERT Solutions for Class 11 Physics Chapter 5 Work Energy and Power

NCERT Solutions for Class 11 Physics Chapter 5 Work Energy and Power

Vishal kumarUpdated on 22 Aug 2025, 12:36 AM IST

Consider a bicycle rolling uphill; you provide force, the bicycle moves against gravity and energy gets transferred in the form of work. The key concepts of Work, Energy, and Power in Class 11 Physics can be demonstrated through this simple activity. Incidentally, when you are holding a heavy bag and feel tired, in physics terms, there is no work done, as the box does not move (the displacement is zero).

This Story also Contains

  1. NCERT Solutions for Class 11 Physics Chapter 5: Download PDF
  2. NCERT Solutions for Class 11 Physics Chapter 5: Exercise Questions
  3. Class 11 Physics NCERT Chapter 5: Higher Order Thinking Skills (HOTS) Questions
  4. NCERT Chapter 5 Work, Energy and Power Topics
  5. NCERT Solutions for Class 11 Chapter 5: Important Formulae
  6. Approach to Solve Questions of Work, Energy, and Power Class 11
  7. What Extra Should Students Study Beyond the NCERT for JEE/NEET?
  8. NCERT Solutions for Class 11 Physics Chapter-Wise

These NCERT Solutions for Class 11 Physics Work, Energy and Power give clear explanations to such real-life examples and also cover the principles of kinetic energy, potential energy, power, and the law of conservation of energy, which says that energy can neither be created nor destroyed, only transformed. Solving these NCERT problems would enable the students to sharpen their skills in tackling problems, and thus they would be exam-prepared to give their CBSE Class 11, Class 12 Boards and JEE-NEET competitive exams. These NCERT solutions are designed to help achieve better scores by involving step-by-step solutions, which will also allow the development of good conceptual clarity.

NCERT Solutions for Class 11 Physics Chapter 5: Download PDF

NCERT Solutions of Class 11 Physics Chapter 5: Law of Motion has been provided in the form of a well arranged PDF document to be easily downloaded. The solutions clarify the concepts of forces, inertia and friction, and motion with solution to all textbook problems explained in their own steps. Its PDF form enables students the opportunity to do a swift revision, homework, and examination preparation such as CBSE, JEE, and NEET.

Download PDF

NCERT Solutions for Class 11 Physics Chapter 5: Exercise Questions

The NCERT Solutions Class 11 Physics Chapter 5: Work, energy and power are extensive and precise discussions to all the exercise questions found in the textbook. These solutions make students understand important concepts such as work done by a force, different forms of energy, power and the work energy theorem, which are important to board exams as well as competitive exams such as JEE and NEET.

Q1(b) 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: work done by gravitational force in the above case

Answer:

In this case, the direction of displacement is upward and the direction of force is downward. Thus work done is negative in nature.

Q1 (c) 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: work done by friction on a body sliding down an inclined plane,

Answer:

We know that friction acts in the direction opposite to the direction of motion. Hence work done by the frictional force is negative.

Q2 (a) 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 work done by the applied force in 10 s,

Answer:

Using Newton's law we can write :

$\begin{aligned}
a & =\frac{F}{m} \\
& =\frac{7}{2}=3.5 \mathrm{~m} / \mathrm{s}^2
\end{aligned}$

The frictional force is given by :

$\begin{aligned}
f & =\mu m g \\
& =0.1 \times 2 \times 9.8=1.96 \mathrm{~N}
\end{aligned}$

Its direction will be opposite of the direction of the motion. Thus acceleration produced will be negative.

$a=\frac{-1.96}{2}=-0.98 \mathrm{~m} / \mathrm{s}^2$

Thus the net acceleration is $=3.5-0.98=2.52 \mathrm{~m} / \mathrm{s}^2$.
The total distance travelled is given by :
or

$\begin{aligned}
s & =u t+\frac{1}{2} a t^2 \\
& =0+\frac{1}{2}(2.52) 10^2=126 \mathrm{~m}
\end{aligned}$

Hence the work done by applied force is given by :

$W=F . s=7 \times 126=882 \mathrm{~J}$

Q2 (b) 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 work done by friction in 10 s,

Answer:

The work done by frictional force will be negative as the force opposes the motion.

W = f.s = −1.96×126 = −247 J

Q2 (d) 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 change in kinetic energy of the body in 10 s

Answer:

It is given that initial velocity is zero. The final velocity can be calculated by the equation of motion:

$\begin{aligned}
v & =u+a t \\
& =0+(2.52) 10 \\
& =25.2 \mathrm{~m} / \mathrm{s}
\end{aligned}$

Thus change is kinetic energy is :

$\begin{aligned}
\Delta K & =\frac{1}{2} m v^2-\frac{1}{2} m u^2 \\
& =\frac{1}{2} \times 2(25.2)^2-0 \\
& =635 \mathrm{~J}
\end{aligned}$

Q3 Given in Fig 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:

Total energy = kinetic energy (KE) + potential energy(PE)

KE > 0 since m and v2 is positive. If KE <0, particles cannot be find. If PE>TE, then KE<0 (now in all graph, check for this condition)

In case 1: kinetic energy is negative for x<a. So at x<a particle cannot be found.

In case 2: for x<a and for x> b, kinetic energy is negative. So the particle cannot be found in these regions.

In the third case, the minimum potential energy is when a<x<b. At this position, the potential energy is negative (- U1 ).

The kinetic energy in this case is given by :

K.E. = E − (−U1) = E + U1

And the minimum energy of particle is - U1 .

In the fourth case, the particle will not exist in the states which will have potential energy greater than the total energy.

Thus, the particle will not exist in −b2<x<b2 and −a2<x<a2.

The minimum energy of the particle will be - U1 as it is the minimum potential energy.

Q4 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=0.5Nm−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

Answer:

The total energy of the particle is given by :

$\begin{aligned}
E & =K \cdot E+P \cdot E \\
& =\frac{1}{2} m v^2+\frac{1}{2} k x^2
\end{aligned}$

At the extreme position, the object's velocity is zero; thus, its kinetic energy at that point is zero.

$\begin{aligned}
& E=\frac{1}{2} k x^2 \\
& 1=\frac{1}{2}(0.5) x^2
\end{aligned}$
$x^2=4$
$x= \pm 2$

Hence, the extreme position is $\pm 2 \mathrm{~m}$.

Q5 (a) Answer the following :
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?

Answer:

The total energy is given by :

E = K.E + P.E

or E = $\frac{1}{2} m v^2 + mgh$

The burning of casing results in a reduction in the mass of the rocket. This leads to a lowering of the total energy.

Thus, the heat required for burning is obtained from the expenses of the rocket.

Q5 (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?

Answer:

This is because the gravitational force is a conservative force. And we know that the work done by a conservative force in a closed path is always zero. That's why the work done by the gravitational force is zero in a complete orbit revolution.

Q5 (c) An artificial satellite orbiting the earth in a 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?

Answer:

The total energy of an artificial satellite remains constant. Thus, when it approaches the Earth, the distance between them decreases. This results in a decrease in the potential energy of the satellite. By energy conservation, the kinetic energy of the satellite increases, and so does the velocity.

Q5 (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?

Answer:

In the first case,

Work done is :

W = F.s = Fscos⁡Θ

or W = mgscos⁡Θ

or W = 15×9.8×2×cos⁡90

or W = 0

In the second case :

W = Fscos⁡Θ

or W = mgscos⁡0

or W = 15×9.8×2 = 294 J

Thus work done in the second case is greater than the first case.

Q6 (a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered.

Answer:

It is given that work done by the conservative force is positive, thus the force acts in the direction of the motion. This results in a decrease in distance between the bodies. Thus it's potential energy decreases .

Q6 (b) Work done by a body against friction always results in a loss of its kinetic/potential energy.

Answer:

Work done by the body against friction results in a decrease in the velocity of the body. Thus the kinetic energy of the body decreases.

Q6 (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.

Answer:

The internal force cannot produce a change in the total momentum as no external force is acting. Thus the change in total momentum is proportional to the external forces acting on the body.

Q6(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:

The conservation of total linear momentum doesn't depend upon the fact whether it is an elastic collision or an inelastic collision.

Q7(a) State if each of the following statements is true or false. Give reasons for your answer.

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

Answer:

False: - The linear momentum and energy will be conserved if both are considered in a system. But for individual bodies, this conservation of momentum and energy doesn't hold. This is because the impact during the collision may transfer energy/momentum of one ball to the other ball.

Q7 (b) State if each of the following statements is true or false. Give reasons for your answer.

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

Answer:

False: - Internal forces will not change the energy of the system but external forces can change the total energy by changing their magnitude or direction.

Q7 (c) State if each of the following statements is true or false. Give reasons for your answer.

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

Answer:

False:- This is true only for conservative forces e.g. gravitational force. For e.g in case of frictional force (non-conservative force), the work done in a closed-loop cannot be zero as energy is wasted throughout.

Q7 (d) State if each of the following statements is true or false. Give reasons for your answer.

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

Answer:

True but not always:- In the case of inelastic collisions, few amounts of energy is converted into other forms of energy such as sound or in deformation. Thus final kinetic energy is always less as compared to initial kinetic energy. But in case of the explosion of a bomb final kinetic energy is greater than the initial kinetic energy

Q8 (a) Answer carefully, with reasons: 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)?

Answer:

No, because at the time of the collision, the kinetic energy is converted to the potential energy. Thus total kinetic energy is not constant at the collision.

Q8 (b) Is the total linear momentum conserved during the short time of an elastic collision of two balls?

Answer:

Yes, in case of elastic collision the total linear momentum of the system remains conserved as no external force is acting on the system of balls.

Q8 (c) What are the answers to (a) and (b) for an inelastic collision?

Answer:

The total kinetic energy of the system cannot be conserved in case of inelastic collision as there is loss of energy in the form of deformation. But the total linear momentum of the system remains constant even in the case of inelastic collision as no external force is acting.

Q8 (d) If the potential energy of two billiard balls depends only on the separation distance between their centres, 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:

Since the potential energy of the system depends upon the separation between the bodies thus the forces acting on the body are conservative in nature. We know that conservative forces produce elastic collisions.

Q 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

Answer:

It is given that acceleration is constant thus force will also be constant (by Newton's law of motion F = ma).

Also,

a = dv/dt = constant

or dv = C dt

Thus v ∝ t

Now, the work done by the force is given by :

P = F.v

Hence power is directly proportional to the time.

Q10 A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to

$(i) t^{1 / 2}(i i) t(i i i) t^{3 / 2}(i v) t^2$

Answer:

We know that the power is given by:

$\begin{aligned}
P & =F \cdot v \\
& =m a \cdot v \\
& =m \frac{d v}{d t} v
\end{aligned}$

It is given that power is constant, thus :

$\begin{aligned}
m v \frac{d v}{d t} & =\text { constant } \\
v d v & =\frac{C}{m} d t
\end{aligned}$

By integrating both sides, we get

$v=\left(\sqrt{\frac{2 C t}{m}}\right)$

Also, we can write :

$\begin{aligned}
& v=\frac{d x}{d t} \\
& \frac{d x}{d t}=\sqrt{\frac{2 C}{m}} t^{\frac{1}{2}}
\end{aligned}$

By integrating we get the relation :

$x \propto t^{\frac{3}{2}}$

Q11 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 $\hat{i}, \hat{j}, \hat{k}$ 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:

Force is given to be :

$\vec{F}=-\hat{i}+2 \hat{j}+3 \hat{k} \ N$

And the displacement is :

$\vec{s}=4 \hat{k} \ m$

Thus the work done is given by :


$W$ =$\vec{F} \cdot \vec{s}$
$W =(-\hat{i}+2 \hat{j}+3 \hat{k}) \cdot(4 \hat{k}) \\$
$W =0+0-3(4)=12 \mathrm{~J}$

Q12 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 $=9.11 \times 10^{-31} \mathrm{Kg}$, proton mass $1.67 \times 10^{-27} \mathrm{Kg}, 1 \mathrm{eV}=1.60 \times 10^{-19} \mathrm{~J}$)

Answer:

The kinetic energy of the electron is given by:

$\begin{gathered}
K_e=\frac{1}{2} m v_e^2 \\
1.6 \times 10^{-15} J=\frac{1}{2} \times 9.11 \times 10^{-31} \times v_e^2
\end{gathered}$

Thus velocity is obtained as:

$\begin{aligned}
v_e & =\sqrt{\frac{2 \times 1.6 \times 10^{-15}}{9.11 \times 10^{-31}}} \\
& =5.93 \times 10^7 \mathrm{~m} / \mathrm{s}
\end{aligned}$

Similarly, we can find the velocity of the proton :

$\begin{aligned}
K_p & =\frac{1}{2} m v_p^2 \\
1.6 \times 10^{-14} J & =\frac{1}{2} \times 1.67 \times 10^{-27} \times v_p^2
\end{aligned}$

Thus velocity is obtained as:

$\begin{aligned}
v_p & =\sqrt{\frac{2 \times 1.6 \times 10^{-14}}{1.67 \times 10^{-27}}} \\
& =4.38 \times 10^6 \mathrm{~m} / \mathrm{s}
\end{aligned}$

Thus the ratio of their velocities is:

$\frac{v_e}{v_p}=\frac{5.93 \times 10^7}{4.38 \times 10^6}=13.54$

Q13 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 10ms−1 ?

Answer:

The volume of the drop is:

$
V=\frac{4}{3} \pi r^3=\frac{4}{3} \times 3.14 \times\left(2 \times 10^{-3}\right)^3
$

Thus the mass of raindrop is :

$
\begin{aligned}
m & =\rho V \\
& =10^3 \times \frac{4}{3} \times 3.14 \times\left(2 \times 10^{-3}\right)^3 \mathrm{Kg}
\end{aligned}
$

Thus the work done is given by:

$
W =F . s \\$
$ \Rightarrow W=m g s \\$
$\Rightarrow W =10^3 \times \frac{4}{3} \times 3.14 \times\left(2 \times 10^{-3}\right)^3 \times 9.8 \times 250\\$

$\Rightarrow W =0.082 \mathrm{~J}$

Now the total energy at the peak point is :

$
\begin{aligned}
E_p=m g h+0 & =m g h \\
= & 10^3 \times \frac{4}{3} \times 3.14 \times\left(2 \times 10^{-3}\right)^3 \times 9.8 \times 500 \\
= & 0.146 \mathrm{~J}
\end{aligned}
$

And the energy at the ground is :

$
\begin{aligned}
E_b=0+\frac{1}{2} m v^2 & =\frac{1}{2} m v^2 \\
& =\frac{1}{2} \times 10^3 \times \frac{4}{3} \times 3.14 \times\left(2 \times 10^{-3}\right)^3 \times(10)^2 \\
& =1.67 \times 10^{-3} \mathrm{~J}
\end{aligned}
$

Thus work done by the resistive force is :

$
=1.67 \times 10^{-3} J-0.164 J=-0.162 J
$

Q14 A molecule in a gas container hits a horizontal wall with speed 200ms−1 and angle 30o with the normal, and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?

Answer:

The momentum is conserved in the collision as no external force is acting on the system. In the given case the rebound velocity is the same as the initial velocity thus the kinetic energy of the molecule initially and finally are same. Hence this is an elastic collision.

Q15 A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m3 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:

Mass of the water is:

$
\begin{aligned}
m & =\rho V \\
& =30 \times 10^3 \mathrm{Kg}
\end{aligned}
$

Thus the output power is given by :

$
\begin{aligned}
\text { Power } & =\frac{\text { Work done }}{\text { Time }} \\
& =\frac{m g h}{t} \\
& =\frac{30 \times 10^3 \times 9.8 \times 40}{900} \\
& =13.067 \times 10^3 \mathrm{~W}
\end{aligned}
$

Also, we are given that efficiency is 30 per cent.

Thus the input power is :

$
\begin{aligned}
P_i & =\frac{13.067}{\frac{30}{100}} \times 10^3 \\
& =43.6 \mathrm{kW}
\end{aligned}
$

Q16 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) is a possible result after collision?

Answer:

The initial kinetic energy of the system is given by :

$
\begin{aligned}
& =\frac{1}{2} m v^2-\frac{1}{2} 2 m(0) \\
& =\frac{1}{2} m v^2
\end{aligned}
$

Case (i):-

The final kinetic energy is :$=\frac{1}{2} m \cdot 0-\frac{1}{2} 2 m\left(\frac{v}{2}\right)^2=\frac{1}{4} m v^2
$

Thus the kinetic energy is not conserved in this case.
Case (ii):-

The final kinetic energy is :$=\frac{1}{2} 2 m \cdot 0-\frac{1}{2} m v^2=\frac{1}{2} m v^2
$

Thus kinetic energy is conserved in this case.
Case (iii):-

The final kinetic energy is:-$=\frac{1}{2} \times 3 m \times\left(\frac{v}{3}\right)^2 \\ =\frac{1}{6} m v^2

$

Thus the kinetic energy is not conserved in this case.

Q17 The bob A of a pendulum released from 30o to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 5.15. How high does the bob A rise after the collision? Neglect the size of
the bobs and assume the collision to be elastic.

Answer:

This is an elastic collision thus the transfer of momentum will take place. It is given that bob B is at rest and bob A has some velocity. So in momentum transfer, bob B will gain the velocity in the left direction whereas bob A will come to rest (complete momentum transfer takes place).

Hence bob A will not rise.

Q18 The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance?

Answer:

Consider the extreme position (horizontal) :-
The kinetic energy at this position is zero as velocity is zero.
Thus total energy is given by: $=m g l+0=m g l$
Now consider the mean position (lowermost point) :
Here the potential energy of Bob is zero.
Whereas kinetic energy is $
=\frac{1}{2} m v^2
$


Further, it is given that 5 per cent of energy is dissipated due to air resistance while coming down.
Thus energy equation becomes (conservation of energy):-

$
\frac{1}{2} m v^2=\frac{95}{100} \times m g l
$

or

$
v=\sqrt{\frac{2 \times 95 \times 1.5 \times 9.8}{100}}=5.28 \mathrm{~m} / \mathrm{s}
$

Q19 A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.005Kgs−1 . What is the speed of the trolley after the entire sandbag is empty?

Answer:

Since the sand is falling in the trolley thus the force generated on the system (trolley and sandbag) is an internal force. There is no external force thus momentum of the system doesn't change. Hence speed remains the same i.e., 27 Km/hr.

Q20 A body of mass 0.5 kg travels in a straight line with velocity $v=ax^{3/2}$ where $a=5m^{−1/2}s^{−1}$ . What is the work done by the net force during its displacement from x = 0 to x = 2 m ?

Answer:

The relation between work done and the kinetic energy is given by :

$
\text { Work }=\frac{1}{2} m v^2-\frac{1}{2} m u^2
$

Using the relation $v=a x^{3 / 2}$ we can write :

$
\text { Initial velocity }=0 \quad(\text { at } x=0)
$

And the final velocity $=10 \sqrt{2} \mathrm{~m} / \mathrm{s} \quad($ at $\mathrm{x}=2)$.
Thus work done is :

$
\begin{aligned}
\text { Work } & =\frac{1}{2} m\left(v^2-u^2\right) \\
& =\frac{1}{2} \times 0.5 \times(10 \sqrt{2})^2 \\
& =50 \mathrm{~J}
\end{aligned}
$

Q21 (a) The blades of a windmill sweep out a circle of area A. If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t?

Answer:

The volume of wind = Av here A is the swept circle and v is the velocity.

Thus the mass of the wind is : - ρAv , ρ is the density of the air.

Hence mass of wind flowing through windmill in time t = ρAvt .

Q21 (b) The blades of a windmill sweep out a circle of area A. What is the kinetic energy of the air?

Answer:

The kinetic energy is given by :

$
\begin{aligned}
& =\frac{1}{2} m v^2 \\
& =\frac{1}{2} \rho A v t v^2 \\
& =\frac{1}{2} \rho A t v^3
\end{aligned}
$

Thus the kinetic energy of wind is $\frac{1}{2} \rho A t v^3 \mathrm{~J}$.

Q21 (c) The blades of a windmill sweep out a circle of area A. Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that A=30m2,v=36Km/h and the density of air is 1.2Kgm−3 What is the electrical power produced?

Answer:

It is given that 25 per cent of wind energy is converted into electrical energy.
Thus electric energy produced is :

$
\begin{aligned}
& =\frac{25}{100} \times \frac{1}{2} \rho A t v^3 \\
&= \frac{1}{8} \rho A t v^3
\end{aligned}
$

Now the electric power is given by :

$
\begin{aligned}
\text { Power } & =\frac{\text { Energy }}{\text { Time }} \\
& =\frac{\frac{1}{8} \rho A t v^3}{t}=\frac{1}{8} \rho A v^3 \\
& =\frac{1}{8} \times 1.2 \times 30 \times(10)^3 \\
& =4.5 \mathrm{kW}
\end{aligned}
$

Q22 (a) A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated. How much work does she do against the gravitational force?

Answer:

The work done against the gravitational force is given by :

= Number of times the weight is lifted × work done in 1 time.

= 1000×mgh

= 1000×10×9.8×0.5

= 49 kJ

Q22 (b) A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated. Fat supplies 3.8×107J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?

Answer:

Efficiency is given to be 20 per cent.

Thus energy supplied by the person:

$
=\frac{20}{100} \times 3.8 \times 10^7
$


Thus the amount of fat lost is :

$
\begin{aligned}
& =\frac{49 \times 10^3}{\frac{20}{100} \times 3.8 \times 10^7} \\
& =6.45 \times 10^{-3} \mathrm{Kg}
\end{aligned}
$

Q23 (a) A family uses 8 kW of power. Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW?

Answer:

It is given that the efficiency of energy conversion is 20 per cent.
According to question, we can write (equating power used by family) :

$
8 \times 10^3=\frac{20}{100} \times A \times 200
$

(Here A is the area required.)

$
A=\frac{8 \times 10^3}{40}
$

Thus required area is $200 \mathrm{~m}^2$.

Q23 (b) A family uses 8 kW of power. Compare this area to that of the roof of a typical house.

Answer:

A typical has dimensions of 14×14 m2 .

The area of the roof of the house is 225 m2 .

This is nearly equal to the area required for the production of the given amount of electricity

Class 11 Physics NCERT Chapter 5: Higher Order Thinking Skills (HOTS) Questions

Higher Order Thinking Skills (HOTS) questions of Class 11 Physics Chapter 5: Work, Energy and Power are set with a view to improving conceptual understanding and analysis skills. Such questions are not a simple exercise in solving problems, but are more of a way in urging the students to use logic and reason in dealing with situations in real life and those that are not simple. Solving HOTS questions is the key to success on any competitive exam and developing a solid background in physics.

Question 1) In an hourglass, approximately 100 grains of sand fall per second(starting from rest), and it takes 2 seconds for each sand particle to reach the bottom of the hourglass. If the average mass of each sand particle 0.2 g is then the average force exerted by the falling sand on the bottom of the hourglass is close to :
1) 0.4 N
2) 0.8 N
3) 1.2 N
4) 1.6 N

Answer:

The velocity of sand particles just before striking the bottom.

$
\begin{aligned}
& v=u+a t \\
& v=0+10 \times 2[u=0 \text { and } t=2 s \text { given }] \\
& v=20 \mathrm{~m} / \mathrm{s}
\end{aligned}
$
For 1 grain (Momentum just before the collision at the bottom)

$
\begin{aligned}
p_i & =m v_i \\
p_i & =\left(0.2 \times 10^{-3}\right) \times 20 \\
& =4 \times 10^{-3}
\end{aligned}
$

(momentum after the collision at the bottom)

$
\begin{aligned}
& p_f=\left(0.2 \times 10^{-3}\right) \times 0 \\
& p_f=0[\text { Final velocity }=0] \\
& |\Delta p|=\left|p_f-p_i\right| \\
& =\left|0-4 * 10^{-3}\right| \\
& =4 * 10^{-3} \mathrm{Kg} \mathrm{~m} / \mathrm{s}
\end{aligned}
$

Force exerted by 100 grains in 1 sec

$
\begin{aligned}
& \left(f_{\text {average }}\right)=\left\{\frac{[\Delta p]}{[\Delta t]}\right\} * 100= \left\{\frac{\left[4 * 10^{-3}\right]}{1}\right\} * 100=0.4 N
\end{aligned}
$
Hence, the answer is the option (1).

Question 2) A constant power P is supplied to a car of mass $\mathrm{m}=3000 \mathrm{~kg}$. The velocity of the car increases from $u=2 \mathrm{~ms}^{-1}$ to $v=5 \mathrm{~ms}^{-1}$ when the car travels a distance of $\mathrm{x}=117 \mathrm{~m}$. Find the value of P in kW (neglect friction).
1) 1
2) 2
3) 3
4) 4

Answer:

Power can be written as $P=F v$, where $F$ is force and $v$ is the velocity.

$
P=F v=m a v \Rightarrow a=\frac{P}{m v}
$
Now $a=\frac{d v}{d t}=\frac{d v}{d x} \cdot \frac{d x}{d t}=\frac{v d v}{d x}$
From equations (i) and (ii),

$v \frac{d v}{d x}=\frac{P}{m v} \Rightarrow v^2 d v=\frac{P}{m} d x $


$ \therefore \quad \int_u^v v^2 d v=\frac{P}{m} \int_0^x d x $


$ \Rightarrow \frac{1}{3}\left(v^3-u^3\right)=\frac{P x}{m}$

$\Rightarrow P=\frac{m\left(v^3-u^3\right)}{3 x}=\frac{3000 \times\left[(5)^3-(2)^3\right]}{3 \times 117}$

$P =1000 \mathrm{~W}=1 \mathrm{~kW}

$
Hence, the answer is 1 kW.

Question 3) A small particle $(\mathrm{m}=2 \mathrm{~kg})$ is hanging from a fixed point by a light inextensible string $(\ell=10 \mathrm{~m})$ is projected horizontally with speed $10 \mathrm{~ms}^{-1}$ from the equilibrium position. Find the correct option(s):

When speed of particle is $\frac{10}{\sqrt{3}} \mathrm{~m} \mathrm{~s}^{-1}$
1) Angle $\theta$ of the string from vertical is $\cos ^{-1} \frac{2}{3}$
2) The value of tension is 20 N
3) Angle $\theta$ of the string from vertical is $\sin ^{-1} \frac{2}{3}$
4) The value of tension is 40 N

Answer:

At point B, the resultant force will be:

$
F_R=T-\mathrm{mg} \cos \theta
$
If $v$ be the speed of the particle at $B$, the resultant force will give centripetal acceleration,

$
F_R=\frac{m v^2}{\ell}
$
From (i) and (ii), we get

$
T-\mathrm{mg} \cos \theta=\frac{m v^2}{\ell}
$
Conserving the energy of the particle at point $A$ and $B$, we have

$
\frac{1}{2} m v_o^2=m g \ell(1-\cos \theta)+\frac{1}{2} m v^2
$
Putting the given values, $v_0=10 \mathrm{~ms}^{-1}$
We get, $v=10 \sqrt{(2 \cos \theta)-1}$

According to the question,

$
v=\frac{10}{\sqrt{3}} \mathrm{~m} \mathrm{~s}^{-1}
$
Hence,

$
\cos \theta=\frac{2}{3}
$
Putiing the value of $\cos \theta$ in equation (iii) we get,

$
\begin{aligned}
& T=m g \cos \theta+\frac{m v^2}{\ell} \\
& T=\frac{40}{3}+\frac{20}{3}=20 \mathrm{~N}
\end{aligned}
$

Hence, the answer are the options (1, 2).

Question 4) Johnny was whirling his yoyo ( 0.5 kg ) in a horizontal circle of radius 1 m at a height of 5 m above the ground. The string of yoyo suddenly breaks and the yoyo finally strikes the ground at a horizontal separation of 20 m . Find out the maximum tension(nearest integer in N ) that the yoyo string can bear.
1) 200
2) 100
3) 300
4) 400

Answer:

After the string breaks the yoyo will fall under gravity in a parabolic path.
Initial vertical velocity $=0$

Hence by vertical motion analysis,
The time of flight will be,

$
t=\sqrt{\frac{2 H}{g}} \Rightarrow t=\sqrt{\frac{2 \times 5}{10}}=1 s
$


The horizontal distance travelled $=20 \mathrm{~m}$
Hence speed of yoyo will be $=20 \mathrm{~ms}^{-1}$
In circular path, $\mathrm{v}=20 \mathrm{~ms}^{-1}$

Hence centripetal acceleration,

$
\begin{aligned}
& a_c=\frac{v^2}{r} \\
& a_c=400 \mathrm{~ms}^{-2}
\end{aligned}
$
The string will be slightly inclined to give centripetal acceleration and balance the weight.

$
\begin{aligned}
& T=\sqrt{\left(\frac{m v^2}{r}\right)^2+(m g)^2} \\
& T=0.5 \sqrt{(400)^2+10^2} \\
& T = 200 N
\end{aligned}
$
This must be the breaking strength of the string.

Hence, the answer is 200.

Question 5) A reservoir of capacity 1000 liters is at a height of 30 m above the ground. A pump of power 1 kW situated on the ground is used to fill the reservoir with water using a pipe of diameter 2.4 cm . The time taken to fill the reservoir is $x \times 10^y s$ What is the value of $\mathrm{x}+\mathrm{y}$ ?

$
(0 \leq x, y \leq 10)
$

Answer:

Work done by the pump against gravity is

$
W_1=m g h=(\rho V) g h=\rho g h V
$
Work done by the pump against pressure difference is

$
W_2=\Delta P \times \Delta V=h \rho g V
$

$\therefore$ Total work done,

$
W=W_1+W_2=2 h \rho g V
$


The power of the pump given is

$
P=\frac{W}{t}
$
Therefore,

$
\begin{aligned}
t & =\frac{W}{P}=\frac{2 h \rho g V}{P} \\
& =\frac{2 \times 30 \times\left(10^3\right) \times(10) \times\left(10^3 \times 10^{-3}\right)}{10^3} \\
& =600 \mathrm{~s}=6 \times 10^2 \mathrm{~s}
\end{aligned}
$
So, $x=6$ and $y=2$
Therefore, $x+y=8$
Hence, the answer is 8.

NCERT Chapter 5 Work, Energy and Power Topics

Chapter 5 of NCERT Class 11 Physics Work, Energy and Power discusses important concepts like work by a force, kinetic energy, and potential energy, power and the work-energy theorem. The chapter also contains the law of conservation of energy and its uses, hence it is important in board exams as well as competitive exams such as JEE and NEET.

5.1 Introduction
5.1.1 The Scalar Product
5.2 Notions Of Work And Kinetic Energy: The Work-Energy Theorem
5.3 Work
5.4 Kinetic Energy
5.5 Work Done By A Variable Force
5.6 The Work-Energy Theorem For A Variable Force
5.7 The Concept Of Potential Energy
5.8 The Conservation Of Mechanical Energy
5.9 The Potential Energy Of A Spring
5.10 Power
5.11 Collisions
5.11.1 Elastic And Inelastic Collisions
5.11.2 Collisions In One Dimension
5.11.3 Collisions In Two Dimensions

NCERT Solutions for Class 11 Chapter 5: Important Formulae

NCERT Solutions for Class 11 Physics Chapter 5: Work, Energy and Power – Important Formulae provide a quick reference to all key equations like work done, kinetic energy, potential energy, power, and conservation of energy. These formulas simplify solving numerical problems and strengthen conceptual understanding for board exams and competitive tests.

1. Work Done by a Constant Force

$
W=\vec{F} \cdot \vec{d}=F d \cos \theta
$
Where:
- $F=$ magnitude of force
- $d=$ displacement
- $\theta=$ angle between force and displacement

2. Work Done by a Variable Force

$W=\int_{x_1}^{x_2} F(x) d x$

3. Kinetic Energy (K.E)

$
\text { K.E. }=\frac{1}{2} m v^2
$

4. Work-Energy Theorem

$W_{\mathrm{net}}=\Delta K . E .=\frac{1}{2} m v^2-\frac{1}{2} m u^2$

5. Potential Energy (U)

$
U=m g h \text { (Near Earth's surface) }
$
For spring: $U=\frac{1}{2} k x^2$

6. Power

$
P=\frac{W}{t}
$
Instantaneous Power: $P=\vec{F} \cdot \vec{v}$

Approach to Solve Questions of Work, Energy, and Power Class 11

In order to answer NCERT Class 11 Chapter: Work, Energy and Power related questions, one must first of all be aware of the meaning of work, energy and power and their differences. Recognise the kind of work: positive, negative or zero and use the appropriate work-energy theorem. Apply the right formulae to kinetic energy, potential energy and mechanical energy, and all of your energies must be in the same units. In case of power questions, concentrate on the rate of doing work or energy conversion. In problems involving conservative and non-conservative forces, track how energy is conserved or dissipated. Always visualise the physical situation with diagrams and apply the law of conservation of energy when possible.

What Extra Should Students Study Beyond the NCERT for JEE/NEET?

Although NCERT is very good preparatory material in Chapter 5: Work, Energy and Power, to achieve competitive scores in JEE, the students must look beyond this textbook. They must pay attention to the correct solutions to highly designed problems, theoretical use and numerical precision. The best practice occurs due to higher-level questions available in the reference books and past year JEE papers, which enhance in-depth learning and speed and accuracy.

NCERT Solutions for Class 11 Physics Chapter-Wise

The NCERT Solutions to Class 11 Physics are available in chapter-wise format to have inexpensive, step-by-step solutions to each chapter. The links are useful in revising major concepts, doing numerical problems, and preparing a sound base on exams such as CBSE, JEE, and NEET. All solutions are designed by professionals according to the new syllabus.

Subject wise NCERT Exemplar solutions

Also Check NCERT Books and NCERT Syllabus here

Subject wise NCERT solutions

Frequently Asked Questions (FAQs)

Q: Is work energy power tough for JEE/NEET?
A:

It is one of the crucial chapters in NEET and you get tricky questions on the basis of this Chapter, so you are advised to practice Questions of this Chapter so much so that you master and understand the concepts in a better way.

Q: How is power related to energy and time?
A:

Power is the rate at which energy is transferred or work is done. It is calculated as the amount of energy used per unit of time.

Q: What are conservative and non-conservative forces? Explain with examples.
A:

Conservative Forces: Forces for which the work done does not depend on the path taken, only on the initial and final positions.

Example: Gravitational force, spring force.

Non-conservative Forces: Forces for which the work done depends on the path taken. They cause energy dissipation (e.g., in the form of heat).

Example: Friction, air resistance.

Q: What is the principle of conservation of mechanical energy?
A:

The principle of conservation of mechanical energy states that the total mechanical energy (sum of kinetic and potential energy) of an isolated system remains constant if only conservative forces (like gravity) are acting on it.

Q: What are collisions covered in class 11 physics work energy and power?
A:

A collision is an event that happens when two objects come into direct contact with each other and exert forces on each other for a relatively short period of time. This can involve two or more bodies interacting with one another, resulting in changes in their motion and momentum.

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