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 17 Sep 2025, 11:36 PM IST

Consider how hard you strain when you are riding a bicycle uphill, the force you apply overcomes your weight, and the energy expended in the task is changed into mechanical work. Conversely, when you carry a heavy bag in your hands and do not move it, physics will count this as no work having been done, as the distance covered is zero. The NCERT Solutions of Class 11 Physics Chapter 5 - Work, Energy and Power make such real-life concepts easier to understand and provide precise explanations of the principle of work, kinetic energy, potential energy, power, and the law of conservation of energy that energy cannot be created or destroyed; instead, it can be transformed into any form as required.

This Story also Contains

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

The NCERT Solutions for Class 11 Physics Chapter 5 - Work, Energy and Power are developed in a step-by-step format containing answers to all NCERT textbook problems, such as numericals, derivations, and conceptual problems. They enable students to know how force, motion, and energy are related to each other in a logical and systematic manner. Regular practice of these NCERT solutions not only makes the conceptual picture clearer, but also trains problem-solving skills that will be needed in the advanced physics level. These NCERT Solutions for Class 11 Physics Chapter 5 - Work, Energy and Power are provided in a detailed form with solved examples and serve to provide a balance between theory and application so that a student can have accuracy and speed when trying questions. With these solutions, students will be able to increase their confidence, improve their analytical power and achieve better marks by revising them.

Work, Energy and Power NCERT Solutions: Download PDF

Work, Energy and Power class 11 Question Answers: 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 solutions 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

Work, Energy and Power NCERT Solutions: Exercise Questions

The Class 11 Physics Chapter 5 - Work, Energy and Power question answers are extensive and precise discussions of 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 :

a=Fm=72=3.5 m/s2

The frictional force is given by :

f=μmg=0.1×2×9.8=1.96 N

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

a=1.962=0.98 m/s2

Thus, the net acceleration is =3.50.98=2.52 m/s2.
The total distance travelled is given by :
or

s=ut+12at2=0+12(2.52)102=126 m

Hence, the work done by the applied force is given by :

W=F.s=7×126=882 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 the frictional force will be negative as the force opposes the motion.

W = f.s = −1.96×126 = −247 J

Q2 (c) 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 net force on the body in 10 s,

Answer:

The net work done will be the sum of the work done by the applied force and the work done by the frictional force.

W = 882 − 247 = 635 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 the initial velocity is zero. The final velocity can be calculated by the equation of motion:

v=u+at=0+(2.52)10=25.2 m/s

Thus change in kinetic energy is :

ΔK=12mv212mu2=12×2(25.2)20=635 J

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 are positive. If KE <0, particles cannot be found. If PE>TE, then KE<0 (now in all graphs, 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 the 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 :

E=KE+PE=12mv2+12kx2

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

E=12kx21=12(0.5)x2
x2=4
x=±2

Hence, the extreme position is ±2 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 = 12mv2+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 the distance between the bodies. Thus, its 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 from 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 example, in the case of a 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 amount of energy is converted into other forms of energy, such as sound or deformation. Thus final kinetic energy is always less as compared to initial kinetic energy. But in the 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 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 the case of an 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 the case of an inelastic collision, as there is a loss of energy in the form of deformation. But the total linear momentum of the system remains constant even in the case of an 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 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)t1/2(ii)t(iii)t3/2(iv)t2

Answer:

We know that the power is given by:

P=Fv=mav=mdvdtv

It is given that power is constant, thus :

mvdvdt= constant vdv=Cmdt

By integrating both sides, we get

v=(2Ctm)

Also, we can write :

v=dxdtdxdt=2Cmt12

By integrating, we get the relation :

xt32

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 =9.11×1031Kg, proton mass 1.67×1027Kg,1eV=1.60×1019 J)

Answer:

The kinetic energy of the electron is given by:

Ke=12mve21.6×1015J=12×9.11×1031×ve2

Thus, velocity is obtained as:

ve=2×1.6×10159.11×1031=5.93×107 m/s

Similarly, we can find the velocity of the proton :

Kp=12mvp21.6×1014J=12×1.67×1027×vp2

Thus, velocity is obtained as:

vp=2×1.6×10141.67×1027=4.38×106 m/s

Thus the ratio of their velocities is:

vevp=5.93×1074.38×106=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=43πr3=43×3.14×(2×103)3

Thus, the mass ofa raindrop is :

m=ρV=103×43×3.14×(2×103)3Kg

Thus, the work done is given by:

W=F.s
W=mgs
W=103×43×3.14×(2×103)3×9.8×250

W=0.082 J

Now the total energy at the peak point is :

Ep=mgh+0=mgh=103×43×3.14×(2×103)3×9.8×500=0.146 J

And the energy at the ground is :

Eb=0+12mv2=12mv2=12×103×43×3.14×(2×103)3×(10)2=1.67×103 J

Thus work done by the resistive force is :

=1.67×103J0.164J=0.162J

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 is the 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:

m=ρV=30×103Kg

Thus, the output power is given by :

Power = Work done Time =mght=30×103×9.8×40900=13.067×103 W

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

Thus, the input power is :

Pi=13.06730100×103=43.6kW

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 :

=12mv2122m(0)=12mv2

Case (i):-

The final kinetic energy is :=12m0122m(v2)2=14mv2

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

The final kinetic energy is:=122m012mv2=12mv2

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

The final kinetic energy is:-=12×3m×(v3)2=16mv2

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 res,t, 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 the velocity is zero.
Thus total energy is given by: =mgl+0=mgl
Now consider the mean position (lowermost point) :
Here, the potential energy of Bob is zero.
Whereas kinetic energy is =12mv2


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):-

12mv2=95100×mgl

or

v=2×95×1.5×9.8100=5.28 m/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, the 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=ax3/2 where a=5m1/2s1 . 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 :

Work =12mv212mu2

Using the relation v=ax3/2 we can write :

Initial velocity =0( at x=0)

And the final velocity =102 m/s( at x=2).
Thus work done is :

Work =12m(v2u2)=12×0.5×(102)2=50 J

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, where 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 the 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 :

=12mv2=12ρAvtv2=12ρAtv3

Thus the kinetic energy of wind is 12ρAtv3 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 :

=25100×12ρAtv3=18ρAtv3

Now the electric power is given by :

Power = Energy Time =18ρAtv3t=18ρAv3=18×1.2×30×(10)3=4.5kW

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

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 the question, we can write (equating power used by the family) :

8×103=20100×A×200

(Here A is the area required.)

A=8×10340

Thus required area is 200 m2.

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

Work, Energy and Power NCERT Solutions: 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 of 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.

v=u+atv=0+10×2[u=0 and t=2s given ]v=20 m/s
For 1 grain (Momentum just before the collision at the bottom)

pi=mvipi=(0.2×103)×20=4×103

(momentum after the collision at the bottom)

pf=(0.2×103)×0pf=0[ Final velocity =0]|Δp|=|pfpi|=|04103|=4103Kg m/s

Force exerted by 100 grains in 1 sec

(faverage )={[Δp][Δt]}100={[4103]1}100=0.4N
Hence, the answer is the option (1).

Question 2) A constant power P is supplied to a car of mass m=3000 kg. The velocity of the car increases from u=2 ms1 to v=5 ms1 when the car travels a distance of x=117 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=Fv, where F is force and v is the velocity.

P=Fv=mava=Pmv
Now a=dvdt=dvdxdxdt=vdvdx
From equations (i) and (ii),

vdvdx=Pmvv2dv=Pmdx


uvv2dv=Pm0xdx


13(v3u3)=Pxm

P=m(v3u3)3x=3000×[(5)3(2)3]3×117

P=1000 W=1 kW
Hence, the answer is 1 kW.

Question 3) A small particle (m=2 kg) is hanging from a fixed point by a light inextensible string (=10 m) and is projected horizontally with speed 10 ms1 from the equilibrium position. Find the correct option(s):

When the speed of the particle is 103 m s1
1) Angle θ of the string from vertical is cos123
2) The value of tension is 20 N
3) Angle θ of the string from vertical is sin123
4) The value of tension is 40 N

Answer:

At point B, the resultant force will be:

FR=Tmgcosθ
If v be the speed of the particle at B, the resultant force will give centripetal acceleration,

FR=mv2
From (i) and (ii), we get

Tmgcosθ=mv2
Conserving the energy of the particle at points A and B, we have

12mvo2=mg(1cosθ)+12mv2
Putting the given values, v0=10 ms1
We get, v=10(2cosθ)1

According to the question,

v=103 m s1
Hence,

cosθ=23
Putting the value of cosθ in equation (iii), we get,

T=mgcosθ+mv2T=403+203=20 N

Hence, the answer is 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 the 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=2Hgt=2×510=1s


The horizontal distance travelled =20 m
Hence speed of the yoyo will be =20 ms1
In circular path, v=20 ms1

Hence, centripetal acceleration,

ac=v2rac=400 ms2
The string will be slightly inclined to give centripetal acceleration and balance the weight.

T=(mv2r)2+(mg)2T=0.5(400)2+102T=200N
This must be the breaking strength of the string.

Hence, the answer is 200.

Question 5) A reservoir of capacity 1000 litres 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×10ys. What is the value of x+y?

(0x,y10)

Answer:

Work done by the pump against gravity is

W1=mgh=(ρV)gh=ρghV
Work done by the pump against the pressure difference is

W2=ΔP×ΔV=hρgV

Total work done,

W=W1+W2=2hρgV


The power of the pump given is

P=Wt
Therefore,

t=WP=2hρgVP=2×30×(103)×(10)×(103×103)103=600 s=6×102 s
So, x=6 and y=2
Therefore, x+y=8
Hence, the answer is 8.

Work, Energy and Power class 11 Question Answers: Topics

Chapter 5 of NCERT Class 11 Physics Work, Energy and Power discusses important concepts like work by a force, kinetic energy, 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

Class 11 Physics Chapter 5 - Work, Energy and Power: Important Formulae

Class 11 Physics Chapter 5 - Work, Energy and Power question answers – 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=Fd=Fdcosθ
Where:
- F= magnitude of force
- d= displacement
- θ= angle between force and displacement

2. Work Done by a Variable Force

W=x1x2F(x)dx

3. Kinetic Energy (K.E)

K.E. =12mv2

4. Work-Energy Theorem

Wnet=ΔK.E.=12mv212mu2

5. Potential Energy (U)

U=mgh (Near Earth's surface)
For spring: U=12kx2

6. Power

P=Wt
Instantaneous Power: P=Fv

Approach to Solve Questions of Class 11 Physics Chapter 5 - Work, Energy and Power

Chapter 5 of Class 11 Physics is about the basics of work done by some force, kinetic and potential energy, work-energy theorem and the idea of power. The chapter gives a solid background to problem-solving in mechanics and involves students to use mathematical reasoning and physical understanding. The correct approach should be developed in order to make both theoretical and numerical questions clear.

  • Get a Clear Understanding of the Problem Statement: In a problem it is important to first determine whether actually the problem is one of work done by a force, or kinetic energy, or potential energy or power. Take the units and conditions that are provided in the question carefully.
  • Break Down Forces and Motion: When there are work related questions then resolve the applied force into components and determine which part of it causes displacement. It is only the force component in the direction of displacement that works.
  • Apply Work-Energy Theorem: Apply the formula: Work done = Change in kinetic energy. This method is commonly easier than going directly to work done by forces.
  • Apply Conservation Principles of Energy: Use the law of conservation of mechanical energy in situations where there is an increase or decrease of height, velocity, or oscillations. This prevents lengthy calculations through balance of energy, kinetic and potential.
  • Power-Related Problems: In quantitative questions on power, the formula is P = Work done/Time or P = Force x Velocity. Select the one that fits better in the provided data.
  • Check Units and Dimensions: Ensure the final answer is in standard SI units (Joule, Watt, etc.). Formulas can also be verified using dimensional analysis.
  • Practice Graph-Based Questions: There can be some questions related to work-energy graphs or force-displacement graphs. b Learn how to get work out of the area under the curve.

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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