**NCERT** **Class 11 Physics Chapter 5 Work, Energy and Power Solutions**,

**NCERT**to each chapter is provided in the list so that you can easily browse throughout different chapters

**Class 11 Physics Chapter 5 Work, Energy and Power**Notes**NCERT**

**and select needs one.**

**Question Answer****Class 11 Physics Chapter 5 Work, Energy and Power****NCERT Class 11 Physics Chapter 5 Work, Energy and Power**

Also, you can read the SCERT book online in these sections **NCERT** **Class 11 Physics Chapter 5 Work, Energy and Power****Solutions** by Expert Teachers as per SCERT (CBSE) Book guidelines. These solutions are part of SCERT All Subject Solutions. Here we have given **NCERT** **Class 11 Physics Chapter 5 Work, Energy and Power****Solutions** for All Subjects, You can practice these here.

**Work, Energy and Power**

**Work, Energy and Power****Chapter: 5**

Part – I |

**EXERCISE**

**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.**

Ans: Positive.

**(b) work done by gravitational force in the above case.**

Ans: Negative.

**(c) work done by friction on a body sliding down an inclined plane.**

Ans: Negative.

**(d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity.**

Ans: Positive.

**(e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest.**

Ans: Negative.

**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: and interpret your results.**

(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.

Ans: m = 2kg

F = 7N

Coefficient of Kinetic friction = 0.1

Time (t) = 10s

Calculate the frictional force

F _{frictional} = u.k N

Where N is normal force for a horizontal surface N = mg

F _{friction} = uk x mg

= 0.1 × 2.98

= 1.96N.

**Net force acting on the body:**

Fnet = Fapplied – Friction

= 7 – 1.96

= 5.04 N.

Calculate the acceleration of the body,

Newton’s second law

(F = ma)

a = Fnet/m

= 5.04/2

= 2.52m/s2

s = ut + ½ at2

= 0 + ½ x 2.52 × (10)2

= 126m.

(a) Work done by applied force in 10s

= 7 × 126 = 882J.

(b) Work done by friction in 10s

= 1.96 × 126 = 246.96 J

(c) Work done by net force in 10s

= 5.04 × 126 = 635.04J

(d) Since the initial Kinetic energy of the body was zero and the final K.E. acquired by the body in 10s under the net force is 635. 04 J.

**3. Given in Fig. 5.11 are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinary 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.**

Ans: (i) Using PE.+ K.E. = constant, i.e. E. in the region x = 0 to x = a; P.E. is zero. In the region x>a the P.E. has a value more than E. Thus, K.E. = E – P.E., i.e. K.E. has a negative value so the particle can not exist in the region x>a.

(ii) Here, P.E.>E the total energy of the object and as such the kinetic energy of the object would be negative. Thus, objects can not be present in any region of the graph.

(iii) Here,x = 0 to x = a and x >b, the P.E. is more than E, so K.E. is negative. The particle cannot exist in these regions.

(iv) The object cannot exist in the region between x = – b/2 to x = – a/2 and x = – a/2 to x = – b/2 because in this region P.E.>E.

**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. 5.12. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m.**

Ans: The potential energy function v (x) = Kx^{2}/2

Force constant K = 0.5 N/m

Total energy of the particle E = 1J

The potential energy v(x)

v(x) = Kx^{2}/2

Substituting K = 0.5 N/m

v (x) = 0.5 x^{2}/2

= 0.25 x^{2}

Total energy and turning point

E = K (x) + v (x)

Given the total energy E = 1J at the turning points of the motion.

The Kinetic energy K (x) is zero.

Thus,

E = v (x)

= 1 = 0.2x^{2}

x^{2} = 1/0.25 = 4

x = ± √4

= ± 2.

**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?**

Ans: The casing of a rocket in flight burns up due to friction. The heat energy required for burning comes from the rocket.The fuel of the rocket burns which provides the energy to resist the gravitation force and the air resistance. As a result, the work is done continuously against the friction.

**(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?**

Ans: This is because gravitational force is a conservative force Work done by the gravitational force of the sun over a closed path in every complete orbit of the comet is zero. However, due to atmospheric friction, the total energy of the satellite decreases by a small amount.

**(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?**

Ans: As a satellite comes closer to earth, its velocity increases but due to loss in energy through friction, the total energy decreases.

**(d) In Fig. 5.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. 5.13(ii), he walks the same distance pulling the rope behind him. 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?**

Ans: In Fig. (i), the force is applied horizontally, and the movement is also along the horizontal plane.

θ = 90°

W = F_{s} cos 90° = zero.

(ii) Here θ = 0°; when mass is lifted upwards with some upward force given to the mass by man through pulley.

In (ii) case, work done is more.

**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.**

Ans: Decreases.

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

Ans: Kinetic.

**(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.**

Ans: External force.

**(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.**

Ans: Total linear momentum.

**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.**

Ans: False. The total energy and momentum of both the bodies is conserved and it is not applicable for each individual body in the case of an elastic collision. However, the momentum and energy of each individual body are not necessarily conserved.

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

Ans: False. The total energy of a system is conserved in the presence of only internal forces, but not necessarily in the presence of external forces. Therefore, the total energy of a system will be changed due to external forces on a body.

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

Ans: False. The work done in the motion of a body over a closed loop is zero for a conservation force only, it is not applicable 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.**

Ans: True. In an inelastic collision, there will always be a loss of energy in the form of sound, heat etc, so the final kinetic energy is always less than the initial kinetic energy of the system.

**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)?**

Ans: At the time of collision of two billiard balls, the kinetic energy of the balls is not conserved and it will get converted into potential energy.

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

Ans: Yes, Linear momentum is always conserved, no matter what. The total momentum is covered according to the law of conservation of momentum.

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

Ans: (i) For an inelastic collision, the total kinetic energy is not conserved during the collision.

(ii) The total linear momentum of the system of billiards balls will remain conserved even in the case of an inelastic collision.

**(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).**

Ans: It is a case of elastic collision because in this case the forces will be of conservative nature.

**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^{1/2}

(ii) t

(iii) t^{3/2}

(iv) t^{2}

Ans: Power = work done/ time taken

= F x d / t = F.v

But, v = u + at, for u = at

Power, P = F.v = F . at = ma^{2}t

= P ∝ t.

**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^{1/2}

(ii) t

(iii) t^{3/2}

(iv) t^{2}

Ans: (c) Power, P = F.v = ma .v

Here, P and m are constant,

= s^{2 }∝ t^{3}

**11. A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by**

**F = i + 2j + 3k 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?**

Ans: Work = F × d

Where

F is the force vector

d is the displacement vector

The force F is given by F = i + 2j + 3kN

The displacement d is along the z – axis, which can be represented d = 0i + 0j + 4km

**Dot product calculation:**

F × d = ( 1.0) + (2 .0) + (3.4)

F × d = 0 + 0 + 12

12J.

**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 = 9.11×10**^{-31}** kg, proton mass = 1.67×10**^{–27}** kg, 1 eV = 1.60×10**^{–19}** J).**

Ans: **Given:**

v_{e} = speed of electron.

v_{p} = speed of proton.

m_{e} = mass of electron.

m_{p} = mass of proton.

Then, ½ me v^{2}_{e} = 10 Ke V

= 10 × 10^{3} x 1.6 x 10^{-19}

= 1.6 × 10^{-15} J

And ½ m_{p}v^{2}_{p}

= 100 Ke V

= 100 × 10^{3} × 1.6 × 10^{-19}

Or

The electron is faster.

**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 m s–1?**

Ans: **Given data:**

Radius of raindrop r= 2 mm = 2 × 10^{-3}m

Height from which the raindrop falls,h = 500m

Terminal speed of the raindrop, vt = 10 m/s

Density of water, pw = 1000 kg/m^{3}

Gravitational acceleration, g = 9.8 m/s^{2}

**Calculation of the raindrops:**

V = 4/3 𝜋r^{3}

V = 4/3 𝜋 ( 2 × 10^{-3})^{3}

V = 4/3𝜋 ( 8 × 10^{-9})

V = 3.35 × 10^{-8} m

**The mass m of the raindrop is then:**

m = P_{w} × V

m = 1000 kg/ m^{3} × 3.35 × 10^{-8} m^{3}

m = 3.35 × 10^{-5} kg

Calculate the work done by the gravitational force

W_{g} = mgh

First half of the journey (h/2 = 250m)

W_{g1} = mg (h/2)

W_{g1} = 3.35 × 10^{-5} kg × 9.8 m/s^{2} × 250m

W_{g1} = 8.21 × 10^{-2}J

Second half of the journey (h/2 = 250m)

W_{g2} = mg (h/2)

W_{g2} = 3.35 × 10^{-5} kg × 9.8 m/s^{2} × 250

W_{g2} = 8.21 × 10^{-2}J

**Total gravitational force:**

W_{g }= W_{g1} + W_{g2}

W_{g} = 8.21 × 10^{-2} J + 8.21 × 10^{-2} J

W_{g} = 1.64 × 10^{-1} J

**Calculate the resistive force:**

The KE of the raindrops when it hits the ground is

KE = ½ mv2

KE = ½ x 3.35 × 10^{-5} kg × (10m/s)^{2}

KE = ½ x 3.35 × 10^{-5} × 100

KE = 1.675 × 10^{-3} J

W_{r} = Wg – KE

W_{r} = 1.64 × 10^{-1} J – 1.675 × 10^{-3} J

W_{r} = 0.1623J.

**14. A molecule in a gas container hits a horizontal wall with speed 200 m s–1 and angle 30° with the normal, and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?**

Ans: Since the molecule approaches the wall of the container at a particular speed and then rebounds from the container at the same speed. Since there is no loss in the velocity due to any factor and the mass is the same. So, the momentum of the system remains conserved.

Yes; Collision is elastic. The momentum of the gas molecule remains conserved whether the collision is elastic or inelastic.

**15. 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?**

Ans: **Given:**

V = 30m3

h = 40m

t = 15 min

n = 30 % = 0.3

Calculate the Work Done to Lift the Water:

W= mgh

g = (9.8 m/s^{2})

W = 30000 kg × 9.8 m/s^{2} × 40m

W = 117600000 J.

**Calculate the input power:**

n = Useful power output/ power input

Power input = Useful power output/n

Power output = w/t = 11760000J/900s

= 13066.67

Power input = 13066.67w/ 0.3

= 43555.56 w

Power input = 43.56kw.

**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 is a possible result after collision?**

Ans: Let m be the mass of each ball

Total KE of the system before collision = ½ mv2

Total KE of the system after collision is

Case (ii) KE of the system after collision

Case (iii) KE of the system after collision

**17. 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.**

Ans: The bob A will not rise because when two bodies of same mass undergo an elastic collision, their velocities are interchanged. After collision, ball A will come to rest and ball B will move with the velocity of A.

It is because of the fact that in a perfectly elastic collision, when a moving object collides against a target object of the same mass, the two exchange their velocities.

**18. 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?**

Ans: PEinitial = mgh

**Given:**

L = 1.5 m

h = 1.5 m

PEinitial = mg × 1.5

KE at the lowest point

KE = PE remaining = 0.95 × mgh

KE = ½ mv2

½ mv^{2} = 0.95 × mgh

Speed v =

= v^{2} = 2 × 0.95 × gh

= v = √ 2 × 0.95 × g × h

Given g = 9.8 m/s^{2} and h = 1.5m

v = √ 2 × 0.95 × 9.8 × 1.5

v = √ 27.93

v = 5.28 m/s

**19. 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 0.05 kg s–1. What is the speed of the trolley after the entire sand bag is empty?**

Ans: When the sand starts leaking, there is no external force acting on the bag hence, the speed of the trolley will not change. Thus, the speed of the trolley after the emptying of the sand is 27 km⋅ h −1. Since the hole is on the floor, that means sand is falling vertically with respect to trolley. Therefore there is no force in horizontal direction hence in horizontal direction momentum is conserved. Therefore speed of the trolley will not changes as the sand leaks out or even after the sandbag becomes empty.

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

Ans: **Here:**

m = 0.5 kg

v = a x^{3/2}

a = 5m^{-1}/2s^{-1}

Displacement, x = 2m

**The work done:**

Calculate the velocity at x = 0 and x = 2m

At x = 0

v (0) = 5 × 0^{3/2} = 0

At x = 2m

v (2) = 5 × (2)^{3/2}

v (2) = 5 × √2^{3}

= 5 × √ 8

= 5 × 2√2

= 10 √2

**Calculate the KE at both points:**

at x = 0

Kinetic energy at x = 2m

KE2 = ½ mv(2)^{2}

v (2) = 10√2

v(2)^{2} = (10√2)^{2} = 100 × 2 = 200

KE_{2} = ½ x 0.5 × 200

= 0.25 × 200

= 50

**The work done:**

W = KE_{2 }– KE_{0 }

= 50 – 0

= 50J.

**21. The blades of a windmill sweep out a circle of area A.**

**(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?**

Ans: Volume of wind blowing per sec = Av

Mass of wind blowing per sec = A v p

Mass of air passing in time t = A v pt

**(b) What is the kinetic energy of the air ?**

Ans: KE = ½ × p x v^{3 }× A x t

= KE = ½ × 1.2 × 10^{3} × 30 x t

= 18000 × tJ

**(c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that A = 30 m2 , v = 36 km/h and the density of air is 1.2 kg m–3. What is the electrical power produced?**

Ans: Electrical Power = 0.25 × KE

Electrical Power = 0.25 × 18000 × t

= 4500 tW.

= 4.5 tkW.

**22. 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.**

**(a) How much work does she do against the gravitational force?**

Ans: **Given that:**

m = 10 kg

h = 0.5m

n = 1000

Work done against the gravitational force,

W = n mgh

= 1000 × (10 × 9.8 × 0.5)

= 49000 J

**(b) 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?**

Ans: The 20% of the energy from fat is converted into mechanical energy.

Energy from fat = work done/ efficiency.

Energy from fat = 49000/ 0.20

Energy from fat = 245000 J

**Fat conversion:**

**Fat supplies 3.8 x 10**^{7}**J/ the amount of fat used:**

Mass of fat = Energy from fat/ Energy per kg of fat

Mass of fat = 245000/ 3.8 × 10^{7}

Mass of fat = 0.0065 kg

Mass of fat = 6.5 g.

**23. A family uses 8 kW of power.**

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

Ans: Let the area be A m^{2}.

Average power per square metre = 200 W.

Useful electrical energy produced per square metre,

Useful energy = 0.20 × 200 = 40 W/m^{2}

Area = Power required/ useful energy per square metre

Area = 8000W/ 40 W/m^{2}

Area = 200m^{2}

**(b) Compare this area to that of the roof of a typical house.**

Ans: This area is comparable to that of a roof of a house of 14m × 14m. (close to 200 sqm).