Work Energy And Power Solution of CBSE Class 11 Physics

Class 11^th Physics Part I CBSE Solution

Exercise

Question 1.

The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:

A. work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.

B. work done by gravitational force in the above case,

C. work done by friction on a body sliding down an inclined plane,

D. work done by an applied force on a body moving on a rough horizontal plane with uniform velocity,

E. work done by the resistive force of air on a vibrating pendulum in bringing it to rest.

Answer:

Work done by a force f on a object which give body a displacement of d

where,
w is the work done on the object by force f
f is the force applied to the object
d is the displacement of the object after application of force
θ is the angle between force and displacement

A. The displacement of the bucket and the force applied on the bucket by the man have the same direction i.e. θ=0. Hence, the work done by a man is positive and the work is done on the bucket.

⇒ (∵ cos 0⁰=1)

Hence, work done is positive.

B. The gravitational force acts downwards on the bucket. But the displacement of the bucket is upwards i.e. θ=180. Hence, the work done is negative and θ=180

(∵ cos 180⁰=-1)

Hence, work done is negative.

C. Frictional force always acts opposite to the direction of motion of object i.e. θ=180. Hence, work done by frictional force is always negative.

(∵ cos 180⁰=-1)

Hence, work done is negative.

D. In this case, frictional force acts on the body due to the roughness of the horizontal plane in a direction opposite to the direction of motion. To obtain a uniform velocity, a uniform force must be applied in the direction of motion i.e. θ =0. Hence, the work done by the force is positive.

⇒ (∵ cos 0⁰=1)

Hence, work done is positive.

E. The resistive force always acts in the direction opposite to that of the motion of pendulum to bring the pendulum to rest i.e. θ =180. Hence, the work done by the resistive force is negative.

(∵ cos 180⁰=-1)

Hence, work done is negative.

Question 2.

A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the

A. work done by the applied force in 10 s,

B. work done by friction in 10 s,

C. work done by the net force on the body in 10 s,

D. change in kinetic energy of the body in 10 s, and interpret your results.

Answer:

Given,

Mass of the body, m = 2kg

Force applied, F = 7N

Coefficient of kinetic friction, μ = 0.1

A. Time, t = 10s

Using Newton’s second law of motion,

F=ma

where, a is the acceleration caused by force F on a body of mass m

∴

⇒ a=3.5 m/s²

Frictional force is given by

f=μmg

Where, μ is the coefficient of friction

m is the mass of the object

g is the acceleration due to gravity

∴ f=0.1 × 2kg × 9.8ms^-2

⇒ f=1.96N

Acceleration due to friction is given by

a_f=f/m

∴ (The negative sign is due to the fact that the frictional force acts opposite to the direction of motion)

⇒ a_f=-0.98ms^-2

Total acceleration = acceleration due to applied force + acceleration due to friction

Total acceleration, a=a_F+a_f

⇒ a=3.5ms^-2+(-0.98ms^-2)

∴ a=2.52ms^-2

The displacement of the object is given by Newton’s second equation of motion as

⇒ s= 0 + (1/2)(2.52ms^-2)(10s)²

⇒ s=126 m

Work done by the applied force is given by

W=F∙s

⇒ W=7 N × 126 m

⇒ W=882 J

B. Work done by friction is given by

W=f∙s

⇒ W= (-1.96 N) × (126 m)

⇒ W= -246.96 J

C. Net force on the body, F_t=F+f

∴ F_t=7 N + (-1.96 N) = 5.04 N

Work done by total force is given by

W_t=F_t∙s

⇒ W_t=5.04 N × 126 m

⇒ W_t=635.04 J

D. Using newton’s First equation of motion, the final velocity is calculated as

v=u+at

⇒ v=0+(2.52 ms^-2)(10 s)

⇒ v=25.2 m/s

Change in kinetic energy,

So, ΔK = ()(2kg)(25.2ms^-1)² – ()(2kg)(0)

⇒ ΔK = 635.04 J

This result is in accordance with Work-Energy theorem which states that the change in kinetic energy of an object is equal to the net work done on the object.

Question 3.

Given in Fig. 6.11 are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.

Answer:

The kinetic energy of a particle of mass m moving with a velocity v is given as

Since m is always positive, K is also always positive.

The total energy (E) of a particle is equal to the sum of Kinetic energy (K) and Potential energy (PE).

⇒ E = K + PE

⇒ K = E – PE

a) For x>a, PE(=V₀)>E.

This means that the kinetic energy is negative. Hence, the particle cannot exist in the region x>a.

b) For x<a and x>b, PE(=V₀)>E.

This means that the kinetic energy is negative. Hence, the particle cannot exist in the region x<a and x>b.

c) In the region a<x<b, the kinetic energy is positive. The potential energy is -V₁. So, K = E-(-V₁) = E + V₁.

In order for K to be positive, E> -V₁. So, the minimum total energy of the particle is -V₁.

d) For the regions -b/2<x<-a/2 and a/2<x<b/2, the potential energy of the particle is greater than the total energy which suggests that the kinetic energy is negative in this region. Hence, the particle cannot exist in this region.

The lowest potential energy is -V₁. So, K = E-(-V₁) = E+V₁.

For K to be positive, E should be greater than -V₁. So, minimum total energy is -V₁.

Question 4.

The potential energy function for a particle executing linear simple harmonic motion is given by V(x) = kx²/2, where k is the force constant of the oscillator. For k = 0.5 N m-1, the graph of V(x) versus x is shown in Fig. 6.12. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = � 2 m.

Answer:

Given,

Total energy of the particle, E = 1 J

Force constant, k = 0.5 N m^-1

The total energy is equal to the sum of kinetic energy and potential energy.

So, E = PE + K

⇒ E = kx² + mv²

So, 1 = kx² + mv²

When the velocity of the particle is zero i.e., at the turning point, the kinetic energy is zero.

∴ 1 = kx²

⇒ x² = 2/k

⇒ x² = 2/0.5

⇒ x² = 4

⇒ x = �2

Hence, the particle turns back at x= �2.

Question 5.

Answer the following:

A. The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere?

B. Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?

C. An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth?

D. In Fig. 6.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. 6.13(ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater?

Answer:

A. The heat energy required for burning is obtained at the expense of the rocket. The mass of the rocket is consumed to supply for the loss in total energy of the rocket.

B. Gravitational force is conservative in nature. The work done by a conservative force in a closed cycle is zero. Hence, work done by the gravitational attraction of sun is zero.

C. When the satellite comes closer to the earth, its potential energy decreases (∵ PE = mgh where h is the height of the object from the earth). To maintain constant total energy, the kinetic energy increases and hence velocity increases.

D. Given,

Mass = 15 kg

Displacement, s = 2 m

In figure 1:

Work done, W = F∙s

⇒ W = F×s×cosθ

⇒ W = F×s× cos 90°

⇒ W = 0 (∵ cos 90° = 0)

In figure 2:

Work done, W = F∙s

⇒ W = F s cosθ

⇒ W = m g s cos 0°

⇒ W = 15 kg ×9.8 m s^-2× 2 m × 1

⇒ W = 294 J

Hence, more work is done in figure 2.

Question 6.

Underline the correct alternative:

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

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

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

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

Answer:

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

Explanation: When a conservative force does positive work on a body, the object is displaced in the direction of force and towards the centre of the force such that the distance between the source and the object decreases. So, potential energy also decreases.

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

Explanation: Friction acts always in the direction opposite to that of motion and results in a decrease in the velocity of the body. Hence, friction decreases the kinetic energy of the body.

C. The rate of change of total momentum of a many-particle system is proportional to the external force on the system.

Explanation: Internal forces cannot change the total energy of a system in order to maintain a constant energy of the system. Hence, only external forces can bring about a change in the total momentum of a many-particle system.

D. In an inelastic collision of two bodies, the quantities which do not change after the collision are total linear momentum of the system of two bodies.

Explanation: The total momentum of a system remains constant in a collision irrespective of the nature of the collision according to law of conservation of linear momentum.

Question 7.

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

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

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

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

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

Answer:

A. False

Explanation: In an elastic collision, the energy and momentum of the entire system is conserved and not of individual bodies.

B. False

Explanation: External forces can do work on the system and hence, they can change the total energy of the system.

C. False

Explanation: Work done in the motion of a body over a closed loop is zero only for conservative forces.

D. True

Explanation: In inelastic collisions, the final kinetic energy is always less than the initial kinetic energy because some of the energy is lost during the process in the form of sound, heat, etc.

Question 8.

Answer carefully, with reasons:

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

B. Is the total linear momentum conserved during the short time of an elastic collision of two balls?

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

D. 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:

A. In an elastic collision of two billiard balls, the total kinetic energy is not conserved during the short time of collision of the balls. During this period, the kinetic energy of the balls is zero. All the energy is stored in the form of potential energy.

B. The total linear momentum is always conserved during the entire duration of collision. The individual linear momenta change but the total linear momentum always remains constant.

C. For an inelastic collision, kinetic energy is not conserved but the total linear momentum is conserved.

D. The collision is elastic in nature. The forces involved are conservative in nature since potential energy depends on the separation of the centres.

NOTE: Elastic collisions are collisions in which both momentum and kinetic energy are conserved. The total system kinetic energy before the collision equals the total system kinetic energy after the collision. If total kinetic energy is not conserved, then the collision is referred to as an inelastic collision.

Question 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) t²

Answer:

The correct answer is (ii).

Explanation:

From Newton’s first equation of motion,

v = u + at

where, v is the final velocity

u is the initial velocity

a is the acceleration of the body

t is the time taken

Given, u = 0.

So, v = at

Power is given by

P = F × v

⇒ P = ma × at

⇒ P = ma²t

Since m and a are constant,

P ∝ t

Question 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) t²

Answer:

The correct answer is (iii).

Explanation:

Power, P = F v = k (constant)

Considering the dimensions,

[P] = [F][v]

⇒ [P] = [MLT^-2][LT^-1]

⇒ [P] = [ML²T^-3]

So, [ML²T^-3] = k

⇒ [L²T^-3] = k

⇒ [L²/T³] = k

⇒ [L²] = k[T³]

⇒ [L²] ∝ [T³]

⇒ [L] ∝ [T^3/2]

Hence, l ∝ t^3/2

Question 11.

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

Where 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:

Given,

Force, N

Displacement, m

Work done, W =

So, W =

⇒ W = 3×4

⇒ W = 12 J

Question 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).

Answer:

For an electron:

Mass of the electron, m_e= 9.11×10^-31kg

Let v_e be the velocity of the electron.

Kinetic energy of the electron, K_e = m_ev_e²

⇒ 10 = m_ev_e²

⇒ m_ev_e²=20 ……………….(1)

For a proton:

Mass of the proton, m_p= 1.67×10^-27kg

Let v_p be the velocity of the proton.

Kinetic energy of the proton, K_p=m_pv_p²

⇒ 100 = m_pv_p²

⇒ m_pv_p²=200 ……………………(2)

Dividing equation (1) by (2),

⇒

⇒ = 13.54

∴ v_e:v_p = 13.54 : 1

Question 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?

Answer:

Given,

Radius of the rain drop, r = 2 mm = 2×10^-3 m

Total displacement of the rain drop, s = 500 m

Assuming the rain drop to be spherical,

Volume of the rain drop, V = (4/3)πr³

∴ V = (4/3)×π×(2×10^-3)³

⇒ V = 3.35×10^-8 m³

We know, Density of water, ρ = 10³ kg m^-3

So, Mass of rain drop, m = ρV

⇒ m = 10³ kg m^-3 × 3.35×10^-8 m³

⇒ m = 3.35×10^-5 kg

Gravitational force on the rain drop, F = mg

⇒ F = 3.35×10^-5 kg × 9.8 m s^-2

⇒ F = 3.28×10^-4 N

Work done by the gravitational force in the first half of the motion, W₁=F(s/2)

⇒ W₁ = 3.28×10^-4× (500/2)

⇒ W₁ = 8.21×10^-2 J

The work done by the gravitational force is equal in both the halves of the motion.

So, work done by the gravitational force during second half of motion, W₂ = 8.21×10^-2 J

If no resistive force is present, then the total energy of the rain drop will remain conserved according to law of conservation of energy.

Total energy at the starting point, E = mgh

⇒ E = 3.35×10^-5 kg × 9.8 m s^-2 × 500 m

⇒ E = 0.164 J

Due to the resistive force, some energy will be lost. The drop hits ground with velocity 10 m s^-1.

Total energy of the drop when it hits the ground, E’ = (1/2)mv²

⇒ E’ = (1/2) × 3.35×10^-5 kg × (10 m s^-1)²

⇒ E’ = 1.675×10^-3 J

So, loss of energy due to resistive force, ΔE = E’-E

⇒ ΔE = 1.675×10^-3 J – 0.164 J

⇒ ΔE = -0.162 J

NOTE: The negative sign indicates that energy is lost due to the resistive forces.

Question 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?

Answer:

The linear momentum is always conserved irrespective of the nature of collision. Since the gas molecule rebounds with the same speed, the rebound velocity of the wall is zero. Hence, the kinetic energy is conserved as there is no change in velocities. Thus, this is an elastic collision.

NOTE: Initial and final velocities of the molecule and wall remain same before and after the collision. Hence, the kinetic energies also remain same.

Question 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?

Answer:

Given,

Volume of the tank, V = 30 m³

Operation time of pump, t = 15 min = 15×60 sec = 900 sec

Height of the tank from ground, h = 40 m

Efficiency of the pump, η = 30%

We know, density of water, ρ = 10³ kg m^-3

So, Mass of water, m = ρV

⇒ m = 10³ m³ × 30 kg m^-3

⇒ m = 3 × 10⁴ kg

Output power, P_o = Work done/Time of operation

∴ P_o = mgh/t

⇒ P_o = (3×10⁴ kg × 9.8 m s^-2 × 40 m)/(900 sec)

⇒ P_o = 1.307×10⁴ W = 13.07 kW

We know,

Efficiency (η) = Output power (P_o)/Input power (P_i)

∴ Input power (P_i)= Output power (Po)/Efficiency (η)

⇒ P_i = (1.307×10⁴ W)/(0.3)

⇒ P_i = 4.355×10⁴ W = 43.55 kW

Question 16.

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

Answer:

From the figure, it is observed that the linear momentum is conserved in all the cases. For an elastic collision, kinetic energy is also conserved.

Let us assume that the mass of each ball is m.

Total kinetic energy before collision,

K_i = (1/2)mV² + (1/2)m(0)² + (1/2)m(0)²

⇒ K_i = (1/2)mV²

Case (i):

Total kinetic energy after collision,

K_f = (1/2)m(0)² + (1/2)m(V/2)² + (1/2)m(V/2)²

⇒ K_f = (1/8)mV² + (1/8)mV²

⇒ K_f = (1/4)mV²

The total kinetic energy of the system before collision is not equal to the total kinetic energy after collision in this case. The total kinetic energy is not conserved. Hence, the collision is not elastic.

Case (ii):

Total kinetic energy after collision,

K_f = (1/2)m(0)² + (1/2)m(0)² + (1/2)mV²

⇒ K_f = (1/2)mV²

The total kinetic energy of the system before collision is equal to the total kinetic energy after collision in this case. The total kinetic energy is conserved. Hence, the collision is elastic.

Case (iii):

Total kinetic energy after collision,

K_f = (1/2)m(V/3)² + (1/2)m(V/3)² + (1/2)m(V/3)²

⇒ K_f = (1/18)mV² + (1/18)mV² + (1/18)mV²

⇒ K_f = (1/6)mV²

Thus, only case (ii) is an elastic collision.

Question 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. 6.15. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.

Answer:

The bob A will not rise after the collision. It will stay at rest because the collision is elastic. In an elastic collision, the velocities are interchanged. Hence, the bob A will come to rest and the bob B will move with the velocity of bob A after collision.

Question 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?

Answer:

Given,

Length of the pendulum, l = 1.5 m

Energy dissipated against air resistance, E = 5%

Let m be the mass of the bob and v be its velocity at the lowermost point.

At the extreme end:

Potential energy of the bob, P = mgl

Kinetic energy of the bob, K = 0 (∵ bob is at rest)

Total energy of bob, E = P + E

⇒ E = mgl ……………(1)

At the lowermost point:

Potential energy of the bob, P = 0

Kinetic energy of the bob, K = (1/2)mv²

Total energy of the bob, E = P + K

⇒ E = (1/2)mv²

During the motion from an extreme end to the lowermost point, the bob loses 5% of its energy, i.e., it has 95% of the energy.

So, (1/2)mv² = (95/100)mgl

⇒ v² = 95gl/50

⇒ v =

⇒ v = 5.28 m s^-1

Question 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?

Answer:

The trolley, at first, is moving with a constant velocity. Hence, there is no net external force acting on the trolley. Even if the sand starts leaking, there is no external force on the trolley. Hence, the speed of the trolley will not change.

Question 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?

Answer:

Given,

Mass of the body, m = 0.5 kg

Velocity of the body is given by

v =a x^3/2 where a = 5 m^–1/2 s^–1

At x = 0, initial velocity, u = 0

At x = 2, final velocity, v = 5 (2)^3/2 = m s^-1

According to work-energy theorem, work done is equal to the change in kinetic energy of the body.

Work done, W = ΔK

⇒ W = (1/2)mv² – (1/2)mu²

⇒ W = (1/2) × 0.5 kg × ( m s^-1)² – (1/2) × 0.5 kg × 0²

⇒ W = 50 J

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

Answer:

Given,

Area swept by the blades = A

Velocity of wind flow = v

Density of air = ρ

(a) Volume of the wind flowing through the windmill per second, V = Av

Mass of the wind flowing through the mill per second,

m = ρAv

Mass of the wind flowing through the mill in time t,

M = ρAvt

(b) Kinetic energy of air, K = (1/2)Mv²

⇒ K = (1/2)(ρAvt)v²

⇒ K = (1/2)ρAv³t

Area swept by the blades, A = 30 m²

Velocity of wind flow, v = 36 km/h

⇒ v = (36×1000 m)/(60×60 sec)

⇒ v = 10 m/s

Density of air, ρ = 1.2 kg m^-3

Electric energy produced, E = 25% of wind energy

⇒ E = (25/100)(1/2) ρAv³t

⇒ E = (25/100)(1/2)× 1.2 kg m^-3 × 30 m² × (10 m s^-1)³ × t

⇒ E = 4500t J

Electric power, P = Electric energy (E)/ Time (t)

⇒ P = 4500t/t

⇒ P = 4500 W = 4.5 kW

Question 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? (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?

Answer:

(a) Given,

Mass of the weight, m = 10 kg

Height upto which the weight is lifted, h = 0.5 m

Number of times the weight is lifted, n = 1000

Work done against the gravitational force,

W = nmgh

⇒ W = 1000 × 10 kg × 9.8 m s^-2 × 0.5

⇒ W = 49000 J = 49 kJ

(b) 1 kg of fat is equivalent to 3.8 × 10⁷ J.

Efficiency of conversion, η = 20%

Mechanical energy supplied by body of the person,

E = (20/100)×3.8×10⁷ J

⇒ E = 7.6 × 10⁶ J

Equivalent mass of fat lost by dieter,

M =

⇒ M = 6.45×10^-3 kg = 6.45 g

Question 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 meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW? (b) Compare this area to that of the roof of a typical house.

Answer:

Given,

Power used by the family, P = 8 kW = 8000 W

Solar power incident per square meter, E = 200 W

Efficiency of energy conversion, η = 20%

(a) Let A be the area required to supply 8 kW power.

So, 8×10³ = 20% × (A × 200)

⇒

⇒ A = 200 m²

(b) 200 m² is equivalent to a rooftop with dimensions 14m×14m.

Question 24.

A bullet of mass 0.012 kg and horizontal speed 70 m s^–1 strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.

Answer:

Given,

Mass of the bullet, m = 0.012 kg

Horizontal speed (initial speed) of bullet, u = 70 m/s

Mass of wood block, M = 0.4 kg

Initial speed of the block, u’ = 0

After collision, the bullet sticks to the block. Let the final velocity of bullet and block after collision be v.

According to law of conservation of linear momentum,

mu + Mu’ = (m+M)v

⇒

⇒ v = 2.039 m/s

For the bullet-wooden block system,

Mass of the system, m_t = 0.012 kg + 0.4 kg = 0.412 kg

Velocity of the system, v = 2.039 m/s

Applying the law of conservation of energy, the potential energy at the highest point is equal to the kinetic energy at the lowest point.

Hence, m_tgh = (1/2)m_tv²

⇒ h = v²/2g

⇒ h = (2.039 ms^-1)²/(2×9.8 ms^-2)

⇒ h = 0.212m

So, the block will rise to a height of 0.212 m or 21.2 cm.

Heat produced (H) = Kinetic energy of the bullet – Kinetic energy of the system

⇒ H = (1/2)mu² – (1/2)m_tv²

⇒ H=(1/2)×0.012 kg×(70 ms^-1)² – (1/2)×0.412 kg×(2.039 ms^-1)²

⇒ H = 29.4 J – 0.856 J

⇒ H = 28.54 J

Question 25.

Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track (Fig. 6.16). Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. Given θ₁ = 30⁰, θ₂ = 60⁰, and h = 10 m, what are the speeds and times taken by the two stones?

Answer:

It is given that AB and AC are two frictionless planes inclined to the horizontal at angles θ₁ and θ₂ respectively.

The kinetic energy at B will be equal to the kinetic energy at C which will be the potential energy at A due to law of conservation of energy since there are no losses.

Let v₁ and v₂ be the final velocities of the stones on the planes AB and AC respectively and m be the mass of each stone.

Then, mgh = (1/2)mv₁² = (1/2)mv₂²

⇒ v₁ = v₂ = v (say)

If a₁ and a₂ are the accelerations of the two stones along the planes AB and AC respectively, then

a₁ = gsinθ₁

a₂ = gsinθ₂

As θ₂>θ₁, a₂>a₁.

Initial velocities of the two stones are zero.

From Newton’s first equation of motion,

v = u + at

⇒ v = 0 + at

⇒ v = at

⇒ t = v/a

Since v is constant for both the stones,

t ∝ 1/a

Since a₂>a₁, t₂<t₁.

So, the stone on plane AC will take less time than the stone on plane AB to reach the bottom.

Question 26.

A 1 kg block situated on a rough incline is connected to a spring of spring constant 100 N m^–1 as shown in Fig. 6.17. The block is released from rest with the spring in the unstretched position. The block moves 10 cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and the pulley is frictionless.

Answer:

Given,

Mass of the block, m = 1 kg

Spring constant, k = 100 N m^-1

Displacement of the block, s = 10 cm = 0.1 m

The forces can be shown in the figure as follows:

At equilibrium,

Surface normal reaction force, N = mgcos37°

Frictional force, f = μN where, μ is the coefficient of friction

⇒ f = mgsin37°

Net force acting on the block, F = mgsin37° - f

⇒ F = mgsin37° - μmgcos37°

⇒ F = mg(sin37° - μcos37°)

At equilibrium, work done by the block is equal to the potential energy of the spring according to law of conservation of energy.

So, mg(sin37° - μcos37°)s = (1/2)ks²

⇒ 1 kg×9.8 m s^-2×(0.602 – μ0.799)×0.1 = (1/2)×100Nm^-1×(0.1)²

⇒ 0.602 - μ × 0.799 = 0.51

⇒ μ × 0.799 = 0.092

⇒ μ = 0.115

Question 27.

A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with a uniform speed of 7 m s^–1. It hits the floor of the elevator (length of the elevator = 3 m) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary?

Answer:

Given,

Mass of the bolt, m = 0.3 kg

The speed of the bolt, v = 7 m s^-1

Length of the elevator, h = 3 m

The relative velocity of the bolt with respect to the elevator is zero. So, the potential energy gets converted into heat energy upon impact according to law of conservation of energy.

Heat produced, H = Loss of potential energy

⇒ H = mgh
where
m is the mass of the bolt
g is the acceleration due to gravity
h is the height of the elevator

Putting the value in the equation, we get

⇒ H = 0.3 kg × 9.8 m s^-2 × 3 m

⇒ H = 8.82 J
The answer will not change because of the fact that the relative velocity of the bolt with reference to the elevator is 0.

Question 28.

A trolley of mass 200 kg moves with a uniform speed of 36 km/h on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of 4 m s^–1 relative to the trolley in a direction opposite to the its motion, and jumps out of the trolley. What is the final speed of the trolley? How much has the trolley moved from the time the child begins to run?

Answer:

Given,

Mass of the trolley, m = 200 kg

Speed of the trolley, v=36km/h=(36×1000 m)/(60×60 s)= 10m/s

Mass of the child, m’ = 20 kg

Initial momentum of the system involving the child and the trolley, p_i = (m+m’)v

⇒ p_i = (200 kg + 20 kg)×10 ms^-1

⇒ p_i = 2200 kg m s^-1

Let v’ be the final velocity of the trolley with respect to the ground.

Final velocity of the child with respect to ground, v’’ = Final velocity of trolley with respect to ground – final velocity of child with respect to trolley.

So, v’’ = v’ – 4

Final momentum of the system, p_f = mv’ + m’v’’

⇒ p_f = mv’ + m’(v’-4)

⇒ p_f = 200v’ + 20(v’-4)

⇒ p_f = 220v’-80

According to the law of conservation of linear momentum,

Initial momentum (p_i) of system = Final momentum (p_f) of system

⇒ 2200 = 220v’-80

⇒ 220v’ = 2280

⇒ v’ = 2280/220

⇒ v’ = 10.36 m/s

Length of the trolley, l = 10 m

Speed of the child with respect to trolley, v’ = 4m/s

Time taken by the boy to cover the distance, t = l/v’

⇒ t = 10m/4 ms^-1

⇒ t = 2.5 s

Distance moved by the trolley in this time, s = v’t

⇒ s = 10.36 ms^-1 × 2.5 s

⇒ s = 25.9 m

Question 29.

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

Answer:

The potential energy of a system containing two masses is inversely proportional to the distance between the two masses. Also, if R is the radius of each billiard ball and r is the separation between them, then the potential energy becomes zero when the two balls touch each other, i.e., r = 2R. Only (v) satisfies these conditions. Hence, (i), (ii), (iii), (iv) and (vi) cannot possibly describe a collision of two billiard balls.

Question 30.

Consider the decay of a free neutron at rest:

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

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

Answer:

Let the masses of electron and proton be m and m’ respectively and their velocities be v and v’ respectively.

Using law of conservation of linear momentum,

Momentum of electron + momentum of proton = momentum of neutron

mv + m’v’ = 0

⇒ v’ = -mv/m’

This shows that the electrons and protons move in the opposite directions. If mass Δm has been converted into energy, then

(1/2)mv² + (1/2)m’v’² = Δmc²

⇒ (1/2)mv² + (1/2)m’(-mv/m’)² = Δmc²

⇒ (1/2)mv²(1+m/m’) = Δmc²

⇒

All the quantities on the RHS of the above equation are constant. Hence, v² is also constant. It shows that the emitted electron must have a constant energy. Hence, we cannot account for the observed continuous energy distribution in the β-decay of a neutron or a nucleus.

Work, Energy And Power

Class 11th Physics Part I CBSE Solution

Exercise

Class 11^th Physics Part I CBSE Solution