Chapter 5 Work, Energy and Power

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

NCERT Solutions for Class 11 Physics Chapter 5 (Work, Energy and Power) explore the relationship between the physical effort applied to a body and the resulting change in its energy state. This chapter introduces the scalar nature of these quantities, simplifying complex motion problems that were previously solved using vector-heavy dynamics in the Laws of Motion.

NCERT Solutions for Class 11 Physics Chapter 5 Allen – Work-Energy Relation NCERT Solutions for Class 11 Physics Chapter 5 by ALLEN is prepared by expert faculty that explains work-energy relation with clarity and accuracy. The step-by-step solutions are based on energy conservation principles which are important for solving JEE and NEET problems.

The concepts of Kinetic Energy, Potential Energy and Power are central in traditional and modern physics. These solutions provide a rigorous logical derivation of the transformation of energy from one form to another and the invariance of total energy in an isolated system.

1.0Class 11 Physics Chapter 5 : Key Concepts

This chapter focuses on the energy-based perspective of mechanics. Key lessons include:

Work Done by a Constant Force: Understanding the dot product of force and displacement $(W = F \cdot d = F d cos θ) .$
Work Done by a Variable Force: Using integration to calculate work when force changes with position $(W = \int F (x) d x) .$
Kinetic Energy: The energy possessed by an object due to its motion $(K = \frac{1}{2} m v^{2}) .$
Work-Energy Theorem: Proving that the work done by the net force on a body is equal to the change in its kinetic energy $(W = Δ K) .$
Potential Energy: Understanding energy stored due to position or configuration.
Gravitational Potential Energy: U = mgh.
Elastic Potential Energy: Stored in a compressed or stretched spring $(U = \frac{1}{2} k x^{2}) .$
Conservative and Non-conservative Forces: Differentiating between forces like Gravity (path-independent) and Friction (path-dependent).
Conservation of Mechanical Energy: The principle that E = K + U remains constant in the presence of only conservative forces.
Power: The rate at which work is done or energy is transferred $(P = \frac{d W}{d t} = F \cdot v) .$
Elastic Collisions: Both momentum and kinetic energy are conserved.
Inelastic Collisions: Only momentum is conserved; kinetic energy is lost (usually as heat or sound).

2.0Detailed NCERT Textbook Solutions : Class 11 Physics Chapter 5

SOLVED EXAMPLES

Find the angle between force $F = (3 \hat{i} + 4 \hat{j} - 5 \hat{k})$ unit and displacement $d = (5 \hat{i} + 4 \hat{j} + 3 \hat{k})$ unit. Also find the projection of $F$ on $d$ . Sol. $F . d = F_{x} d_{x} + F_{y} d_{y} + F_{z} d_{z}$ $= 3 (5) + 4 (4) + (- 5) (3) = 16$ unit Hence $F . d = Fd cos θ = 16$ unit Now, $F . F = F^{2} = F_{x}^{2} + F_{y}^{2} + F_{z}^{2}$ $= 9 + 16 + 25 = 50$ unit $F = 50$ unit and $d . d = d^{2} = d_{x}^{2} + d_{y}^{2} + d_{z}^{2}$ $= 25 + 16 + 9 = 50$ unit $d = 50$ unit $∴ cos θ = \frac{16}{50 50} = \frac{16}{50} = 0.32$ $θ = cos^{- 1} 0.32$ Projection of $F$ onto $d = d cos θ$ $= 50 \times 0.32 = 7.07 \times 0.32 = 2.26$ unit

Figure :

(a) The scalar product of two vectors $A$ and $B$ is a scalar : $A \cdot B = AB cos θ$ (b) $B cos θ$ is the projection of $B$ onto $A$ . (c) A $cos θ$ is the projection of $A$ onto $B$ .

It is well known that a raindrop falls under the influence of the downward gravitational force and the opposing resistive force. The latter is known to be proportional to the speed of the drop but is otherwise undetermined. Consider a drop of mass 1.00 g falling from a height 1.00 km. It hits the ground with a speed of $50.0 ms^{- 1}$ . (a) What is the work done by the gravitational force? What is the work done by the unknown resistive force? Sol. (a) The change in kinetic energy of the drop is,

$Δ K = \frac{1}{2} mv^{2} - 0 = \frac{1}{2} \times 1 0^{- 3} \times 50 \times 50 = 1.25 J$

where we have assumed that the drop is initially at rest. Assuming that g is a constant with a value $10 m / s^{2}$ , the work done by the gravitational force is.

$W_{g} = mgh = 1 0^{- 3} \times 10 \times 1 0^{3} = 10.0 J$

(b) From the work energy theorem,

$Δ K = W_{g} + W_{r}$

where, $W_{r}$ is the work done by the resistive force on the raindrop. Thus $W_{r} = Δ K - W_{g} = 1.25 - 10 = - 8.75 J$ is negative.

A cyclist comes to a skidding stop in 10 m. During this process, the force on the cycle due to the road is 200 N and is directly opposed to the motion. (a) How much work does the road do on the cycle? (b) How much work does the cycle do on the road? Sol. Work done on the cycle by the road is the work done by the stopping (frictional) force on the cycle due to the road. (a) The stopping force and the displacement make an angle of $18 0^{\circ} (π rad)$ with each other. Thus, work done by the road,

$W_{r} = Fd cos θ = 200 \times 10 \times cos π = - 2000 J$

It is this negative work that brings the cycle to a halt in accordance with WE theorem. (b) From Newton's Third Law an equal and opposite force acts on the road due to the cycle. Its magnitude is 200 N. However, the road undergoes no displacement. Thus, work done by cycle on the road is zero.

In a ballistics demonstration a police officer fires a bullet of mass 50.0 g with speed 200 $ms^{- 1}$ on soft plywood of thickness 2.00 cm. The bullet emerges with only 10% of its initial kinetic energy. What is the emergent speed of the bullet? Sol. The initial kinetic energy of the bullet is, $\frac{1}{2} mv^{2} = 1000 J$ . It has a final kinetic energy of $0.1 \times 1000 = 100 J$ . If $v_{f}$ , is the emergent speed of the bullet,

$\frac{1}{2} mv_{f}^{2} = 100 J v_{f} = \frac{2 \times 100 J}{0.05 kg} = 63.2 ms^{- 1}$

The speed is reduced by approximately 68% (not 90%).

A woman pushes a trunk on a railway platform which has a rough surface. She applies a force of 100 N over a distance of 10 m. Thereafter, she gets progressively tired and her applied force reduces linearly with distance to 50 N. The total distance through which the trunk has been moved is 20 m. Plot the force applied by the woman and the frictional force, which is 50 N versus displacement. Calculate the work done by the two forces over 20 m. Sol. The plot of the applied force is shown in figure. At $x = 20 m . F = 50 N (\neq = 0)$ . We are given that the frictional force f is $∣ f ∣ = 50 N$ . If opposes motion and acts in direction opposite to F . It is therefore, shown on the negative side of the force axis.

Plot of the force F applied by the woman and the opposing frictional force f versus displacement.

The work done by the woman is $W_{F} \to$ area of the rectangle ABCD + area of the trapezium CEID

$W_{F} = 100 \times 10 + \frac{1}{2} (100 + 50) \times 10 = 1000 + 750 = 1750 J$

The work done by the frictional force is, $W_{f} \to$ area of the rectangle AGHI

$W_{f} = (- 50) \times 20 = - 1000 J$

The area of the negative side of the force axis has a negative sign.

A block of mass $m = 1 kg$ , moving on a horizontal surface with speed $v_{i} = 2 ms^{- 1}$ enters a rough patch ranging from $x = 0.10 m$ to $x = 2.01 m$ . The retarding force $F_{r}$ on the block in this range is inversely proportional to x over this range,

$F_{r} = \frac{- k}{x} for 0.1 < x < 2.01 m$

$= 0$ for $x < 0.1 m$ and $x > 2.01 m$ , where $k = 0.5 J$ . What is the final kinetic energy and speed $v_{f}$ of the block as it crosses this patch? Sol. $K_{f} = K_{t} + \int_{0.1}^{2.01} \frac{( - k )}{x} dx = \frac{1}{2} mv_{i}^{2} - k [ℓ n (x)]_{0.1}^{2.01}$

$= \frac{1}{2} mv_{i}^{2} - k ℓ n (\frac{2.01}{0.1}) = 2 - 0.5 ℓ n (20.1) = 2 - 1.5 = 0.5 J v_{f} = \frac{2 K _{f}}{m} = 1 ms^{- 1}$

Here, note that $ℓ n$ is a symbol for the natural logarithm to the base e and not the logarithm to the base $10 [ℓ nX = lo g_{e} X = 2.303 lo g_{10} X]$

A bob of mass m is suspended by a light string of length L. It is imparted a horizontal velocity $v_{0}$ at the lowest point A such that it completes a semi-circular trajectory in the vertical plane with the string becoming slack only on reaching the topmost point C. This is shown in figure. Obtain an expression for (i) $v_{0}$ (ii) the speeds at points B and C; (iii) the ratio of the kinetic energies $(\frac{K _{B}}{K _{C}})$ at B and C. Comment on the nature of the trajectory of the bob after it reaches the point C .

Sol. (i) There are two external force on the bob (a) : gravity and (b) the tension (T) in the string. The tension latter does no work since the displacement of the bob is always normal to the string. The potential energy of the bob is thus associated with the gravitational force only. The total mechanical energy E of the system is conserved. We take the potential energy of the system to be zero at the lowest point A. Thus, at A:

$E = \frac{1}{2} mv_{0}^{2}$

$T_{A} - mg = \frac{mv _{0}^{2}}{L}$ [Newton's Second Law] where $T_{A}$ is the tension in the string at A. At the highest point C, the string slackens, as the tension in the string $(T_{c})$ becomes zero.

Thus, at C

$E = \frac{1}{2} mv_{c}^{2} + 2 mgL$

$mg = \frac{mv _{c}^{2}}{L}$ [Newton's Second Law].. where $v_{C}$ is the speed at C. From equation (ii) and (iii)

$E = \frac{5}{2} mgL$

Equating this to the energy at A

$\frac{5}{2} mgL = \frac{m}{2} v_{0}^{2} or, v_{0} = 5 gL$

(ii) It is clear from equation (iii)

$v_{c} = gL$

At B, the energy is

$E = \frac{1}{2} mv_{B}^{2} + mgL$

Equating this to the energy at A and employing the result from (i), namely

$v_{0}^{2} = 5 gL . \frac{1}{2} mv_{B}^{2} + mgL = \frac{1}{2} mv_{0}^{2} = \frac{5}{2} mg ∴ v_{B} = 3 gL$

(iii) The ratio of the kinetic energies at B and C is :

$\frac{K _{B}}{K _{C}} = \frac{\frac{1}{2} mv _{B}^{2}}{\frac{1}{2} mv _{c}^{2}} = \frac{3}{1}$

At point C , the string becomes slack and the velocity of the bob is horizontal and to the left. If the connecting string is cut at this instant, the bob will execute a projectile motion with horizontal projection akin to a rock kicked horizontally from the edge of a cliff. Otherwise the bob will continue on its circular path and complete the revolution.

To simulate car accidents, auto manufacturers study the collisions of moving cars with mounted springs of different spring constants. Consider a typical simulation with a car of mass 1000 kg moving with a speed 18.0 km/h on a smooth road and colliding with a horizontally mounted spring of spring constant $5.25 \times 1 0^{3} N m^{- 1}$ . What is the maximum compression of the spring? Sol. At maximum compression the kinetic energy of the car is converted entirely into the potential energy of the spring. The kinetic energy of the moving car is

$K = \frac{1}{2} mv^{2} = \frac{1}{2} \times 1 0^{3} \times 5 \times 5 K = 1.25 \times 1 0^{4} J [18 kmh^{- 1} = 5 m / s]$

At maximum compression $x_{m}$ , the potential energy $U$ of the spring is equal to the kinetic energy K of the moving car from the principle of conservation of mechanical energy.

$U = \frac{1}{2} kx_{m}^{2} = 1.25 \times 1 0^{4} J$

We obtain

$x_{m} = 2.00 m$

We note that we have idealised the situation. The spring is considered to the massless. The surface has been considered to possess negligible friction.

Consider Q. 8 taking the coefficient of friction, $μ$ , to be 0.5 and calculate the maximum compression of the spring. Sol. In presence of friction, both the spring force and the frictional force act so as to oppose the compression of the spring as shown in Figure. We invoke the work-energy theorem, rather than the conservation of mechanical energy.

The change in kinetic energy is

The forces acting on the car

$Δ K = K_{f} - K_{i} = 0 - \frac{1}{2} mv^{2}$

The work done by the net force is

$W = - \frac{1}{2} kx_{m}^{2} - μ mg x_{m}$

Equating we have

$\frac{1}{2} m v^{2} = \frac{1}{2} k x_{m}^{2} + μ m g x_{m}$

Now $μ mg = 0.5 \times 1 0^{3} \times 10 = 5 \times 1 0^{3} N$ (taking $g = 10.0 m s^{- 2}$ ). After rearranging the above equation we obtain the following quadratic equation in the unknown $x_{m}$ .

$kx_{m}^{2} + 2 μ mg x_{m} - mv^{2} = 0 x_{m} = \frac{- μ mg + [ μ ^{2} m ^{2} g ^{2} + mkv ^{2} ] ^{1/2}}{k}$

where we take the positive square root since $x_{m}$ is positive. Putting in numerical values we obtain

$x_{m} = 1.43 m$

which, as expected, is less than the result in Q.8.

If the two forces on the body consist of a conservative force $F_{c}$ and a non-conservative force $F_{nc}$ , the conservation of mechanical energy formula will have to be modified. By the WE theorem

$(F_{c} + F_{nc}) Δ x = Δ k$

But

$F_{c} Δ x = - Δ U$

Hence,

$Δ (K + U) = F_{nc} Δ x Δ E = F_{nc} Δ x$

where E is the total mechanical energy. Over the path this assumes the form

$E_{f} - E_{i} = W_{n c}$

where $W_{nc}$ is the total work done by the nonconservative forces over the path. Note that unlike the conservative force, $W_{nc}$ depends on the particular path i to f.

An elevator can carry a maximum load of 1800 kg (elevator + passengers) is moving up with a constant speed of $2 ms^{- 1}$ . The frictional force opposing the motion is 4000 N. Determine the minimum power delivered by the motor to the elevator in watts as well as in horse power. Sol. The downward force on the elevator is

$F = mg + F_{f} = (1800 \times 10) + 4000 = 22000 N$

The motor must supply enough power to balance this force. Hence

$p = F . v = 22000 \times 2 = 44000 W = 59 h p . [1 h p = 746 W]$

Slowing down of neutrons: In a nuclear reactor a neutron of high speed (typically $1 0^{7}$ $m s^{- 1}$ ) must be slowed to $1 0^{3} m s^{- 1}$ so that it can have a high probability of interacting with isotope $_{92}^{235} U$ and causing it to fission. Show that a neutron can lose most of its kinetic energy in an elastic collision with a light nuclei like deuterium or carbon which has a mass of only a few times the neutron mass. The material making up the light nuclei, usually heavy water $(D_{2} O)$ or graphite, is called a moderator.

Sol. The initial kinetic energy of the neutron is

$k_{li} = \frac{1}{2} m_{1} v_{li}^{2}$

final velocity of neutron is given by

$v_{1 f} = (\frac{m _{1} - m _{2}}{m _{1} + m _{2}}) v_{1 i}$

Now, its final kinetic energy

$K_{1 f} = \frac{1}{2} m_{1} v_{1 f}^{2} = \frac{1}{2} m_{1} (\frac{m _{1} - m _{2}}{m _{1} + m _{2}})^{2} v_{1 i}^{2}$

The fractional kinetic energy lost is

$f_{1} = \frac{K _{1 f}}{K _{1 i}} = (\frac{m _{1} - m _{2}}{m _{1} + m _{2}})^{2}$

while the fractional kinetic energy gained by the moderating nuclei $\frac{K _{2 f}}{K _{1 i}}$ is

$f_{2} = 1 - f_{1} (elastic collision) = \frac{4 m _{1} m _{2}}{( m _{1} + m _{2} ) ^{2}}$

For deuterium $m_{2} = 2 m_{1}$ and we obtain $f_{1} = \frac{1}{9}$ while $f_{2} = \frac{8}{9}$ . Almost 90% of the neutron's energy is transferred to deuterium. For carbon $f_{1} = 71.6%$ and $f_{2} = 28.4%$ .

Consider the collision depicted in figure to be between two billiard balls with equal masses $m_{1} = m_{2}$ . The first ball is called the cue while the second ball is called the target. The billiard player wants to 'sink' the target ball in a corner pocket, which is at an angle $θ_{2} = 3 7^{\circ}$ . Assume that the collision is elastic and that friction and rotational motion are not important. Obtain $θ_{1}$ .

Collision of mass $m_{1}$ , with a stationary mass $m_{2}$ . Sol. From momentum conservation, since the masses are equal

$v_{1 i} = v_{1 f} + v_{2 f}$

$v_{1 i}^{2} = (v_{1 f} + v_{2 f}) \cdot (v_{1 f} + v_{2 f}) = v_{1 f}^{2} + v_{2 f}^{2} + 2 v_{1 f} \cdot v_{2 f} + v_{2 f}^{2} + 2 v_{1 f} v_{2 f} cos (θ_{1} + 3 7^{\circ})}$

Since the collision is elastic and $m_{1} = m_{2}$ it follows from conservation of kinetic energy that $v_{li}^{2} = v_{1 f}^{2} + v_{2 f}^{2}$ Comparing equation (i) and (ii), we get

$cos {θ_{1} + 37} = 0$

$θ_{1} + 3 7^{\circ} = 9 0^{\circ}$

Thus,

$θ_{1} = 5 3^{\circ}$

This proves the following result : when two equal masses undergo a glancing elastic collision with one of them at rest, after the collision, they will move at right angles to each other.

EXERCISE QUESTION WITH SOLUTIONS

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.

Sol. (a) Positive : In the given case, force and displacement are in the same direction. Hence, the sign of work done is positive. In this case, the work is done on the bucket. (b) Negative: In the given case, the direction of force (vertically downward) and displacement (vertically upward) are opposite to each other. Hence, the sign of work done is negative. (c) Negative : Since the direction of frictional force is opposite to the direction of motion, the work done by frictional force is negative in this case. (d) Positive : Here the body is moving on a rough horizontal plane. Frictional force opposes the motion of the body. Therefore, in order to maintain a uniform velocity, a uniform force must be applied to the body. Since the applied force acts in the direction of motion of the body, the work done is positive. (e) Negative : The resistive force of air acts in the direction opposite to the direction of motion of the pendulum. Hence, the work done is negative in this case.

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 10s, (b) work done by friction in 10s, (c) work done by the net force on the body in 10s, (d) change in kinetic energy of the body in 10s, and interpret your results. Sol. Mass of the body, $m = 2 kg$ Applied force, $F = 7 N$ Coefficient of kinetic friction, $μ = 0.1$ Initial velocity, $u = 0$ Time, $t = 10 s$ The acceleration produced in the body by the applied force is given by Newton's second law of motion as: $a^{'} = \frac{F}{m} = \frac{7}{2} = 3.5 ms^{- 2}$ . Frictional force is given as:

$f = μ mg = 0.1 \times 2 \times 9.8 = - 1.96 N$

The acceleration produced by the frictional force:

$a^{''} = - \frac{1.96}{2} = - 0.98 ms^{- 2}$

Total acceleration of the body (a) : $a^{'} + a^{''}$

$= 3.5 + (- 0.98) = 2.52 ms^{- 2}$

The distance travelled by the body is given by the equation of motion:

$s = ut + (1/2) at^{2} = 0 + (\frac{1}{2}) \times 2.52 \times (10)^{2} = 126 m$

(a) Work done by the applied force,

$W_{a} = F \times s = 7 \times 126 = 882 J$

(b) Work done by the frictional force,

$W_{f} = f \times s = - 1.96 \times 126 = - 247 J$

$W_{net} = 5.04 \times 126 = 635 J$

(d) From the first equation of motion, final velocity can be calculated as:

$v = u + at = 0 + 2.52 \times 10 = 25.2 m / s$

Change in kinetic energy

$= (\frac{1}{2}) mv^{2} - (\frac{1}{2}) mu^{2} = (\frac{1}{2}) \times 2 (v^{2} - u^{2}) = (25.2)^{2} - 0^{2} = 635 J$

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

exercise-ques-3-(i)-chap-5-physics-class-11

class-11-chap-5-physics-exercise-ques-3-(ii)

Sol. Total energy of a system is given by the relation:

$E = P.E. + K. E . ∴ K.E. = E - P.E.$

Kinetic energy of a body is a positive quantity. It cannot be negative. Therefore, the particle will not exist in a region where K.E. becomes negative. (i) For $x > a$ , P.E. $(U_{0}) > E$ ∴ K.E. becomes negative. Hence, the object cannot exist in the region $x > a$ . (ii) For $x < a$ and $x > b$ , P.E. $(U_{0}) >$ E. ∴ K.E. becomes negative. Hence the object cannot be present in the region $x < a$ and $x > b$ . (c) $x > a$ and $x < b; - U_{1}$

In the given case, the condition regarding the positivity of K.E. is satisfied only in the region between $x > a$ and $x < b$ . The minimum potential energy in this case is $- U_{1}$ . Therefore, K.E. $= E - (- U_{1}) = E + U_{1}$ . Therefore, for the positivity of the kinetic energy, the total energy of the particle must be greater than $- U_{1}$ . So, the minimum total energy the particle must have is $- U_{1}$ .

(d) $- \frac{b}{2} < x < \frac{a}{2}; \frac{a}{2} < x < \frac{b}{2}; - U_{1}$

In the given case, the potential energy $(U_{0})$ of the particle becomes greater than the total energy

(e) for $- \frac{b}{2} < x < \frac{b}{2}$ and $- \frac{a}{2} < x < \frac{a}{2}$ .

Therefore, the particle will not exist in these regions. The minimum potential energy in this case is $- U_{1}$ . Therefore, K.E. $= E - (- U_{1})$ $= E + U_{1}$ . Therefore, for the positivity of the kinetic energy, the total energy of the particle must be greater than $- U_{1}$ . So, the minimum total energy of the particle must have is $- U_{1}$ .

The potential energy function for a particle executing linear simple harmonic motion is given by $U (x) = \frac{kx ^{2}}{2}$ , where k is the force constant of the oscillator. For $k = 0.5 N m^{- 1}$ , the graph of U(x) versus x is shown in Figure. Show that a particle of total energy 1 J moving under this potential must 'turn back' when it reaches $x = \pm 2 m$ .

Sol. Total energy of the particle, $E = 1 J$ Force constant, $k = 0.5 N m^{- 1}$ Kinetic energy of the particle, $K = \frac{1}{2} mv^{2}$ According to the conservation law:

$E 1 = U + K = \frac{1}{2} kx^{2} + \frac{1}{2} mv^{2}$

At the moment of 'turn back', velocity (and hence K) becomes zero.

$∴ 1 = \frac{1}{2} kx^{2} \frac{1}{2} \times 0.5 x^{2} = 1 x^{2} = 4 x = \pm 2$

Hence, the particle turns back when it reaches $x = \pm 2 m$ .

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 Figure (i) the man walks 2m carrying a mass of 15 kg on his hands. In Figure (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?

Sol. (a) Rocket : The burning of the casing of a rocket in flight (due to friction) results in the reduction of the mass of the rocket. According to the conservation of energy: Total energy = Potential energy+Kinetic energy $E = mgh + (1/2) mv^{2}$ The reduction in the rocket's mass causes a drop in the total energy. Therefore, the heat energy required for the burning is obtained from the rocket. (b) Gravitational force is a conservative force. Since the work done by a conservative force over a closed path is zero, the work done by the gravitational force over every complete orbit of a comet is zero. (c) When an artificial satellite, orbiting around earth, moves closer to earth, its potential energy decreases because of the reduction in the height. Since the total energy of the system remains constant, the reduction in P.E. results in an increase in K.E. Hence, the velocity of the satellite increases. However, due to atmospheric friction, the total energy of the satellite decreases by a small amount. (d) Work done in figure (i) Mass, $m = 15 kg$ Displacement, $s = 2 m$ Work done, $W = Fs cos θ$ Where, $θ =$ Angle between force and displacement $= mgs cos θ = 15 \times 2 \times 9.8 cos 9 0^{\circ} = 0$ Work done in figure (ii) Mass, $m = 15 kg$ Displacement, $s = 2 m$ Here, the direction of the force applied on the rope and the direction of the displacement of the rope are same. Therefore, the angle between them, $θ = 0^{\circ}$ Since $cos 0^{\circ} = 1$ Work done, $W = Fs cos θ = mgs$ $= 15 \times 9.8 \times 2 = 294 J$ Hence, more work is done in the Figure (ii).

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. Sol. (a) Decreases : A conservative force does a positive work on a body when it displaces the body in the direction of force. As a result, the body advances toward the centre of force. It decreases the separation between the two, thereby decreasing the potential energy of the body. (b) Kinetic energy : The work done against the direction of friction reduces the velocity of a body. Hence, there is a loss of kinetic energy of the body.

(c) External force : Internal forces, irrespective of their direction, cannot produce any change in the total momentum of a body. Hence, the total momentum of a many- particle system is proportional to the external forces acting on the system. (d) Total linear momentum : The total linear momentum always remains conserved whether it is an elastic collision or an inelastic collision.

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. Sol. (a) False : In an elastic collision, the total energy and momentum of both the bodies, and not of each individual body, is conserved. (b) False : The external forces on the body may change the total energy of the body. (c) False : The work done in the motion of a body over a closed loop is zero for a conservation force only. (d) True : In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system. This is because in such collisions, there is always a loss of energy in the form of heat, sound, etc.
Answer carefully, with reasons: (a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact)? (b) Is the total linear momentum conserved during the short time of an elastic collision of two balls? (c) What are the answers to (a) and (b) for an inelastic collision? (d) If the potential energy of two billiard balls depends only on the separation distance between their 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). Sol. (a) No : K.E. is not conserved during the given elastic collision, K.E. before and after collision is the same. Infact, during collision, K.E. of the balls gets converted into potential energy. (b) Yes : In an elastic collision, the total linear momentum of the system always remains conserved. (c) No; Yes : In an inelastic collision, there is always a loss of kinetic energy, i.e., the total kinetic energy of the billiard balls before collision will always be greater than that after collision. The total linear momentum of the system of billiards balls will remain conserved even in the case of an inelastic collision. (d) Elastic : In the given case, the forces involved are conservative. This is because they depend on the separation between the centres of the billiard balls. Hence, the collision is elastic.
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) $t^{2}$ Sol. From,

$v = u + at v = 0 + at = at$

As power, $P = Fxv$

$∴ P = (ma) \times at = ma^{2} t$

As m and a are constants, therefore, $P \propto t$ Ans. (ii) 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) $t^{2}$ Sol. As power, $P =$ force × velocity

$P = [MLT^{- 2}] [LT^{- 1}] = [ML^{2} T^{- 3}] = constant As, P = [ML^{2} T^{- 3}]$

or, $L^{2} / T^{3} =$ constant

$∴ L^{2} \propto T^{3} or L \propto T^{3/2}$

Ans. (iii) $t^{3/2}$

A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by $F = - \hat{i} + 2 \hat{j} + 3 \hat{k} N$ 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? Sol. Force exerted on the body,

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

Displacements, $s = 4 \hat{k} m$ Work done, $W = F . s$

$= (- \hat{i} + 2 \hat{j} + 3 k) \cdot (4 \hat{k}) = 0 + 0 + 3 \times 4 = 12 J$

Hence, 12 J of work is done by the force on the body.

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 $= 9.11 \times 1 0^{- 31} kg$ , proton mass

$= 1.67 \times 1 0^{- 27} kg, 1 eV = 1.60 \times 1 0^{- 19} J) .$

Sol. Mass of the electron, $m = 9.11 \times 1 0^{- 31} kg$ Mass of the proton, $m_{p} = 1.67 \times 1 0^{- 27} kg$ Kinetic energy of the electron,

$E_{Ke} = 10 keV = 1 0^{4} eV = 1 0^{4} \times 1.60 \times 1 0^{- 19} = 1.60 \times 1 0^{- 15} J$

Kinetic energy of the proton, $E_{KP} = 100 keV = 1 0^{5} eV = 1.60 \times 1 0^{- 14} J$ For the velocity of an electron $v_{e}$ , its kinetic energy is given by the relation:

$E_{Ke} = (\frac{1}{2}) mv_{e}^{2}$

$∴ v_{e} = (\frac{2 E _{ke}}{m})^{\frac{1}{2}} = (\frac{2 \times 1.60 \times 1 0 ^{- 15}}{9.11 \times 1 0 ^{- 31}})^{\frac{1}{2}} = 5.93 \times 1 0^{7} m / s$

For the velocity of a proton $v_{p}$ , its kinetic energy is given by the relation:

$E_{K_{p}} = (\frac{1}{2}) mv_{p}^{2}$

$v_{p} = (\frac{2 \times 1.60 \times 1 0 ^{- 14}}{1.67 \times 1 0 ^{- 27}})^{\frac{1}{2}} = 4.38 \times 1 0^{6} m / s$ Hence, the electron is moving faster than the proton. The ratio of their speeds

$\frac{v _{e}}{v _{p}} = \frac{5.93 \times 1 0 ^{7}}{4.38 \times 1 0 ^{6}} = 13.54 : 1$

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 $10 m s^{- 1}$ ? Sol. Radius of the rain drop, $r = 2 mm = 2 \times 1 0^{- 3} m$ Volume of the rain drop, $V = (4/3) π r^{3}$

$V = \frac{4}{3} \times 3.14 \times (2 \times 1 0^{- 3})^{3} m^{- 3}$

Density of water, $ρ = 1 0^{3} kg m^{- 3}$ Mass of the rain drop, $m = ρ V$

$m = \frac{4}{3} \times 3.14 \times (2 \times 1 0^{- 3})^{3} \times 1 0^{3} kg = 33.5 \times 1 0^{- 6} kg$

Gravitational force, $F = mg, g = 9.8 m / s^{2}$

$F = 328.3 \times 1 0^{- 6} N$

The work done by the gravitational force on the drop in the first half of its journey:

$W_{I} = Fs = 328.3 \times 1 0^{- 6} \times 250 = 0.082 J$

This amount of work is equal to the work done by the gravitational force on the drop in the second half of its journey,

$i.e. W_{II} = 0.082 J$

As per the law of conservation of energy, if no resistive force is present, then the total energy of the rain drop will remain the same. ∴ Total energy at the top:

$E_{T} = mgh + 0 = 33.5 \times 1 0^{- 6} \times 9.8 \times 500 = 0.164 J$

Due to the presence of a resistive force, the drop hits the ground with a velocity of 10 m/s. ∴ Total energy at the ground :

$E_{G} = \frac{1}{2} mv^{2} + 0 = \frac{1}{2} \times 33.5 \times 1 0^{- 6} \times (10)^{2} = 1.675 \times 1 0^{- 3} J$

$∴$ work done by resistive force

$= E_{G} - E_{T} = - 0.162 J$

A molecule in a gas container hits a horizontal wall with speed $200 ms^{- 1}$ and angle 30° with the normal, and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic? Sol. The momentum of the gas molecule remains conserved whether the collision is elastic or inelastic. The gas molecule moves with a velocity of 200 m/s and strikes the stationary wall of the container, rebounding with the same speed. It shows that the rebound velocity of the wall remains zero. Hence, the total kinetic energy of the molecule remains conserved during the collision. The given collision is an example of an elastic collision.
A pump on the ground floor of a building can pump up water to fill a tank of volume $30 m^{3}$ in 15 min . If the tank is 40 m above the ground, and the efficiency of the pump is $30%$ , how much electric power is consumed by the pump? Sol. Volume of the tank, $V = 30 m^{3}$

$Time of operation, t = 15 min = 15 \times 60 = 900 s$

Height of the tank, $h = 40 m$ Efficiency of the pump, $η = 30%$ Density of water, $ρ = 1 0^{3} kg / m^{3}$ Mass of water, $m = ρ V = 30 \times 1 0^{3} kg$ Output power can be obtained as:

$P_{0} = Work done / Time = mgh / t = 30 \times 1 0^{3} \times 9.8 \times 40/900 = 13.067 \times 1 0^{3} W$

For input power $P_{i}$ , efficiency $η$ , is given by the relation:

$η = \frac{P _{0}}{P _{i}} = 30% P_{i} = 13.067 \times 100 \times 1 0^{3} /30 = 0.436 \times 1 0^{3} W = 43.6 kW$

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 figure is a possible result after collision?

exercise-ques-16-physics-chap-5-class-11

Sol. It can be observed that the total momentum before and after collision in each case is constant. For an elastic collision, the total kinetic energy of a system remains conserved before and after collision. For mass of each ball bearing m, we can write: Total kinetic energy of the system before collision:

$= \frac{1}{2} m v^{2} + \frac{1}{2} (2 m) \times 0^{2} = \frac{1}{2} m v^{2}$

Case (i) Total kinetic energy of the system after collision:

$= \frac{1}{2} m \times 0 + \frac{1}{2} (2 m) (\frac{v}{2})^{2} = \frac{1}{4} mv^{2}$

Hence, the kinetic energy of the system is not conserved in case (i). Case (ii) Total kinetic energy of the system after collision:

$= \frac{1}{2} (2 m) \times 0 + \frac{1}{2} mv^{2} = \frac{1}{2} mv^{2}$

Hence, the kinetic energy of the system is conserved in case (ii). Case (iii) Total kinetic energy of the system after collision:

$= \frac{1}{2} (3 m) (\frac{v}{3})^{2} = \frac{1}{6} m v^{2}$

Hence, the kinetic energy of the system is not conserved in case (iii). Hence, Case II is the only possibility.

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

physics-class-11-maths-chap-5-exercise-ques-17

Sol. 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 the ball B would move with the velocity of A. Thus the bob A will not rise after the collision.

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 ? Sol. Length of the pendulum, $ℓ = 1.5 m$

$Mass of the bob = m$

Energy dissipated $= 5%$

According to the law of conservation of energy, the total energy of the system remains constant.

At the horizontal position: Potential energy of the bob, $E_{p} = mg ℓ$ Kinetic energy of the bob, $E_{K} = 0$ Total energy $= mg ℓ$

At the lowermost point (mean position): Potential energy of the bob, $E_{p} = 0$ Kinetic energy of the bob, $E_{K} = \frac{1}{2} mv^{2}$

$Total energy E_{k} = \frac{1}{2} m v^{2}$

As the bob moves from the horizontal position to the lowermost point, 5% of its energy gets dissipated.

The total energy at the lowermost point is equal to 95% of the total energy at the horizontal point, i.e.

$\frac{1}{2} m v^{2} = (\frac{95}{100}) m g ℓ ∴ v = (\frac{2 \times 95 \times 1.5 \times 9.8}{100})^{\frac{1}{2}} = 5.28 m / s$

A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27km/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 $= 5.28 m / s$ . What is the speed of the trolley after the entire sand bag is empty?

Sol. As the trolley carrying the sand bag is moving uniformly, therefore, external force on the system $= 0$ .

When the sand leaks out, it does not lead to the application of any external force on the trolley. Hence, the speed of the trolley shall not change.

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

Sol. Mass of the body, $m = 0.5 kg$ Velocity of the body is governed by the equation, $v = ax^{3/2}$ and $a = 5 m^{1/2} s^{- 1}$ . Initial velocity, $u ($ at $x = 0) = 0$ Final velocity $v ($ at $x = 2 m) = 102 m / s$ Work done, $W =$ Change in kinetic energy

$= \frac{1}{2} m (v^{2} - u^{2}) = \frac{1}{2} \times 0.5 [(102)^{2} - 0^{2}] = \frac{1}{2} \times 0.5 \times 10 \times 10 \times 2 = 50 J$

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 $A = 30$ $m^{2}, v = 36 km / h$ and the density of air is 1.2 kg $m^{- 3}$ . What is the electrical power produced?

Sol. Area of the circle swept by the windmill,

$A = 30 cm^{2}$

Velocity of the wind $= v = 36 km / h = 10 m / s$ Density of air $= ρ = 1.2 kg m^{- 3}$

(a) Volume of the wind flowing through the windmill per sec =Av Mass of the wind flowing through the windmill per sec = $ρ Av$ Mass m, of the wind flowing through the windmill in time $t = ρ Avt$ (b) Kinetic energy of air $= \frac{1}{2} mv^{2}$

$= \frac{1}{2} (ρ A v t) v^{2} = \frac{1}{2} ρ A v^{3} t$

(c) Area of the circle swept by the windmill $= A = 30 m^{2}$ Velocity of the wind $= v = 36 km / h$ Density of air, $ρ = 1.2 kg m^{- 3}$ Electrical energy produced = 25% of the wind energy

$= \frac{25}{100} \times Kinetic energy of air = \frac{1}{8} ρ Av^{3} t$

Electrical power = Electrical energy/Time

$= \frac{1}{8} ρ A v^{3} t / t = \frac{1}{8} ρ A v^{3} = \frac{1}{8} \times 1.2 \times 30 \times (10)^{3} = 4.5 kW$

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 $3.8 \times 1 0^{7} J$ of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up? Sol. (a) Mass of the weight, $m = 10 kg$ Height to which the person lifts the weight, $h = 0.5 m$ Number of times the weight is lifted, $n = 1000$ ∴ Work done against gravitational force:

$= n (m g h) = 1000 \times 10 \times 9.8 \times 0.5 = 49 kJ$

(b) Energy equivalent of 1 kg of

$fat = 3.8 \times 1 0^{7} J Efficiency rate = 20%$

Mechanical energy supplied by the person's body:

$= \frac{20}{100} \times 3.8 \times 1 0^{7} J = \frac{1}{5} \times 3.8 \times 1 0^{7} J = 0.76 \times 1 0^{7} J$

Equivalent mass of fat lost by the dieter:

$= \frac{49 \times 1 0 ^{3}}{0.76 \times 1 0 ^{7}} = 6.45 \times 1 0^{- 3} kg$

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. Sol. (a) Power used by the family.

$P = 8 kW = 8 \times 1 0^{3} W$

Solar energy received per square metre $= 200 W$ Efficiency of conversion from solar to electricity energy = 20% Area required to generate the desired electricity = A As per the information given in the question, we have:

$8 \times 1 0^{3} = 20% \times (A \times 200) = \frac{20}{100} \times A \times 200$

$∴ A = \frac{8 \times 1 0 ^{3}}{40} = 200 m^{2}$

(b) The area of a solar plate required to generate 8 kW of electricity is almost equivalent to the area of the roof of a building having dimensions $14 m \times 14 m$ , to 15 m × 15m. This result in a roof area of approximately $200 m^{2}$ to $225 m^{2}$ .

3.0NCERT Class 11 Physics – Chapter-Wise Solutions & Explanations

Here you will get NCERT Solutions for Class 11 chapter wise Physics solved with simple explanations for each question asked in the text book. Learn the basics, solve problems correctly and build a strong foundation in Physics.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Units and Measurements
Chapter 2	Motion in a Straight Line
Chapter 3	Motion in a Plane
Chapter 4	Laws of Motion
Chapter 6	System of Particles and Rotational Motion
Chapter 7	Gravitation
Chapter 8	Mechanical Properties of Solids
Chapter 9	Mechanical Properties of Fluids
Chapter 10	Thermal Properties of Matter
Chapter 11	Thermodynamics
Chapter 12	Kinetic Theory
Chapter 13	Oscillations
Chapter 14	Waves

4.0Key Features of NCERT Solutions for Class 11 Physics Chapter 5

Clear Work Calculations: Understand how the scalar product is used to calculate work done when force and displacement act in different directions.
Easy Understanding of Spring Mechanics: Learn the work done by a spring force and elastic potential energy through step-by-step explanations and derivations.
Systematic Collision Solutions: Understand one-dimensional elastic collisions using the laws of conservation of linear momentum and mechanical energy.
Real-Life Numerical Problems: Solve questions based on stopping distance, power, water pumps, work, and energy with proper formulas, calculations, and units.
Prepared by ALLEN Subject Experts: Get accurate and detailed NCERT Solutions for Class 11 Physics Chapter 5, prepared according to the latest NCERT syllabus.
Simple and Concept-Based Explanations: Understand Work, Energy and Power, kinetic energy, potential energy, conservation of energy, and collisions through clear explanations and step-by-step methods.
Complete NCERT Exercise Coverage: Get detailed solutions to NCERT Class 11 Physics Chapter 5 examples and exercise questions, covering important concepts of Work, Energy and Power for CBSE, JEE, and NEET preparation.

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