Question 1

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.

Accepted Answer

Solution:
(a) Work done by the man in lifting the bucket is positive because the force applied by the man and the displacement of the bucket are in the same direction (upwards).

(b) Work done by gravitational force is negative because gravity acts downward while the displacement is upward.

(c) Work done by friction on a body sliding down an inclined plane is negative because friction opposes the motion.

(d) Work done by an applied force on a body moving with uniform velocity on a rough horizontal plane is positive because the applied force and displacement are in the same direction.

(e) Work done by the resistive force of air on a vibrating pendulum is negative because the resistive force opposes the motion, removing energy from the system. — Work done is positive if force and displacement are in the same direction, negative if opposite. Friction and resistive forces oppose motion, so their work is negative. Gravity does negative work when object moves upward against it.

Question 2

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.

Accepted Answer

Given:
Mass, m = 2 kg
Applied force, F = 7 N
Coefficient of kinetic friction, μ = 0.1
Time, t = 10 s

Step 1: Calculate friction force:
Normal force, N = mg = 2 × 9.8 = 19.6 N
Friction force, f = μN = 0.1 × 19.6 = 1.96 N

Step 2: Calculate net force:
Net force, F_net = Applied force - friction = 7 - 1.96 = 5.04 N

Step 3: Calculate acceleration:
a = F_net / m = 5.04 / 2 = 2.52 m/s²

Step 4: Calculate displacement in 10 s:
s = ut + 0.5 a t² = 0 + 0.5 × 2.52 × 100 = 126 m

(a) Work done by applied force:
W_applied = F × s = 7 × 126 = 882 J

(b) Work done by friction:
W_friction = -f × s = -1.96 × 126 = -246.96 J

(c) Work done by net force:
W_net = F_net × s = 5.04 × 126 = 635.04 J

(d) Change in kinetic energy:
ΔKE = 0.5 m v²
First find final velocity:
v = u + at = 0 + 2.52 × 10 = 25.2 m/s
ΔKE = 0.5 × 2 × (25.2)² = 0.5 × 2 × 635.04 = 635.04 J

Interpretation:
The work done by the net force equals the change in kinetic energy, confirming the work-energy theorem. The applied force does positive work, friction does negative work, and the net work results in kinetic energy gain. — Calculated friction force and net force, then displacement using kinematics. Work done is force times displacement. Change in kinetic energy matches net work done, confirming work-energy theorem.

Question 3

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 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.

Accepted Answer

Solution:
For each potential energy graph:
- The particle cannot be found in regions where potential energy V(x) > total energy E, because kinetic energy K = E - V(x) would be negative, which is impossible.
- Minimum total energy is the lowest point on the potential energy curve.

Physical contexts:
- Parabolic potential (harmonic oscillator) corresponds to mass on a spring.
- Potential wells correspond to bound states in quantum mechanics or classical particles trapped in a potential.
- Potential barriers correspond to situations like tunneling or particles overcoming energy barriers.

Exact regions depend on the shape of each graph in Fig. 5.11. — Particle's kinetic energy is E - V(x). Where V(x) > E, kinetic energy is negative, so particle cannot be there. Minimum total energy is minimum of V(x). Physical contexts relate to classical or quantum systems with similar potentials.

Question 4

4 The potential energy function for a particle executing linear simple harmonic motion is given by  $V(x) = kx^2 / 2$ , where  $k$  is the force constant of the oscillator. For  $k = 0.5\mathrm{Nm}^{-1}$ , 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  $x = \pm 2\mathrm{m}$ .

Accepted Answer

Given:
Potential energy, V(x) = (k x^2)/2
k = 0.5 N/m
Total energy, E = 1 J

At turning points, kinetic energy K = 0, so total energy E = potential energy V(x).

Calculate V(x) at x = 2 m:
V(2) = 0.5 × (2)^2 / 2 = 0.5 × 4 / 2 = 1 J

Since V(2) = E, the particle cannot go beyond x = ±2 m because beyond this point potential energy would exceed total energy, making kinetic energy negative.

Hence, the particle turns back at x = ±2 m. — Turning points occur where kinetic energy is zero, so total energy equals potential energy. Calculating V(x) at x=2 m equals total energy, confirming turning points.

Question 5

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. 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?

Accepted Answer

(a) The heat energy required for burning comes at the expense of the rocket's kinetic energy. Friction converts some of the rocket's kinetic energy into heat.

(b) Although gravitational force is not always perpendicular to velocity, it is a conservative force. Over a complete orbit, the net work done by gravity is zero because the comet returns to its initial position and speed, so total mechanical energy remains constant.

(c) As the satellite loses energy due to atmospheric drag, it moves to a lower orbit where gravitational potential energy is less negative, and to conserve total energy, its kinetic energy (and thus speed) increases.

(d) The work done is greater when the man carries the mass because he applies an upward force over the distance. When pulling the rope, the force exerted by the man is horizontal and does no work against gravity on the hanging mass. — Friction converts kinetic energy to heat (a). Gravity is conservative, net work zero over orbit (b). Loss of potential energy converts to kinetic energy, increasing speed (c). Work done depends on force direction and displacement (d).

Question 6

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.

Accepted Answer

(a) decreases
When a conservative force does positive work, potential energy decreases.

(b) kinetic energy
Work done against friction reduces kinetic energy.

(c) external force
Rate of change of total momentum equals external force.

(d) total linear momentum and total energy
In inelastic collisions, total linear momentum and total energy are conserved, but kinetic energy is not. — Conservative force positive work reduces potential energy (a). Friction reduces kinetic energy (b). Momentum change due to external forces (c). Inelastic collision conserves momentum and energy, not kinetic energy (d).

Question 7

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.

Accepted Answer

(a) False. In elastic collisions, total momentum and total kinetic energy of the system are conserved, but not necessarily for each body individually.

(b) False. Total energy is conserved only if the system is closed and isolated; external forces can add or remove energy.

(c) False. Work done over a closed loop is zero only for conservative forces, not for all forces.

(d) True. In inelastic collisions, some kinetic energy is converted to other forms, so final kinetic energy is less. — Momentum and energy conservation apply to system totals, not individual bodies (a). Energy conservation requires isolated system (b). Work over closed loop zero only for conservative forces (c). Inelastic collisions dissipate kinetic energy (d).

Question 8

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) 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).

Accepted Answer

(a) Yes, total kinetic energy is conserved during the short time of elastic collision because no energy is lost to deformation or heat.

(b) Yes, total linear momentum is conserved during the collision because no external forces act on the system.

(c) For inelastic collisions, kinetic energy is not conserved (some energy is lost), but total linear momentum is conserved.

(d) If potential energy depends only on separation distance and no energy is lost, the collision is elastic. — Elastic collisions conserve kinetic energy and momentum (a,b). Inelastic collisions conserve momentum but not kinetic energy (c). Potential energy depending only on separation and no loss implies elastic collision (d).

Question 9

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)  $t^2$

Accepted Answer

Answer: (iii) t^{3/2}

Explanation:
Given constant acceleration a, initial velocity u=0.
Velocity v = at
Displacement s = (1/2) a t^2
Power P = dW/dt = F × v = m a × v = m a^2 t
Kinetic energy K = (1/2) m v^2 = (1/2) m a^2 t^2
Power is rate of change of kinetic energy:
P = dK/dt = m a^2 t
So power is proportional to t.

But the question asks power delivered to the body at time t.
Since velocity v = a t, power P = F × v = m a × a t = m a^2 t
Therefore, power is proportional to t.

Hence correct option is (ii) t.

Note: The question options and explanation in the book may have a typo; based on physics, power ∝ t. — Power = force × velocity = m a × a t = m a^2 t, so power ∝ t.

Question 10

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)  $t^2$

Accepted Answer

Answer: (iii) t^{3/2}

Explanation:
Power P is constant.
Power P = dW/dt = F × v
Work done W = change in kinetic energy = (1/2) m v^2
Since P is constant, P = d/dt (1/2 m v^2) = m v dv/dt
But acceleration a = dv/dt = F/m
So P = m v a
Since P is constant, v a = constant
But a = dv/dt = (dv/dt)
So v dv/dt = constant
Multiply both sides by dt:
v dv = constant × dt
Integrate:
(1/2) v^2 = constant × t + C
Assuming initial velocity zero, C=0
v^2 ∝ t
v ∝ t^{1/2}

Displacement s = ∫ v dt ∝ ∫ t^{1/2} dt ∝ t^{3/2}

Hence displacement is proportional to t^{3/2}. — Constant power implies v a = constant, leading to v ∝ t^{1/2} and displacement s ∝ t^{3/2}.

Question 11

11 A body constrained to move along the  $z$ -axis of a coordinate system is subject to a constant force  $\mathbf{F}$  given by

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

where  $\dot{\mathbf{i}},\dot{\mathbf{j}},\dot{\mathbf{k}}$  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  $4\mathrm{m}$  along the  $z$ -axis?

Accepted Answer

Given:
Force, $ \mathbf{F} = -\hat{i} + 2 \hat{j} + 3 \hat{k} $ N
Displacement along z-axis, $ \mathbf{d} = 4 \hat{k} $ m

Work done, $ W = \mathbf{F} \cdot \mathbf{d} = (-1)(0) + 2(0) + 3(4) = 12 $ J

Since displacement is only along z-axis, only z-component of force does work.

Answer: Work done = 12 J. — Work done = dot product of force and displacement vectors. Only z-components contribute as displacement is along z-axis.

Question 12

What is the scalar product (dot product) of two vectors $\mathbf{A}$ and $\mathbf{B}$ separated by an angle $	heta$?

Accepted Answer

AB \cos 	heta — The scalar product of two vectors $\mathbf{A}$ and $\mathbf{B}$ is defined as $\mathbf{A} \cdot \mathbf{B} = AB \cos 	heta$, where $	heta$ is the angle between them. This product is a scalar quantity.

Question 13

If $\mathbf{A} = 3\hat{\mathbf{i}} + 4\hat{\mathbf{j}} - 5\hat{\mathbf{k}}$ and $\mathbf{B} = 5\hat{\mathbf{i}} + 4\hat{\mathbf{j}} + 3\hat{\mathbf{k}}$, what is the value of $\mathbf{A} \cdot \mathbf{B}$?

Accepted Answer

16 — Using the formula $\mathbf{A} \cdot \mathbf{B} = A_x B_x + A_y B_y + A_z B_z$, we get: $3 	imes 5 + 4 	imes 4 + (-5) 	imes 3 = 15 + 16 - 15 = 16$.

Question 14

Find the angle $	heta$ between the vectors $\mathbf{F} = 3\hat{\mathbf{i}} + 4\hat{\mathbf{j}} - 5\hat{\mathbf{k}}$ and $\mathbf{d} = 5\hat{\mathbf{i}} + 4\hat{\mathbf{j}} + 3\hat{\mathbf{k}}$.

Accepted Answer

71.3 degrees — Given: $\mathbf{F} = (3,4,-5)$ units, $\mathbf{d} = (5,4,3)$ units

Find: Angle $	heta$ between $\mathbf{F}$ and $\mathbf{d}$

Formula: $\cos 	heta = \frac{\mathbf{F} \cdot \mathbf{d}}{|\mathbf{F}||\mathbf{d}|}$

Solution:
Step 1: Calculate dot product
$\mathbf{F} \cdot \mathbf{d} = 3	imes5 + 4	imes4 + (-5)	imes3 = 16$

Step 2: Calculate magnitudes
$|\mathbf{F}| = \sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 7.07$

$|\mathbf{d}| = \sqrt{5^2 + 4^2 + 3^2} = \sqrt{25 + 16 + 9} = \sqrt{50} = 7.07$

Step 3: Calculate $\cos 	heta$
$\cos 	heta = \frac{16}{7.07 	imes 7.07} = \frac{16}{50} = 0.32$

Step 4: Calculate $	heta$
$	heta = \cos^{-1}(0.32) = 71.3^{\circ}$

Answer: $71.3^{\circ}$

Question 15

State the work-energy theorem for a particle under a net force.

Accepted Answer

The work-energy theorem states that the change in kinetic energy of a particle is equal to the work done on it by the net force. For example, if a force moves a particle through a displacement, the work done by the force changes the particle's kinetic energy. — The theorem relates the net work done on a particle to its kinetic energy change, providing a scalar relation between force and motion.

WORK, ENERGY AND POWER

WORK, ENERGY AND POWER — Study Notes

5.1 Introduction

5.1.1 The Scalar Product

5.2 Notions of Work and Kinetic Energy: The Work-Energy Theorem

Practice Questions — WORK, ENERGY AND POWER

All 7 Chapters in Physics Part-I