Unit 5: Work, Energy and Power

Introduction

In physics, work, energy, and power are interrelated quantities that describe how forces cause motion and how systems exchange energy. This chapter builds a rigorous foundation for these ideas, presenting definitions, key theorems, and illustrative examples suitable for Class 11 students.

1. Work Done

Constant Force

When a constant force F acts on a body that undergoes a displacement s at an angle θ between the force and displacement vectors, the work done is given by the dot product:

W = F · s = F s cosθ

Where:

F = magnitude of the force (newtons, N)
s = magnitude of the displacement (metres, m)
θ = angle between F and s

If the force is parallel to the displacement (θ = 0°), W = Fs; if perpendicular (θ = 90°), no work is done.

Variable Force

When the force varies with position, the infinitesimal work done over a small displacement ds is dW = F · ds. The total work is obtained by integrating along the path:

W = ∫ F · ds

This integral is evaluated over the initial and final positions of the object.

Units

The SI unit of work is the joule (J), defined as one newton‑metre:

1 J = 1 N·m

Larger units include the kilojoule:

1 kJ = 1000 J

2. Power

Definition

Power is the rate at which work is done or energy is transferred:

P = W / t

Using the definition of work for a constant force, power can also be expressed as:

P = F · v

where v is the instantaneous velocity of the point of application of the force.

Units

The SI unit of power is the watt (W):

1 W = 1 J/s

Another common unit is the horsepower (hp):

1 hp = 746 W ≈ 0.746 kW

Average and Instantaneous Power

Average power over a time interval Δt is:

P_avg = ΔW / Δt

Instantaneous power is the limit as Δt → 0:

P = dW/dt = F · v

3. Work‑Energy Theorem

Statement

The net work done by all forces acting on a particle equals the change in its kinetic energy:

W_net = ΔK = K_f – K_i

Derivation from Newton’s Second Law

Starting from F = ma and using the kinematic relation v^2 = u^2 + 2as (where u and v are initial and final speeds, a is constant acceleration, and s is displacement):

Multiply both sides of F = ma by ds: F·ds = m a·ds
Recognise that a·ds = v dv (since a = dv/dt and ds = v dt): F·ds = m v dv
Integrate from initial state (v = u) to final state (v = v): ∫ F·ds = ∫_u^v m v dv
Left side is the net work W_net; right side evaluates to ½ m v^2 – ½ m u^2.
Thus W_net = ½ m v^2 – ½ m u^2 = ΔK.

Example

A 2 kg block is pulled across a horizontal surface by a constant force of 10 N acting at 30° to the horizontal. The block moves 5 m. Find the work done and the final speed if it starts from rest.

Work: W = F s cosθ = 10 × 5 × cos30° = 50 × (√3/2) ≈ 43.3 J
Using work‑energy theorem: W = ΔK = ½ m v^2 – 0 → v = √(2W/m) = √(2×43.3/2) ≈ √43.3 ≈ 6.58 m/s

4. Kinetic and Potential Energy

Kinetic Energy (KE)

The energy possessed by a body due to its motion:

KE = ½ m v^2

Where m is mass (kg) and v is speed (m/s).

Potential Energy (PE)

Potential energy arises from the position or configuration of a system in a conservative force field.

Gravitational PE (near Earth’s surface): PE_g = m g h

Elastic PE (spring): PE_s = ½ k x^2

Definitions:

g = acceleration due to gravity (≈9.8 m/s²)

h = vertical height above a reference level (m)

k = spring constant (N/m)

x = displacement from the spring’s natural length (m)

Total Mechanical Energy

The sum of kinetic and potential energies:

E = KE + PE

In the absence of non‑conservative forces, E remains constant.

5. Conservation of Energy

Principle

When only conservative forces act, the total mechanical energy of an isolated system is conserved:

KE_i + PE_i = KE_f + PE_f

Energy may change form (e.g., kinetic → potential) but the total remains unchanged.

Applications

Simple Pendulum: At the highest point, energy is all potential (PE = mgh); at the lowest point, it is all kinetic (KE = ½ mv^2).

Roller Coaster: Energy transforms between gravitational potential and kinetic as the coaster climbs and descends hills, assuming negligible friction.

Falling Body: A body dropped from height h converts potential energy mgh into kinetic energy ½ mv^2 just before impact (ignoring air resistance).

6. Conservative and Non‑Conservative Forces

Conservative Forces

A force is conservative if the work done by it on a particle moving between two points is independent of the path taken. Equivalently, the work done over any closed loop is zero.

Examples: gravitational force, electrostatic force, spring force.

Key property: a scalar potential energy function U(r) can be defined such that F = –∇U.

Non‑Conservative Forces

For these forces, work depends on the trajectory; energy is dissipated as heat, sound, or other internal forms.

Examples: kinetic friction, air resistance, viscous drag.

No potential energy can be assigned to non‑conservative forces.

Implications for Energy Conservation

When non‑conservative forces are present:

ΔKE + ΔPE = W_nc

where W_nc is the work done by non‑conservative forces (usually negative, representing energy loss).

7. Elastic and Inelastic Collisions

Elastic Collisions

Both momentum and kinetic energy are conserved.

For two bodies of masses m1 and m2 with initial velocities u1, u2 and final velocities v1, v2:

Momentum: m1 u1 + m2 u2 = m1 v1 + m2 v2

Kinetic Energy: ½ m1 u1^2 + ½ m2 u2^2 = ½ m1 v1^2 + ½ m2 v2^2

Inelastic Collisions

Momentum is conserved, but kinetic energy is not; some kinetic energy is transformed into internal energy (heat, sound, deformation).

Perfectly Inelastic Collision

The colliding objects stick together after impact, moving with a common velocity V.

Momentum conservation gives:

(m1 + m2) V = m1 u1 + m2 u2 → V = (m1 u1 + m2 u2)/(m1 + m2)

Coefficient of Restitution (e)

Defined as the ratio of relative speed of separation to relative speed of approach along the line of impact:

e = (v2 – v1)/(u1 – u2)

Values:

e = 1 → perfectly elastic

0 < e < 1 → partially elastic (inelastic)

e = 0 → perfectly inelastic

Example: Elastic Collision

A 0.5 kg ball moving at 4 m/s strikes a stationary 0.3 kg ball elastically. Find the final velocities.

Using conservation of momentum and kinetic energy (or the derived formulas for elastic collisions):

v1 = (m1 – m2)/(m1 + m2) u1 + (2 m2)/(m1 + m2) u2

v2 = (2 m1)/(m1 + m2) u1 + (m2 – m1)/(m1 + m2) u2

Substituting u2 = 0:

v1 = (0.5 – 0.3)/(0.5 + 0.3) × 4 = (0.2/0.8)×4 = 1 m/s

v2 = (2×0.5)/(0.5+0.3) × 4 = (1.0/0.8)×4 = 5 m/s

Thus after collision, the 0.5 kg ball continues forward at 1 m/s, while the 0.3 kg ball moves ahead at 5 m/s.

Example: Perfectly Inelastic Collision

The same two balls now stick together after impact. Find their common velocity.

V = (m1 u1 + m2 u2)/(m1 + m2) = (0.5×4 + 0.3×0)/(0.5+0.3) = 2.0/0.8 = 2.5 m/s

Kinetic energy before: ½×0.5×4^2 = 4 J. After: ½×(0.5+0.3)×2.5^2 = 0.4×6.25 = 2.5 J. The loss of 1.5 J appears as heat and deformation.

Summary

This chapter has linked the concepts of work, energy, and power through precise definitions, mathematical derivations, and practical examples. Mastery of these principles enables the analysis of a wide range of mechanical systems, from simple particles to complex machines.