Olive Defence

Physics · CDS

PC03 — Work, Energy & Power

📖 PC03 · CDS General Science — Physics 🎯 CDS Level : High Priority

Energy is the capacity to do work. This chapter connects mechanics to the real world — it explains why a moving car is hard to stop (kinetic energy), why things fall (potential energy), and why machines are never 100% efficient (work-energy theorem). CDS consistently tests the relationship between KE and momentum, and the concept of zero work.

📌 CDS regularly tests: Zero work conditions (F ⊥ displacement); work done on incline; KE-momentum relation (if KE doubles, momentum multiplies by √2); elastic vs inelastic collision distinction; power = Fv; practical applications (who does more work between two people lifting same load in different times).

1. Work

1.1

Definition, Types & Zero-Work Conditions

⚡ Work Formula

W = F · s · cosθ F = applied force (N); s = displacement (m); θ = angle between F and s W is POSITIVE when F and s are in same direction (0° < θ < 90°) W is NEGATIVE when F and s are in opposite directions (90° < θ ≤ 180°) W is ZERO when F ⊥ s, i.e., θ = 90° → cosθ = 0 Examples of ZERO work: • Carrying a load horizontally (gravity is vertical; displacement is horizontal) • Uniform circular motion (centripetal force ⊥ velocity at all times) • A satellite in circular orbit (gravity ⊥ velocity) • Magnetic force on a moving charge (always ⊥ velocity)

1 Joule = 1 Newton × 1 metre. Work is a scalar — it can be positive, negative, or zero. Negative work means the force is opposing the motion (e.g. friction does negative work on a sliding object).

2. Kinetic Energy, Potential Energy & Conservation

2.1

Forms of Energy & the Work-Energy Theorem

⚡ Energy Formulae & Key Relations

Kinetic energy: KE = ½mv² [Joule] Potential energy: PE = mgh [Joule] (gravitational) Spring PE: PE = ½kx² (k = spring constant; x = extension) Work-Energy Theorem: Net work done = Change in KE = ½mv² − ½mu² Conservation of Mechanical Energy (no friction): KE + PE = constant At top of swing: max PE, min KE (KE=0 at highest point) At bottom: max KE, min PE KE — Momentum relationship: p = mv → KE = p²/(2m) If KE doubles → momentum increases by √2 (≈ 1.414) If KE × 4 → momentum doubles If momentum doubles → KE increases by 4

The KE-momentum relation (KE = p²/2m) is consistently tested in CDS. Learn the table: KE × 4 → p × 2; KE × 2 → p × √2; p × 2 → KE × 4.

Fig. 1 — Energy conservation on a track. The total mechanical energy (KE + PE) remains constant at every point when there is no friction. Speed is maximum at the lowest point and minimum (zero) at the highest point.

📍 Types of Collision

Elastic: both momentum AND kinetic energy conserved. Balls bounce back (billiard balls, atomic collisions). Coefficient of restitution e = 1
Inelastic: momentum conserved; KE partially lost to heat/sound. e < 1
Perfectly inelastic: bodies stick together; maximum KE lost; e = 0
Momentum is conserved in ALL collisions (Newton's 3rd law)

📍 KE-Momentum Relation Quick Table

Same p, different m → lighter body has more KE
Same KE, different m → heavier body has more momentum
KE doubles → p multiplies by √2
KE × 4 → p × 2
p doubles → KE × 4

3. Power

3.1

Rate of Doing Work

⚡ Power Formulae

Power: P = W/t = F·v [unit: Watt (W) = J/s] F = force; v = velocity of the body 1 horsepower (HP) = 746 W ≈ 750 W (British unit of power) 1 kilowatt (kW) = 1000 W If two people lift the same load to the same height: Work done is equal (same W = mgh) The one who takes less time has MORE power (P = W/t)

Power = Fv is the most useful form when force and velocity are given. A car engine provides more power at high speeds not because force increases, but because P = Fv — if F stays the same, P increases with v.

📝 CDS PYQ

Work, Energy & Power

Q1. A person carries a 20 kg suitcase horizontally. The work done against gravity is:

(a) 200 J
(b) Zero
(c) 20 J
(d) Depends on distance

Answer: (b) Zero
Work done against gravity = mgh. Since the person moves horizontally, there is no change in height (h = 0). Also, from W = Fs cosθ: gravity acts downward, displacement is horizontal → θ = 90° → cos90° = 0 → W = 0. This is one of the most repeated zero-work examples in CDS.

Q2. If the kinetic energy of a body is increased by 44%, the percentage increase in momentum is:

(a) 44%
(b) 22%
(c) 20%
(d) 72%

Answer: (c) 20%
p = √(2m·KE). New KE = 1.44 × old KE. New p = √(2m × 1.44·KE) = 1.2 × old p. Increase = 20%. √1.44 = 1.2 — remember this. This precise problem appears in CDS with slight variations — the answer is always √(KE_ratio) − 1 expressed as a percentage.

Q3. A body is moving in a circle at constant speed. The work done by the centripetal force is:

(a) Positive
(b) Negative
(c) Zero
(d) Maximum at the top

Answer: (c) Zero
In uniform circular motion, the centripetal force always acts perpendicular (90°) to the velocity (direction of motion). Since W = Fs cosθ and θ = 90°, cos90° = 0, work done = 0. The centripetal force changes the direction of motion but not the speed, so it does no work.

📚 Formula Sheet — PC03

⚡ Work

W = Fs cosθ
W = 0 when θ = 90°
W = 0: carrying load horizontally, UCM
Negative work: friction, opposing force
1 J = 1 N·m

⚡ Energy

KE = ½mv²; PE = mgh
W-E theorem: W_net = ΔKE
KE = p²/2m
KE×4 → p×2; KE×2 → p×√2
Elastic: KE + p conserved

⚡ Power

P = W/t = Fv
Unit: Watt (W) = J/s
1 HP = 746 W ≈ 750 W
Same work in less time → more power
Efficiency = useful output / input × 100%

📍 Collisions

All collisions: momentum conserved
Elastic: KE also conserved; e=1
Inelastic: KE partially lost; e<1
Perfectly inelastic: stick together; e=0
Max KE loss: perfectly inelastic

⚡ Quick Revision — PC03

🚨 Zero Work Conditions

Carrying load horizontally (gravity ⊥ motion)
UCM — centripetal force ⊥ velocity
Satellite in circular orbit
Magnetic force on charge
Normal force on a horizontal surface

⚡ KE-Momentum Table

KE × 2 → p × √2 (≈1.41)
KE × 4 → p × 2
p × 2 → KE × 4
p × √2 → KE × 2
KE = p²/2m (fundamental link)

📍 Power Facts

P = W/t = Fv
Same work, less time → more power
1 HP = 746 W
Power is rate of energy use
kW × hours = kWh (electricity bill)

📝 Practice Exercise

E-01

A spring is compressed by 0.1 m with spring constant 200 N/m. The PE stored is:

(a) 20 J
(b) 10 J
(c) 1 J
(d) 2 J

E-02

Two bodies have momenta in ratio 1:2. Their kinetic energies are in ratio:

(a) 1:2
(b) 1:4
(c) 2:1
(d) 4:1

E-03

A man lifts a 10 kg mass to 2 m height in 4 seconds. Power exerted (g=10 m/s²):

(a) 200 W
(b) 50 W
(c) 80 W
(d) 400 W

Answers: E-01: (c) 1 J [PE = ½kx² = ½×200×0.01 = 1 J] | E-02: (b) 1:4 [KE = p²/2m; same mass: KE ∝ p²; ratio 1²:2² = 1:4] | E-03: (b) 50 W [W = mgh = 10×10×2 = 200 J; P = W/t = 200/4 = 50 W]

📝 Mock Tests 🎯 Subject Quizzes ✈️ Telegram