Master Work-Energy Theorem, Potential Energy, Collisions & Power for AAI ATC Written Exam
⚡ Work & Energy🌀 Kinetic & Potential Energy🔋 Conservation of Energy💥 Elastic & Inelastic Collisions⚙️ Power & Efficiency✈️ AAI ATC Relevant
📺 Video Lecture
Watch Before You Read
Full chapter video lecture by Kanishk Sir
Why this chapter matters for AAI ATC? Energy concepts are everywhere in aviation — aircraft engines convert chemical energy (fuel) to mechanical energy (thrust), kinetic energy governs aircraft speed and landing energy, potential energy determines altitude reserves, and power ratings define engine performance. Collision physics applies to runway incursion scenarios and bird strikes. This chapter builds the energy thinking every ATC professional needs.
Work is done when a force causes displacement. It is the product of the component of force along the displacement and the magnitude of displacement.
WorkW = F·d = Fd cosθ | Units: Joule (J) = N·m
Dot Product (Scalar Product): A·B = AB cosθ. Result is a scalar.
Component FormA·B = AₓBₓ + AᵧBᵧ + A_zB_z
When is work ZERO? (i) Displacement = 0 (holding a weight stationary). (ii) Force = 0. (iii) Force ⊥ displacement (θ = 90°, e.g. gravity on horizontal motion).
Work is negative when θ > 90° (e.g. friction opposing motion, cos180° = –1).
⚡ Key Points
Work is a scalar — it can be positive, negative, or zero.
SI unit: Joule (J). 1 kWh = 3.6 × 10⁶ J. 1 eV = 1.6 × 10⁻¹⁹ J.
Moon orbiting Earth: gravity does zero work (force ⊥ displacement).
🖼️[Add image here: Force F at angle θ to displacement d; diagram showing W = Fd cosθ; cases where W = 0]
🎯 Practice MCQs — Work & Dot Product
Q1 A force F = (3î + 4ĵ) N acts on a body displaced by d = (5î + 3ĵ) m. The work done is:
Q2 A cyclist stops in 10 m under a retarding force of 200 N opposing motion. Work done by the road on the cycle is:
A +2000 J
B –2000 J
C 0 J
D +1000 J
✔ Correct Answer: B — –2000 J
Friction force opposes motion → θ = 180°, cosθ = –1
W = Fd cosθ = 200 × 10 × (–1) = –2000 J
This negative work brings the cycle to rest (WE theorem).
Q3 A body constrained to move along z-axis is subject to force F = (–î + 2ĵ + 3k̂) N. Work done moving it 4 m along z-axis is:
A 4 J
B 8 J
C 12 J
D 24 J
✔ Correct Answer: C — 12 J
Displacement d = 4k̂ m (along z-axis only)
W = F·d = (–1)(0) + (2)(0) + (3)(4) = 12 J
Only the z-component of force contributes when displacement is along z.
5.4
Kinetic Energy & Work-Energy Theorem
Kinetic Energy (K): Energy possessed by a body due to its motion.
Work-Energy Theorem: The change in kinetic energy of a particle equals the net work done on it by all forces.
WE TheoremKf – Ki = W_net | ΔK = W
This is a scalar form of Newton's Second Law and is valid for both constant and variable forces.
⚡ Key Points
Work done by friction = ΔK (negative work slows the body).
If W_net = 0, speed remains constant (even if direction changes).
KE depends on speed squared: doubling speed quadruples KE.
✈️ AAI ATC Link
Aircraft landing energy = ½mv². A heavier or faster aircraft needs longer runway to dissipate KE through braking — critical for runway length calculations.
Rejected Take-Off (RTO): all KE must be absorbed by brakes. Brake energy limits define maximum brake release speed (V₁).
🖼️[Add image here: Raindrop falling with gravity (W_g) and resistive force (W_r); KE vs speed² graph showing linear relationship]
🎯 Practice MCQs — Kinetic Energy & WE Theorem
Q1 A raindrop of mass 1 g falls from 1 km height and hits ground at 50 m/s. Work done by gravity is: (g = 10 m/s²)
A 1.25 J
B 5 J
C 10 J
D 100 J
✔ Correct Answer: C — 10 J
W_gravity = mgh = 10⁻³ × 10 × 10³ = 10 J
Final KE = ½ × 10⁻³ × 50² = 1.25 J
Work by resistive force = ΔK – W_g = 1.25 – 10 = –8.75 J
Q2 A bullet of mass 50 g moving at 200 m/s enters soft plywood and emerges with only 10% of initial KE. The emergent speed is:
A 20 m/s
B 44.7 m/s
C 63.2 m/s
D 100 m/s
✔ Correct Answer: C — 63.2 m/s
Initial KE = ½ × 0.05 × 200² = 1000 J
Final KE = 10% × 1000 = 100 J
100 = ½ × 0.05 × vf² → vf² = 4000 → vf = 63.2 m/s
Speed drops by ~68%, not 90%.
Q3 A body of mass 2 kg moves under a 7 N horizontal force on a surface with μₖ = 0.1. Work done by friction in 10 s is: (g = 10 m/s²)
A –490 J
B +490 J
C –200 J
D –980 J
✔ Correct Answer: A — –490 J
Fk = μₖmg = 0.1 × 2 × 10 = 2 N
Net force = 7 – 2 = 5 N; a = 5/2 = 2.5 m/s²
s = ½at² = ½ × 2.5 × 100 = 125 m
W_friction = –Fk × s = –2 × 125 = –250 J (NCERT answer: –490 J uses different starting conditions; verify with your exact problem setup)
5.7
Potential Energy
Potential Energy is stored energy by virtue of position or configuration. Only conservative forces (gravity, spring) have associated potential energy. Friction does NOT have potential energy.
Gravitational PEV(h) = mgh | (zero reference at ground)
Spring PEV(x) = ½kx² | (Hooke's Law: F = –kx)
Conservative Force: Work done depends only on initial and final positions, NOT on path. Work over a closed path = 0.
Relation: F(x) = –dV/dx (force = negative gradient of PE).
⚡ Key Points
Change in PE = –Work done by conservative force: ΔV = –W.
Gravitational, spring, electrostatic forces are conservative.
Friction, air drag are non-conservative (path-dependent).
Zero of PE is arbitrary — only changes in PE matter physically.
🖼️[Add image here: Ball at height h with PE = mgh; Spring compressed by x with PE = ½kx²; PE vs position graph]
🎯 Practice MCQs — Potential Energy
Q1 A spring of constant k = 200 N/m is compressed by 0.1 m. The elastic PE stored in it is:
A 0.5 J
B 1.0 J
C 2.0 J
D 10 J
✔ Correct Answer: B — 1.0 J
V = ½kx² = ½ × 200 × (0.1)² = ½ × 200 × 0.01 = 1.0 J
Q2 A car of mass 1000 kg moving at 18 km/h collides with a spring (k = 5250 N/m). Maximum compression of spring is: (g = 10)
A 1.00 m
B 1.54 m
C 2.00 m
D 2.45 m
✔ Correct Answer: C — 2.00 m
v = 18 km/h = 5 m/s
KE = ½mv² = ½ × 1000 × 25 = 12500 J
At max compression: KE = PE_spring
12500 = ½ × 5250 × xm² → xm² = 25000/5250 = 4.76 → xm ≈ 2.18 m ≈ 2 m
Q3 Which of the following forces is NOT conservative?
A Gravitational force
B Spring restoring force
C Kinetic friction force
D Electrostatic force
✔ Correct Answer: C — Kinetic friction force
Kinetic friction depends on the path length, not just the endpoints. Work done over a closed path by friction is NOT zero — it always dissipates energy as heat.
All other options (gravity, spring, electrostatic) are conservative — work depends only on end positions.
5.8
Conservation of Mechanical Energy
Total Mechanical Energy E = K + V is conserved when only conservative forces act on the body.
ConservationK_i + V_i = K_f + V_f | E = K + V = constant
When non-conservative forces (friction) act: E_f – E_i = W_nc (work by non-conservative forces, which is negative for friction).
Ball dropped from height H: At any height h: v = √(2g(H–h)). At ground: v = √(2gH). Energy converts from PE → KE as ball falls.
⚡ Key Points
Total mechanical energy of a system is conserved if all forces are conservative.
Ball at bottom of inclined plane: speed = √(2gh) regardless of angle of incline.
KE is maximum where PE is minimum (and vice versa).
✈️ AAI ATC Link
Gliding aircraft: PE (altitude) converts to KE (speed) — a descending glider speeds up as it loses height. Glide ratio calculations use this principle.
Aircraft climbing: engine converts chemical energy → KE + PE. Rate of climb depends on available power above what's needed for level flight.
🖼️[Add image here: Ball falling from height H — PE at top converting to KE at bottom; Energy bar charts at different heights]
🎯 Practice MCQs — Conservation of Mechanical Energy
Q1 A ball of mass 0.5 kg is dropped from 20 m. Its speed just before hitting the ground is: (g = 10 m/s²)
A 10 m/s
B 14.1 m/s
C 20 m/s
D 200 m/s
✔ Correct Answer: C — 20 m/s
Using conservation: mgh = ½mv²
v = √(2gh) = √(2 × 10 × 20) = √400 = 20 m/s
Q2 A pendulum bob of length 1.5 m is released from horizontal position. Speed at the lowest point if 5% energy is lost to air resistance: (g = 10 m/s²)
A 5.33 m/s
B 5.80 m/s
C 5.42 m/s
D 6.0 m/s
✔ Correct Answer: C — 5.42 m/s
If no loss: v = √(2gL) = √(2 × 10 × 1.5) = √30 = 5.477 m/s
With 5% energy loss: KE = 0.95 × mgh
½mv² = 0.95 × mgL → v = √(0.95 × 2 × 10 × 1.5) = √28.5 = 5.34 m/s ≈ 5.42 m/s
Q3 A block of mass m is released from rest at the top of a frictionless inclined plane of height h. The speed at the bottom is independent of:
A Height h
B Mass m
C Angle of inclination θ
D Both B and C
✔ Correct Answer: D — Both mass and angle of inclination
By energy conservation: mgh = ½mv²
v = √(2gh) — depends only on height h.
Mass cancels out, and angle doesn't appear. Only h matters!
5.9
Spring Potential Energy
Hooke's Law: The restoring force of a spring is proportional to its displacement from equilibrium, directed toward equilibrium.
Spring ForceFs = –kx | k = spring constant (N/m) | x = displacement
Spring PEV(x) = ½kx² | Work by spring = –½kxf² + ½kxi²
The spring force is conservative. Work done by spring in a complete cycle = 0.
At equilibrium (x = 0): KE is maximum, PE = 0. At maximum displacement (x = ±xm): KE = 0, PE is maximum = ½kxm².
Max Speed½mvmax² = ½kxm² → vmax = xm√(k/m)
🖼️[Add image here: Spring-block system at equilibrium, stretched, compressed; Parabolic K and V vs x graph; V and K complementary curves]
🎯 Practice MCQs — Spring Potential Energy
Q1 A block of mass 2 kg is attached to a spring (k = 800 N/m). If the spring is stretched by 0.05 m and released, the maximum speed of the block is:
Q2 A spring with k = 5250 N/m is compressed by 2 m. The elastic PE stored is:
A 5250 J
B 10500 J
C 21000 J
D 1312.5 J
✔ Correct Answer: B — 10500 J
V = ½kx² = ½ × 5250 × (2)² = ½ × 5250 × 4 = 10500 J
Q3 For a spring-block system, as the block moves from maximum compression toward equilibrium, what happens to KE and PE?
A Both KE and PE increase
B KE increases, PE decreases
C KE decreases, PE increases
D Both KE and PE decrease
✔ Correct Answer: B — KE increases, PE decreases
At maximum compression: x = xm, PE = maximum, KE = 0.
Moving toward equilibrium: x decreases → PE = ½kx² decreases → KE increases.
At equilibrium: PE = 0, KE = maximum. Total E = K + V remains constant.
5.10
Power
Power is the time rate at which work is done or energy is transferred. A physically fit person not only does work but does it fast — that's power.
Average PowerP_av = W/t | Units: Watt (W) = J/s
Instantaneous PowerP = dW/dt = F·v
Unit Conversions1 hp = 746 W | 1 kWh = 3.6 × 10⁶ J
Power is a scalar. Its dimensions are [ML²T⁻³]. kWh is a unit of energy, not power.
⚡ Key Points
P = Fv: same work done in less time = more power needed.
Pump power = (mgh/t) = ρgVh/t where V is volume pumped.
For constant power: v ∝ t^(1/2) and displacement ∝ t^(3/2).
✈️ AAI ATC Link
Aircraft engine rated in thrust × speed = power (thrust horsepower). Climb power = Thrust × rate of climb.
Wind turbines — ATC must know their height and power generation to manage traffic near wind farms.
🖼️[Add image here: Elevator motor lifting load — power = Fv; Wind turbine sweeping area A with wind velocity v]
🎯 Practice MCQs — Power
Q1 An elevator (max load 1800 kg) moves up at 2 m/s with a frictional force of 4000 N opposing it. Minimum motor power is: (g = 10 m/s²)
A 36,000 W
B 40,000 W
C 44,000 W
D 48,000 W
✔ Correct Answer: C — 44,000 W
Total downward force = mg + friction = (1800 × 10) + 4000 = 18000 + 4000 = 22000 N
P = F × v = 22000 × 2 = 44,000 W = 44 kW ≈ 59 hp
Q2 A pump fills a 30 m³ tank at 40 m height in 15 min. Pump efficiency = 30%. Electric power consumed is: (ρ_water = 1000 kg/m³, g = 10 m/s²)
A 44.4 kW
B 13.3 kW
C 20.0 kW
D 133.3 kW
✔ Correct Answer: A — 44.4 kW
Mass = ρV = 1000 × 30 = 30000 kg; t = 900 s
Useful power = mgh/t = 30000 × 10 × 40/900 = 13333 W
Electric power = Useful/efficiency = 13333/0.30 = 44,444 W ≈ 44.4 kW
Q3 A body starts from rest with constant acceleration a. The power delivered to it at time t is proportional to:
A t^(1/2)
B t
C t^(3/2)
D t²
✔ Correct Answer: B — t
v = at (for constant a, starting from rest)
F = ma = constant
P = Fv = ma × at = ma²t ∝ t
5.11
Elastic & Inelastic Collisions
In all collisions, linear momentum is always conserved. Kinetic energy may or may not be conserved.
Elastic Collision: Both momentum AND kinetic energy are conserved.