Work. Power

General physics

🎯 Learning Objectives

By the end of this lesson, you will be able to:
• Understand the concept of work and apply the equation W = F × s cos θ
• Calculate work done when force and displacement are in different directions
• Define power as the rate of doing work using P = W/t
• Derive the relationship P = Fv from fundamental definitions
• Solve complex problems involving work, power, and energy in real-world scenarios
• Apply work and power concepts to analyze mechanical systems and efficiency

🗣️ Language Objectives

Students will develop their physics communication skills by:
• Using precise scientific terminology when describing work and power concepts
• Explaining the difference between everyday and scientific meanings of «work»
• Reading and interpreting work-energy problems written in English with confidence
• Communicating mathematical derivations and solutions clearly in written English
• Understanding and using power units (watts, horsepower) accurately in engineering contexts

📚 Key Terms and Translations

English Term	Russian Translation	Kazakh Translation
Work	Работа	Жұмыс
Power	Мощность	Қуат
Force	Сила	Күш
Displacement	Перемещение	Орын ауыстыру
Energy	Энергия	Энергия
Efficiency	Эффективность	Тиімділік
Joule	Джоуль	Джоуль
Watt	Ватт	Ватт

🃏 Vocabulary Study Cards

Work

Definition: Energy transferred when a force moves through a distance

Formula: W = F × s × cos θ

Units: Joule (J) = N⋅m

Key Point: Only force component in direction of motion does work

Power

Definition: Rate of doing work or transferring energy

Formula: P = W/t = Fv

Units: Watt (W) = J/s

Example: 100W light bulb uses 100J of energy per second

Positive vs Negative Work

Positive work: Force and displacement in same direction (θ < 90°)

Negative work: Force opposes motion (θ > 90°)

Zero work: Force perpendicular to motion (θ = 90°)

Example: Friction always does negative work

Efficiency

Definition: Ratio of useful output power to input power

Formula: η = P_out/P_in × 100%

Range: 0% to 100% (perfect efficiency impossible)

Application: Motors, engines, machines

📖 Glossary of Terms

Work

The energy transferred to or from an object when a force acts on it over a distance. Work is only done when there is movement in the direction of the applied force. The scientific definition differs from everyday usage — holding a heavy object stationary does no work in physics terms.

Translation

Russian: Работа — это энергия, передаваемая объекту или от объекта, когда на него действует сила на определенном расстоянии. Работа совершается только при наличии движения в направлении приложенной силы. Научное определение отличается от повседневного использования — удержание тяжелого объекта неподвижно не совершает работу в физическом смысле.

Kazakh: Жұмыс — бұл затқа күш әсер еткенде белгілі бір қашықтықта затқа немесе заттан берілетін энергия. Жұмыс тек қолданылған күш бағытында қозғалыс болған кезде ғана жасалады. Ғылыми анықтама күнделікті қолданыстан ерекшеленеді — ауыр затты қозғалыссыз ұстау физикалық мағынада жұмыс жасамайды.

Power

The rate at which work is done or energy is transferred. Power indicates how quickly energy conversion occurs, not how much total energy is involved. A high-power device can do the same amount of work as a low-power device, but in less time.

Translation

Russian: Мощность — это скорость совершения работы или передачи энергии. Мощность показывает, как быстро происходит преобразование энергии, а не сколько общей энергии задействовано. Устройство высокой мощности может выполнить ту же работу, что и устройство низкой мощности, но за меньшее время.

Kazakh: Қуат — бұл жұмыс жасау немесе энергия беру жылдамдығы. Қуат энергия түрлену жылдамдығын көрсетеді, жалпы энергияның қанша болатынын емес. Жоғары қуатты құрылғы төмен қуатты құрылғымен бірдей жұмысты істей алады, бірақ аз уақытта.

Mechanical Efficiency

The ratio of useful mechanical work output to the total energy input, expressed as a percentage. Efficiency is always less than 100% due to energy losses through friction, heat, sound, and other non-useful forms. It measures how well a machine converts input energy into desired output.

Translation

Russian: Механическая эффективность — это отношение полезной механической работы на выходе к общей энергии на входе, выраженное в процентах. Эффективность всегда меньше 100% из-за потерь энергии через трение, тепло, звук и другие неполезные формы. Она измеряет, насколько хорошо машина преобразует входную энергию в желаемый выход.

Kazakh: Механикалық тиімділік — бұл пайдалы механикалық жұмыстың шығысының жалпы кіріс энергиясына қатынасы, пайызбен көрсетілген. Тиімділік үйкеліс, жылу, дыбыс және басқа пайдасыз формалар арқылы энергия жоғалуына байланысты әрқашан 100%-дан аз болады. Ол машинаның кіріс энергиясын қажетті шығысқа қаншалықты жақсы түрлендіретінін өлшейді.

Joule

The SI unit of work and energy, defined as the work done when a force of one newton acts through a distance of one meter. Named after James Prescott Joule, it represents a relatively small amount of energy — roughly the energy needed to lift an apple one meter high.

Translation

Russian: Джоуль — единица измерения работы и энергии в СИ, определяемая как работа, совершаемая при действии силы в один ньютон на расстоянии одного метра. Названная в честь Джеймса Прескотта Джоуля, она представляет относительно небольшое количество энергии — примерно энергию, необходимую для подъема яблока на один метр в высоту.

Kazakh: Джоуль — SI жүйесіндегі жұмыс пен энергияның өлшем бірлігі, бір ньютон күшінің бір метр қашықтықта әрекет етуі кезінде жасалатын жұмыс ретінде анықталады. Джеймс Прескотт Джоульдің атымен аталған, ол салыстырмалы түрде аз энергия мөлшерін білдіреді — алманы бір метр биіктікке көтеру үшін қажетті энергияға жуық.

🔬 Theory: Understanding Work and Power

The Concept of Work

In physics, work has a very specific meaning that differs from everyday usage. Work is only done when a force causes displacement in the direction of that force.

Key points about work:

If there is no movement, no work is done
Force must have a component in the direction of motion
Work can be positive or negative

Work = Force × Displacement × cos θ
W = F × s × cos θ
where θ is the angle between force and displacement
Units: Joule (J) = N⋅m

When force and displacement are in the same direction (θ = 0°): cos 0° = 1, so W = F × s

When force is perpendicular to displacement (θ = 90°): cos 90° = 0, so W = 0

Translation

Russian: В физике работа имеет очень специфическое значение, которое отличается от повседневного использования. Работа совершается только тогда, когда сила вызывает перемещение в направлении этой силы.

Kazakh: Физикада жұмыстың күнделікті қолданыстан ерекшеленетін өте нақты мағынасы бар. Жұмыс тек күш сол күш бағытында орын ауыстыруға себеп болған кезде ғана жасалады.

Defining Power

measures how quickly is or work is done. It is the of energy conversion.

Power = Work done / Time taken
P = W / t
Units: Watt (W) = J/s

A watt is quite a small unit. Common multiples include:

Kilowatt (kW) = 1000 W
Megawatt (MW) = 1,000,000 W
Horsepower (hp) ≈ 746 W

Translation

Russian: Мощность измеряет, как быстро передается энергия или совершается работа. Это скорость преобразования энергии.

Kazakh: Қуат энергияның қаншалықты тез берілетінін немесе жұмыстың қаншалықты тез жасалатынын өлшейді. Бұл энергия түрлену жылдамдығы.

Deriving P = Fv

We can derive an alternative expression for power when an object moves at constant velocity:

Starting with: P = W/t

Since W = F × s (when force and displacement are parallel):

P = (F × s)/t

Since velocity v = s/t:

P = Fv
Power = Force × Velocity

This relationship is particularly useful for analyzing vehicles, engines, and other mechanical systems.

Translation

Russian: Мы можем вывести альтернативное выражение для мощности, когда объект движется с постоянной скоростью. Это соотношение особенно полезно для анализа транспортных средств, двигателей и других механических систем.

Kazakh: Зат тұрақты жылдамдықпен қозғалған кезде қуат үшін балама өрнекті шығаруға болады. Бұл қатынас көлік құралдарын, қозғалтқыштарды және басқа механикалық жүйелерді талдау үшін өте пайдалы.

Theory Questions

Easy Question: A person pushes a box with a force of 50N for a distance of 10m. Calculate the work done if the force is applied horizontally.

Answer

Since the force is applied horizontally and the box moves horizontally, θ = 0°
W = F × s × cos θ = 50N × 10m × cos(0°) = 50N × 10m × 1 = 500 J
The work done is 500 joules.

Medium Question: A 2000W motor lifts a 500kg mass vertically at constant speed. Calculate: (a) the lifting force required, (b) the speed at which the mass is lifted.

Answer

(a) For constant speed (equilibrium): Lifting force = Weight
F = mg = 500kg × 9.8 m/s² = 4900N

(b) Using P = Fv:
v = P/F = 2000W / 4900N = 0.408 m/s
The mass is lifted at 0.408 m/s (about 41 cm/s).

Medium Question: A car engine produces 150 kW of power. If the car travels at 30 m/s against a total resistive force, calculate the magnitude of this resistive force.

Answer

At constant speed, driving force equals resistive force.
Using P = Fv:
F = P/v = 150,000W / 30 m/s = 5000N
The resistive force is 5000N (5 kN).

Hard Question (Critical Thinking): A cyclist pedals up a 10° incline at constant speed. The cyclist and bike have a combined mass of 80kg, and the cyclist maintains a power output of 300W against gravity and air resistance. If air resistance provides 50N of opposing force, calculate: (a) the speed of the cyclist, (b) the coefficient of rolling resistance between the tires and road, (c) analyze how the power requirement would change if the cyclist doubled their speed on the same incline.

Answer

Part (a): Calculate cyclist speed
Forces opposing motion:
— Component of weight down the slope: mg sin(10°) = 80 × 9.8 × sin(10°) = 136.1N
— Air resistance: 50N
— Rolling resistance: μmg cos(10°) = μ × 80 × 9.8 × cos(10°) = 772.5μ N

Total resistive force: F_total = 136.1 + 50 + 772.5μ

Using P = Fv: v = P/F_total = 300/(186.1 + 772.5μ)

We need another equation. Assuming typical rolling resistance μ ≈ 0.01:
F_total = 136.1 + 50 + 7.725 = 193.8N
v = 300/193.8 = 1.55 m/s

Part (b): Coefficient of rolling resistance
If we assume the calculated speed is correct:
193.8 = 186.1 + 772.5μ
772.5μ = 7.7
μ = 0.01 (confirms our assumption)

Part (c): Analysis of doubled speed
At doubled speed (3.1 m/s):
— Gravitational component unchanged: 136.1N
— Air resistance increases quadratically: F_air = k × v²
If F_air = 50N at 1.55 m/s, then k = 50/(1.55)² = 20.8
At 3.1 m/s: F_air = 20.8 × (3.1)² = 200N
— Rolling resistance unchanged: 7.7N

New total force = 136.1 + 200 + 7.7 = 343.8N
Required power = Fv = 343.8 × 3.1 = 1066W

Critical Analysis:
Doubling speed increases power requirement by factor of 1066/300 = 3.55, not just 2, because:
1. Air resistance increases as v², requiring quadratic power increase
2. Gravity and rolling resistance components increase linearly with speed
3. At higher speeds, air resistance dominates, making further speed increases very expensive energetically

This explains why cycling efficiency decreases dramatically at high speeds and why competitive cyclists focus on aerodynamic positioning.

💪 Memorization Exercises for Key Terms

Complete the Definitions

1. Work is done when a _______ causes _______ in the direction of the force.

Answer

force causes displacement

2. The formula for work when force and displacement are at angle θ is: W = _______

Answer

W = F × s × cos θ

3. Power is defined as _______ done per unit _______

Answer

work done per unit time

4. The relationship between power, force, and velocity is: P = _______

Answer

P = Fv

5. The SI unit of work is the _______, which equals one _______ meter.

Answer

joule, newton meter (N⋅m)

🎥 Educational Video Resource

Additional Video Resources:

• Khan Academy: Work and Energy

• Crash Course Physics: Work and Power

• Physics Girl: What is Work?

📐 Worked Problem Examples

Example 1: Work at an Angle

Problem: A person pulls a suitcase with a force of 40N at an angle of 30° above the horizontal. The suitcase moves 20m horizontally along the ground. Calculate: (a) the work done by the pulling force, (b) the work done against friction if the coefficient of friction is 0.2 and the suitcase weighs 200N.

Force diagram showing angled pulling force

Step-by-Step Solution

Given:
— Applied force: F = 40N at 30° above horizontal
— Displacement: s = 20m (horizontal)
— Weight of suitcase: W = 200N
— Coefficient of friction: μ = 0.2

Part (a): Work done by pulling force
The displacement is horizontal, so we need the horizontal component of the force.
Horizontal component of force: F_x = F cos(30°) = 40 × cos(30°) = 40 × 0.866 = 34.6N

Work done by pulling force: W_pull = F_x × s = 34.6N × 20m = 692J

Part (b): Work done against friction
First, find the normal force:
Vertical forces: N + F sin(30°) = Weight
N = 200 — 40 sin(30°) = 200 — 40 × 0.5 = 180N

Friction force: f = μN = 0.2 × 180 = 36N
Work done against friction: W_friction = f × s = 36N × 20m = 720J

Note: The work done against friction (720J) is greater than the work done by the horizontal component of the pulling force (692J), indicating the suitcase would decelerate or require additional force to maintain constant speed.

Example 2: Power and Efficiency

Problem: An electric motor rated at 5kW is used to lift water from a well. The motor lifts 200 liters of water through a height of 30m in 45 seconds. Calculate: (a) the useful power output, (b) the efficiency of the motor, (c) the energy wasted as heat per minute of operation.

Complete Solution with Efficiency Analysis

Given:
— Motor power rating: P_input = 5kW = 5000W
— Volume of water: V = 200 liters = 0.2 m³
— Height lifted: h = 30m
— Time taken: t = 45s
— Density of water: ρ = 1000 kg/m³

Part (a): Useful power output
Mass of water: m = ρV = 1000 × 0.2 = 200kg
Weight of water: W = mg = 200 × 9.8 = 1960N
Work done against gravity: W_useful = mgh = 1960 × 30 = 58,800J
Useful power output: P_output = W_useful/t = 58,800/45 = 1307W ≈ 1.31kW

Part (b): Efficiency of the motor
Efficiency: η = (P_output/P_input) × 100%
η = (1307/5000) × 100% = 26.1%

Part (c): Energy wasted per minute
Power wasted: P_waste = P_input — P_output = 5000 — 1307 = 3693W
Energy wasted per minute: E_waste = P_waste × 60s = 3693 × 60 = 221,580J ≈ 222kJ

Analysis:
The low efficiency (26.1%) indicates significant energy losses, typical for:
— Motor internal losses (resistance, magnetic losses)
— Pump inefficiencies (fluid friction, turbulence)
— Mechanical friction in bearings and seals

Improvement suggestions:
— Use variable frequency drive for optimal speed control
— Regular maintenance to reduce mechanical friction
— Upgrade to high-efficiency motor design
— Optimize pump selection for the specific operating point

Example 3: Vehicle Power Analysis

Problem: A car with mass 1500kg accelerates from rest to 25 m/s in 10 seconds while traveling up a 5° incline. During this time, air resistance averages 400N and rolling resistance is 200N. Calculate: (a) the total work done by the engine, (b) the average power output, (c) the instantaneous power required at 25 m/s to maintain constant speed on the same incline.

Advanced Vehicle Dynamics Analysis

Given:
— Mass: m = 1500kg
— Initial velocity: u = 0 m/s
— Final velocity: v = 25 m/s
— Time: t = 10s
— Incline angle: θ = 5°
— Air resistance: F_air = 400N (average)
— Rolling resistance: F_roll = 200N

Step 1: Calculate distance traveled
Average velocity = (u + v)/2 = (0 + 25)/2 = 12.5 m/s
Distance: s = average velocity × time = 12.5 × 10 = 125m

Part (a): Total work done by engine
Work components:
1. **Kinetic energy change:** ΔKE = ½mv² — ½mu² = ½ × 1500 × 25² = 468,750J
2. **Potential energy change:** ΔPE = mgh = mg(s sin θ) = 1500 × 9.8 × 125 × sin(5°) = 125,747J
3. **Work against air resistance:** W_air = F_air × s = 400 × 125 = 50,000J
4. **Work against rolling resistance:** W_roll = F_roll × s = 200 × 125 = 25,000J

Total work by engine: W_total = 468,750 + 125,747 + 50,000 + 25,000 = 669,497J ≈ 669.5kJ

Part (b): Average power output
P_average = W_total/t = 669,497/10 = 66,950W ≈ 67.0kW

Part (c): Instantaneous power at constant speed
At constant 25 m/s on the incline, engine must overcome:
— Gravitational component: F_gravity = mg sin(5°) = 1500 × 9.8 × sin(5°) = 1281N
— Air resistance: F_air = 400N (assumed constant at this speed)
— Rolling resistance: F_roll = 200N

Total force needed: F_total = 1281 + 400 + 200 = 1881N
Instantaneous power: P = Fv = 1881 × 25 = 47,025W ≈ 47.0kW

Engineering insights:
1. **Peak vs sustained power:** Average power during acceleration (67kW) exceeds steady-state power (47kW)
2. **Energy distribution:**
— Kinetic energy: 70% of total work
— Potential energy: 19% of total work
— Losses (air + rolling): 11% of total work
3. **Real-world considerations:**
— Air resistance increases with v², so actual power curve would be non-linear
— Engine efficiency varies with RPM and load
— Transmission losses not included in this analysis

This analysis demonstrates why cars need high power for acceleration but can cruise efficiently at constant speed.

🧪 Interactive Investigation - PhET Simulation

Explore work and energy concepts using this interactive simulation:

Investigation Tasks:

Task 1: Set up a ramp and observe how work done by gravity converts to kinetic energy. Measure the relationship between height and speed.

Task 2: Add friction to the system and observe how energy is «lost.» Calculate the work done against friction.

Task 3: Use the bar charts to track energy transformations. Verify that total energy is conserved in frictionless conditions.

Investigation Answers and Analysis

Task 1 Analysis:
Students should observe that gravitational potential energy (mgh) converts to kinetic energy (½mv²). The relationship follows energy conservation: mgh = ½mv², giving v = √(2gh). This demonstrates work-energy theorem in action.

Task 2 Friction Effects:
With friction present, mechanical energy decreases over time. The «lost» energy equals work done against friction (W_friction = f × distance). Students can calculate friction coefficient from energy loss rates.

Task 3 Energy Conservation:
Bar charts show energy transformations clearly:
— Frictionless: PE + KE = constant
— With friction: PE + KE + Thermal = constant
— Work done by friction converts mechanical energy to thermal energy

This visualization reinforces that energy is never truly «lost,» only transformed between different forms.

👥 Collaborative Group Activity

Work with your team to complete this interactive work and power challenge:

Group Design Challenge:

Design an Efficient Lifting System

Challenge: Your team must design a pulley system to lift a 50kg mass 5m high using minimum input work and maximum efficiency.

Requirements:

Calculate theoretical work required
Design pulley configuration for mechanical advantage
Estimate efficiency including friction losses
Compare input power for different lifting speeds

Deliverables:

Technical drawing with force analysis
Complete calculations showing work and power requirements
Efficiency analysis with loss mechanisms identified
Demonstration or simulation (5 minutes maximum)

Alternative Group Activities:

• Power Investigation: Measure power output of students climbing stairs at different speeds

• Efficiency Analysis: Compare efficiency of different simple machines (lever, inclined plane, pulley)

• Energy Transformation: Design Rube Goldberg machine demonstrating multiple work-energy conversions

📝 Individual Assessment - Structured Questions

Question 1: Analysis and Application

A delivery drone with mass 5kg carries a 2kg package and flies horizontally at constant velocity 15 m/s for 10 minutes against a headwind providing 8N of resistance. The propellers also work against gravity. Calculate: (a) the work done against air resistance, (b) the work done against gravity, (c) the minimum power output of the motors, (d) analyze how the power requirement would change if the drone climbed at 2 m/s while maintaining horizontal speed.

Answer

Given:
— Drone mass: 5kg
— Package mass: 2kg
— Total mass: m = 7kg
— Horizontal velocity: v = 15 m/s
— Flight time: t = 10 min = 600s
— Air resistance: F_air = 8N

Part (a): Work done against air resistance
Horizontal distance: s = vt = 15 × 600 = 9000m
Work against air resistance: W_air = F_air × s = 8 × 9000 = 72,000J = 72kJ

Part (b): Work done against gravity
For horizontal flight at constant altitude, no work is done against gravity.
W_gravity = 0J (displacement perpendicular to gravitational force)

Part (c): Minimum power output
Power to overcome air resistance: P_air = F_air × v = 8 × 15 = 120W
Power to maintain altitude: P_gravity = Weight × 0 = 0W (no vertical displacement)
Total minimum power: P_min = 120W

Part (d): Power during climbing
During climb at 2 m/s while maintaining 15 m/s horizontal:
— Power against air resistance: P_air = 8 × 15 = 120W (unchanged)
— Power against gravity: P_gravity = mg × v_climb = 7 × 9.8 × 2 = 137.2W

Total power during climb: P_total = 120 + 137.2 = 257.2W

Analysis: Climbing increases power requirement by 114% (257.2/120), demonstrating why aircraft climb performance is critical for design and why drones have limited payload capacity when climbing.

Question 2: Synthesis and Critical Thinking

A hydroelectric power plant uses water falling from a height of 200m to generate electricity. The water flow rate is 1000 m³/min, and the overall efficiency from water kinetic energy to electrical output is 85%. Calculate: (a) the theoretical power available from the falling water, (b) the electrical power output, (c) analyze the energy losses and suggest improvements. If electricity sells for $0.10 per kWh, determine the revenue per day and discuss the economic factors affecting hydroelectric viability.

Answer

Given:
— Height: h = 200m
— Flow rate: Q = 1000 m³/min = 16.67 m³/s
— Overall efficiency: η = 85% = 0.85
— Electricity price: $0.10/kWh
— Water density: ρ = 1000 kg/m³

Part (a): Theoretical power available
Mass flow rate: ṁ = ρQ = 1000 × 16.67 = 16,670 kg/s
Theoretical power: P_theoretical = ṁgh = 16,670 × 9.8 × 200 = 32.63 MW

Part (b): Electrical power output
P_electrical = η × P_theoretical = 0.85 × 32.63 = 27.74 MW

Part (c): Energy loss analysis
Power losses: P_loss = 32.63 — 27.74 = 4.89 MW (15% of input)

Sources of energy loss:
1. **Hydraulic losses (5-8%):** Friction in penstocks, turbulence, flow separation
2. **Turbine losses (3-5%):** Mechanical friction, imperfect blade design, cavitation
3. **Generator losses (2-3%):** Electrical resistance, magnetic losses, bearing friction
4. **Transformer losses (1-2%):** Copper losses, iron losses in power transmission

Improvement suggestions:
— Optimize penstock design to minimize friction losses
— Use advanced turbine blade profiles for better hydraulic efficiency
— Implement variable speed generators for optimal efficiency across flow ranges
— Regular maintenance to prevent efficiency degradation
— Modern power electronics for improved electrical conversion

Revenue calculation:
Daily energy output: E = P × 24h = 27.74 MW × 24h = 665.8 MWh = 665,800 kWh
Daily revenue: R = 665,800 × $0.10 = $66,580
Annual revenue: ≈ $24.3 million

Economic factors affecting viability:
1. **Capital costs:** High initial investment ($2-5 million per MW)
2. **Environmental regulations:** Fish ladders, minimum flow requirements
3. **Grid integration:** Transmission infrastructure costs
4. **Capacity factor:** Seasonal water availability affects revenue
5. **Competing technologies:** Solar and wind cost reductions
6. **Government incentives:** Renewable energy credits, tax benefits

Long-term considerations:**strong>
— Low operational costs (no fuel required)
— 50-100 year operational life
— Climate change effects on water availability
— Energy storage potential through pumped storage

Hydroelectric remains economically viable due to low operating costs and long operational life, despite high initial capital requirements.

Question 3: Complex Analysis

A mountain railway uses a rack and pinion system to climb a 25° gradient. The train has mass 150 tonnes and climbs at constant speed 5 m/s. The efficiency of the drive system is 75%, and additional resistance forces total 50 kN. Calculate: (a) the power required at the wheels, (b) the motor power input, (c) the fuel consumption if the diesel engine has 40% efficiency and diesel has energy content 35 MJ/L. Analyze the trade-offs between speed, power, and efficiency for mountain railway design.

Answer

Given:
— Train mass: m = 150 tonnes = 150,000 kg
— Gradient angle: θ = 25°
— Speed: v = 5 m/s
— Drive system efficiency: η_drive = 75% = 0.75
— Additional resistance: F_resistance = 50 kN = 50,000 N
— Engine efficiency: η_engine = 40% = 0.40
— Diesel energy content: E_diesel = 35 MJ/L
Part (a): Power required at wheels
Forces to overcome:
1. **Gravitational component:** F_gravity = mg sin(25°) = 150,000 × 9.8 × sin(25°) = 621,558 N
2. **Additional resistance:** F_resistance = 50,000 N
3. **Total force:** F_total = 621,558 + 50,000 = 671,558 N
Power at wheels: P_wheels = F_total × v = 671,558 × 5 = 3,357,790 W ≈ 3.36 MW
Part (b): Motor power input
P_motor = P_wheels / η_drive = 3,357,790 / 0.75 = 4,477,053 W ≈ 4.48 MW
Part (c): Fuel consumption
Power from fuel: P_fuel = P_motor / η_engine = 4,477,053 / 0.40 = 11,192,633 W ≈ 11.19 MW
Fuel consumption rate:
Energy rate = 11.19 MW = 11.19 MJ/s
Volume rate = 11.19 MJ/s ÷ 35 MJ/L = 0.32 L/s = 19.2 L/min = 1,152 L/h
Mountain railway design trade-offs analysis:
Speed vs Power:
— Power requirement scales linearly with speed: P = Fv
— Higher speeds require exponentially more power due to increased air resistance
— Steep gradients make speed increases very expensive energetically
— Optimal speed balances journey time with energy consumption
Efficiency considerations:
1. **Rack and pinion advantages:**
— Positive engagement prevents slipping on steep grades
— High reliability and safety
— Relatively high efficiency (75-85%)
2. **Alternative systems comparison:**
— Cable systems: Higher efficiency but limited flexibility
— Adhesion railways: Limited to ~6% grades
— Cog railways: Can handle steeper grades but lower efficiency
Design optimization strategies:
1. **Gradient optimization:** Reduce maximum gradients through longer routes
2. **Weight reduction:** Lightweight materials, optimal consist length
3. **Regenerative braking:** Recover energy during descent
4. **Variable speed operation:** Optimize for efficiency rather than schedule
5. **Hybrid propulsion:** Electric assist for peak power demands
Economic considerations:**strong>
— **High fuel costs:** 1,152 L/h × operating hours = significant expense
— **Infrastructure costs:** Rack rail systems expensive to build and maintain
— **Capacity limits:** Power requirements limit maximum train weight/length
— **Environmental impact:** High fuel consumption vs tourism/transport benefits
Engineering solutions:**
— **Electric propulsion:** Use grid electricity or regenerative systems
— **Distributed traction:** Multiple smaller motors for better efficiency
— **Advanced materials:** Reduce train weight while maintaining strength
— **Smart scheduling:** Coordinate up/down trains for energy optimization
Mountain railways represent extreme engineering challenges where power requirements dominate design decisions, making efficiency optimization critical for economic viability.

Question 4: Engineering Application and Synthesis

Design a human-powered aircraft for a competition requiring sustained flight for 1 hour while carrying a 50kg pilot. The aircraft must maintain level flight at 40 km/h against air resistance that varies as F = 0.02v² (where F is in Newtons and v in m/s). Calculate: (a) the continuous power output required from the pilot, (b) compare this to typical human power capabilities, (c) design solutions to minimize power requirements, (d) analyze the engineering constraints and trade-offs for achieving sustained human-powered flight.

Answer

Design Requirements Analysis:
Given specifications:
— Flight duration: 1 hour
— Pilot mass: 50kg
— Flight speed: 40 km/h = 11.11 m/s
— Air resistance: F = 0.02v²
— Level flight (no climbing required)
Part (a): Required power output
Air resistance force: F = 0.02 × (11.11)² = 0.02 × 123.4 = 2.47 N
Power required: P = Fv = 2.47 × 11.11 = 27.4 W
Wait — this seems too low! Let’s reconsider…
The given air resistance formula likely has incorrect units. For realistic aircraft:
Typical drag equation: F = ½ρv²C_DA
For a streamlined human-powered aircraft: F ≈ 0.02v² is unrealistically low.
Realistic analysis:
Assume proper drag coefficient for human-powered aircraft:
F = 0.5 × 1.225 × v² × C_D × A
With C_D ≈ 0.03 and wing area A ≈ 20 m²:
F = 0.5 × 1.225 × (11.11)² × 0.03 × 20 = 27.8 N
Required power: P = 27.8 × 11.11 = 309 W
Part (b): Comparison with human capabilities
Human power output capabilities:
— **Peak power:** 1000-1500 W (professional athletes, <10 seconds)
— **Sustained high power:** 400-500 W (30 minutes, trained cyclist)
— **Sustainable power:** 200-300 W (1-2 hours, trained athlete)
— **Recreational level:** 100-150 W (sustained)
**Analysis:** 309 W for 1 hour is at the upper limit of human capability, requiring exceptional physical conditioning.
Part (c): Design solutions to minimize power
Aerodynamic optimization:
1. **Streamlined fuselage:** Minimize frontal area and drag coefficient
2. **High aspect ratio wings:** Reduce induced drag (AR > 20)
3. **Smooth surfaces:** Minimize skin friction drag
4. **Optimal wing design:** Laminar flow airfoils for low Reynolds numbers
Weight reduction strategies:
1. **Advanced materials:** Carbon fiber construction
2. **Structural optimization:** Computer-aided design for minimum weight
3. **Component integration:** Eliminate unnecessary parts
4. **Target weight:** <30kg total aircraft weight
Efficiency improvements:
1. **Propeller optimization:** Large diameter, low RPM for high efficiency
2. **Transmission efficiency:** Minimize drivetrain losses
3. **Pilot position:** Optimize for both aerodynamics and power output
Part (d): Engineering constraints and trade-offs
Primary constraints:
1. **Structural strength:** Must withstand flight loads while being lightweight
2. **Stability and control:** Maintain controllability with minimal pilot input
3. **Human factors:** Pilot position for sustained power output and visibility
4. **Manufacturing:** Practical construction with available materials/methods
Critical trade-offs:
**Weight vs Strength:**
— Lighter structure reduces drag but may compromise safety margins
— Solution: Advanced composites and topology optimization
**Span vs Structural complexity:**
— Longer wings reduce induced drag but increase weight and complexity
— Solution: Optimal aspect ratio analysis (typically AR = 15-25)
**Speed vs Power:**
— Power increases as v³ for overcoming drag
— Slower flight reduces power but increases total flight time energy
— Optimal speed minimizes total energy expenditure
**Stability vs Performance:**
— Inherent stability reduces pilot workload but adds weight/drag
— Relaxed stability requires more pilot attention but improves efficiency
Real-world examples analysis:
**Successful designs:**
— **Gossamer Albatross:** 96 kg, 3.6 kW (peak), crossed English Channel
— **MIT Daedalus:** 31 kg, flew 115 km in 3h 54min
— **AeroVelo Atlas:** Won Sikorsky Prize for human-powered helicopter
Key design principles:**
1. **Minimize drag coefficient:** Achieve C_D 80% of weight in wing structure
3. **Optimize for sustained power:** Design around 250W continuous output
4. **Redundant safety systems:** Multiple backup controls and landing options
Feasibility conclusion:
Sustained 1-hour flight at 40 km/h is achievable but requires:
— Elite athlete pilot (>300W sustainable power)
— Extremely efficient aircraft design (total drag <30N)
— Optimal weather conditions (no headwinds, smooth air)
— Backup landing strategies for pilot fatigue
The engineering challenge lies in achieving the necessary balance between minimum weight, maximum efficiency, and adequate safety margins while operating at the absolute limits of human power capability.

Question 5: Advanced Critical Analysis

A renewable energy advocate claims that «electric vehicles are always more efficient than gasoline cars because electric motors are 95% efficient while gasoline engines are only 25% efficient.» Critically evaluate this statement by analyzing the complete energy chain from source to wheels for both technologies. Consider power plant efficiency, transmission losses, battery charging/discharging efficiency, and fuel refining/distribution. Calculate the overall «well-to-wheels» efficiency for both systems and discuss the implications for environmental impact and energy policy.

Answer

Critical Evaluation: The statement is OVERSIMPLIFIED and potentially misleading.
Complete energy chain analysis reveals the full picture:
Electric Vehicle Energy Chain:
Electricity generation efficiency:
— **Natural gas plants:** 50-60% efficient
— **Coal plants:** 33-40% efficient
— **Nuclear plants:** 33-35% efficient
— **Renewable sources:** 95-99% (solar panel ~20%, but direct conversion)
— **Average US grid mix:** ~38% overall efficiency
Transmission and distribution losses:** 8-15% (average ~10%)
Battery charging efficiency:** 85-95% (average ~90%)
Battery discharging efficiency:** 85-95% (average ~90%)
Electric motor efficiency:** 90-95% (average ~92%)
Overall electric vehicle efficiency calculation:
η_EV = η_generation × η_transmission × η_charging × η_discharging × η_motor
**Best case (renewable source):**
η_EV = 0.95 × 0.90 × 0.90 × 0.90 × 0.92 = 64%
**Average US grid mix:**
η_EV = 0.38 × 0.90 × 0.90 × 0.90 × 0.92 = 25%
**Coal-heavy grid:**
η_EV = 0.35 × 0.90 × 0.90 × 0.90 × 0.92 = 23%
Gasoline Vehicle Energy Chain:
Crude oil extraction efficiency:** 95-98% (average ~96%)
Refining efficiency:** 85-90% (average ~87%)
Transportation and distribution:** 95-98% (average ~96%)
Gasoline engine efficiency:** 20-35% (average ~25%)
Overall gasoline vehicle efficiency:
η_ICE = η_extraction × η_refining × η_distribution × η_engine
η_ICE = 0.96 × 0.87 × 0.96 × 0.25 = 20%
Comparative Analysis:
Well-to-wheels efficiency comparison:
— **EV (renewable grid):** 64%
— **EV (US average grid):** 25%
— **EV (coal-heavy grid):** 23%
— **Gasoline vehicle:** 20%
Critical findings:
1. **Grid composition matters crucially:** EVs are only significantly more efficient with clean electricity sources
2. **Regional variations:** EV efficiency varies dramatically by location:
— **Norway (hydro-heavy):** EV efficiency ~60%
— **West Virginia (coal-heavy):** EV efficiency ~23%
— **California (mixed renewables):** EV efficiency ~35%
3. **Time-of-charging effects:** Grid efficiency varies by time of day and season
Environmental Impact Analysis:
Carbon emissions (kg CO₂/kWh):
— **Coal power:** 0.82-1.05
— **Natural gas:** 0.35-0.49
— **US grid average:** ~0.40
— **Renewable sources:** 0.01-0.05
**Carbon footprint comparison (per 100 km):**
— **Gasoline car:** ~20 kg CO₂ (8L/100km × 2.3 kg CO₂/L)
— **EV (US average grid):** ~8 kg CO₂ (20 kWh/100km × 0.40 kg CO₂/kWh)
— **EV (renewable grid):** ~0.5 kg CO₂
Additional Considerations:
Manufacturing impact:
— **EV batteries:** High embodied energy and rare earth requirements
— **ICE vehicles:** Lower manufacturing energy but ongoing fuel extraction impact
— **Lifecycle analysis:** EVs typically break even after 20,000-50,000 km
Infrastructure requirements:
— **EV charging:** Requires grid upgrades and charging infrastructure
— **Gasoline:** Extensive existing refining and distribution network
Energy Policy Implications:
Short-term considerations:
1. **Grid cleaning priority:** Decarbonizing electricity generation maximizes EV benefits
2. **Regional strategies:** EV incentives most effective in regions with clean grids
3. **Charging timing:** Smart charging can utilize lowest-carbon electricity
Long-term strategies:
1. **Renewable energy expansion:** Essential for realizing EV environmental potential
2. **Grid modernization:** Smart grids enable optimal EV integration
3. **Battery technology:** Improved efficiency and reduced environmental impact
4. **Alternative fuels:** Hydrogen and synthetic fuels for applications unsuited to batteries
Conclusion:
**The advocate’s statement is misleading because:**
1. **Ignores upstream efficiency:** Electric motor efficiency is only one component
2. **Assumes clean electricity:** Most grids still rely heavily on fossil fuels
3. **Oversimplifies comparison:** Real-world efficiency depends on complete energy chain
**More accurate statement:** «Electric vehicles can be significantly more efficient than gasoline cars, but only when powered by clean electricity sources. In regions with coal-heavy grids, the efficiency advantage may be minimal.»
**Policy recommendations:**
— Prioritize renewable energy development alongside EV adoption
— Implement regional EV incentives based on local grid carbon intensity
— Develop smart charging systems to optimize grid efficiency
— Continue research into all low-carbon transportation technologies
The analysis demonstrates that technology comparisons must consider complete systems, not isolated components, and that environmental benefits depend critically on energy source composition and regional factors.

🔗 Additional Learning Resources

📚 Comprehensive Study Resources:

🤔 Lesson Reflection and Self-Assessment

💭 Knowledge Self-Check

Conceptual Understanding (Rate 1-5):

□ I understand the scientific definition of work and how it differs from everyday usage

□ I can apply W = F × s × cos θ correctly for forces at different angles

□ I can define power and use both P = W/t and P = Fv appropriately

□ I understand when work is positive, negative, or zero

□ I can analyze efficiency in mechanical systems and energy transformations

Problem-Solving Skills Assessment:

Which problem-solving strategies worked best for you today?

Identifying the correct angle between force and displacement vectors
Choosing between P = W/t and P = Fv based on given information
Breaking complex scenarios into individual work components
Using energy conservation principles to check answers
Converting between different power units (W, kW, hp)

What challenges did you encounter?

Distinguishing between distance traveled and displacement in work calculations
Understanding when forces do no work (perpendicular motion)
Analyzing efficiency in multi-stage energy conversion systems
Working with variable forces or changing speeds

Real-World Connections:

How can you apply work and power concepts in everyday life?

Understanding why pushing a car uphill requires more power than on level ground
Appreciating energy efficiency ratings for appliances and vehicles
Recognizing why electric tools have different power ratings for different tasks
Understanding fuel consumption in vehicles at different speeds
Analyzing the efficiency of renewable energy systems

Language Development Reflection:

New physics vocabulary mastered:

□ Can distinguish between «work» in physics vs everyday language

□ Understand power units and can convert between W, kW, and hp

□ Can explain efficiency concepts clearly in written English

□ Comfortable reading and interpreting energy and power problems in English

Communication goals for next lesson:

• Practice explaining work-energy relationships using real examples

• Use more precise scientific language when describing energy transformations

• Develop confidence in presenting power calculations with proper units

Future Learning Goals:

What aspects of work and power would you like to explore further?

Advanced energy conservation and transformation principles
Rotational work and power in mechanical systems
Applications in renewable energy and sustainability
Power analysis in electrical circuits and systems
Thermodynamic efficiency and heat engines

How will this knowledge help in future physics topics?

Understanding kinetic and potential energy relationships
Analyzing oscillatory motion and wave energy
Studying thermodynamics and energy conversion
Connecting to electromagnetic induction and electrical power

🎯 Action Plan for Continued Learning:

This week I will:

□ Practice identifying work and power in everyday mechanical systems

□ Calculate efficiency for various household appliances and devices

□ Review and strengthen understanding of vector components in work calculations

□ Explore the additional simulations and online practice problems

□ Prepare questions about energy applications for next class

Work

Power

Positive vs Negative Work

Efficiency

Work

Power

Mechanical Efficiency

Joule

The Concept of Work

Defining Power

Deriving P = Fv

Theory Questions

Complete the Definitions

Additional Video Resources:

Example 1: Work at an Angle

Example 2: Power and Efficiency

Example 3: Vehicle Power Analysis

Investigation Tasks:

Group Design Challenge:

Design an Efficient Lifting System

Alternative Group Activities:

Question 1: Analysis and Application

Question 2: Synthesis and Critical Thinking

Question 3: Complex Analysis

Question 4: Engineering Application and Synthesis

Question 5: Advanced Critical Analysis

Electric Vehicle Energy Chain:

Gasoline Vehicle Energy Chain:

Comparative Analysis:

Environmental Impact Analysis:

Additional Considerations:

Energy Policy Implications:

Conclusion:

📚 Comprehensive Study Resources:

📖 Theory Resources

🎥 Video Tutorials

🧪 Interactive Tools

📝 Practice Problems

💭 Knowledge Self-Check

Conceptual Understanding (Rate 1-5):

Problem-Solving Skills Assessment:

Real-World Connections:

Language Development Reflection:

Future Learning Goals:

🎯 Action Plan for Continued Learning: