Work, Energy and Power

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Work — The Foundation of Energy:

Work is defined as the product of a force and the displacement of its point of application in the direction of the force: W = F × d × cosθ. SI unit: Joule (J). The angle θ is critical — work is maximum (Fd) when force and displacement are in the same direction, zero when perpendicular, and negative when opposed.

Practical examples for WAEC level:

• Lifting a 5 kg bag of rice through 1.5 m: W = 5 × 10 × 1.5 = 75 J
• Pushing a desk 4 m across a floor with a force of 20 N parallel to the floor: W = 20 × 4 = 80 J
• Dragging a sledge 10 m with a force of 30 N at 30° to the horizontal: W = 30 × 10 × cos30° = 300 × 0.866 = 259.8 J

Energy — The Capacity to Do Work:

Kinetic Energy (KE): Energy possessed by a body due to its motion. KE = ½mv². SI unit: Joule. A cyclist of mass 80 kg cycling at 5 m/s has KE = ½ × 80 × 25 = 1000 J.

Potential Energy (PE): Energy possessed by a body due to its position. Gravitational PE = mgh, where h is height above the chosen zero level. A 3 kg bag of oranges placed on a shelf 2 m high has PE = 3 × 10 × 2 = 60 J relative to the floor.

⚡ WAEC Tip: For WAEC calculations, g = 10 m/s² is standard (use 10, not 9.8, unless specified otherwise). The WASSCE often requires you to show your working — write the formula first, substitute values with units, then give the final answer with correct units.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Power — Rate of Working:

Power = Work done / Time taken = Energy transformed / Time. SI unit: Watt (W). Average power = Total work / Total time. Instantaneous power = F × v (when force and velocity are in the same direction).

Common power units and conversions:

• 1 Watt = 1 Joule per second
• 1 kilowatt (kW) = 1000 W
• 1 megawatt (MW) = 10⁶ W
• 1 horsepower (hp) ≈ 746 W (used for engines and motors in Nigeria/Africa context)

A 2 kW electric kettle can do 2000 J of work in 1 second. To heat 1 kg of water by 1°C (specific heat capacity = 4200 J/kg°C), it takes 4200 J. A 2 kW kettle would do this in 4200/2000 = 2.1 seconds (ignoring heat losses).

Conservation of Energy:

Energy cannot be created or destroyed, only transformed from one form to another. For mechanical energy conservation (no non-conservative forces): Total mechanical energy at any point = KE + PE = constant.

For a ball thrown vertically upward from ground level with speed v₀: maximum height h_max = v₀²/(2g). At any intermediate height y: v² = v₀² - 2gy. At maximum height: v = 0, so v₀² = 2gh_max.

⚡ WAEC Tip: When friction is present, mechanical energy is NOT conserved. Energy lost = work done against friction = μ × N × d = μmg × d (for horizontal surface). This energy appears as heat. A common WAEC question: “A stone of mass 2 kg falls from 5 m height onto sand. It penetrates 0.1 m into the sand before stopping. Find the average resistive force.” Using work-energy: loss in PE = gain in KE (at sand surface) = mgh = 2 × 10 × 5 = 100 J. This 100 J is used to do work against sand resistance over 0.1 m: F × 0.1 = 100 → F = 1000 N.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Elastic and Inelastic Collisions:

A perfectly elastic collision conserves both momentum AND kinetic energy. For a mass m₁ moving at u₁ colliding head-on with a stationary mass m₂:

• v₁ = ((m₁ - m₂)/(m₁ + m₂)) × u₁
• v₂ = (2m₁/(m₁ + m₂)) × u₁

A perfectly inelastic collision: objects stick together and move as one mass. Momentum is conserved but kinetic energy is lost (converted to heat, sound, deformation energy). Final velocity v = (m₁u₁ + m₂u₂)/(m₁ + m₂).

The coefficient of restitution e measures elasticity: e = v_separation/v_approach. For perfectly elastic: e = 1. For perfectly inelastic: e = 0. For real collisions: 0 < e < 1.

Efficiency of Machines:

Efficiency η = (Useful output energy / Input energy) × 100%. Since some energy is always dissipated (friction, air resistance), η < 100% always. For a machine lifting a load L using effort E with velocity ratio VR and mechanical advantage MA:

• Velocity ratio VR = distance moved by effort / distance moved by load (for ideal machine)
• Mechanical advantage MA = Load / Effort
• Efficiency η = (MA/VR) × 100%

For a block and tackle pulley system with n supporting strands: VR = n. If the load is L and effort needed is E (including friction), MA = L/E, and η = (L/(E × n)) × 100% × VR… no: η = (MA/VR) × 100%.

⚡ WAEC WASSCE Pattern: WAEC questions frequently involve Nigerian everyday contexts: “A hawker sells oranges in Lagos. She lifts a basket of 20 kg from the ground onto her head, 1.8 m high. Calculate the work done.” The physics is identical — W = mgh = 20 × 10 × 1.8 = 360 J. Other common WAEC questions include calculating the power of a water pump lifting water to a tank on a storey building, and determining the efficiency of a pulley system used in construction. Always show the formula, substitution, and result with units for full marks.