Work, Energy, Power and Pressure
This domain connects mechanical energy transfer with pressure in solids and fluids. The exam tests work done by a force, kinetic and gravitational potential energy, conservation of energy, power, efficiency, and pressure in liquids and hydraulic systems.
Core rules and concepts
- Work done W = Fs cos θ, where θ is the angle between force and displacement. Work is zero when force is perpendicular to motion.
- Kinetic energy KE = ½mv²; gravitational potential energy change ΔGPE = mgh.
- Conservation of energy: energy is transferred, not created or destroyed. Useful output plus wasted energy equals input energy.
- Power P = W/t = energy transferred/time. Efficiency = useful output power/input power × 100%.
- Pressure p = F/A for solids. Liquid pressure p = ρgh depends on depth, density and g, not on the shape or area of the container.
- In a hydraulic system, pressure is transmitted equally: F₁/A₁ = F₂/A₂. A small force on a small piston can produce a large force on a large piston, but the pistons move different distances.
Worked example
A pump raises 30 kg of water through 6.0 m in 12 s. The input power is 200 W. Find the useful output power and efficiency (g = 10 N/kg). Useful work done = mgh = 30 × 10 × 6.0 = 1800 J. Useful power = 1800/12 = 150 W. Efficiency = 150/200 × 100% = 75%. Reasoning: the pump’s input power is greater than the useful output because some energy is wasted in friction and sound.
Traps that cost marks
- Calculating work when the force is perpendicular to displacement.
- Reporting efficiency above 100% by mixing input and output values.
- Using mass instead of weight in pressure calculations.
- Thinking liquid pressure depends on the total volume or container shape.
- Confusing gauge pressure with absolute pressure; atmospheric pressure must be added when absolute pressure is asked.