A motor lifts a 100 kg load 5 meters in 10 seconds. What is the motor's power output? (g = 10 m/s² for simplicity)

Power = Work / Time. Work = Force × Distance = (mg) × h = 100 × 10 × 5 = 5000 J (joules). Power = 5000 J / 10 s = 500 W (watts). Power units: WATT = Joule per second (SI standard); 1 kW = 1000 W; 1 MW = 1,000,000 W; HORSEPOWER (hp) = 746 W (or 550 ft-lb/s); 1 hp ≈ 0.75 kW. Common comparisons: human walking power ≈ 100 W; cyclist sustained ≈ 200 W; cyclist sprint ≈ 1000-2000 W; horse sustained ≈ 750 W (defining 1 hp); car engine ≈ 100,000-300,000 W (135-400 hp); jet engine takeoff ≈ 50-100 million W; nuclear power plant ≈ 1-3 billion W (1-3 GW). Power = Force × Velocity for constant force scenarios: a car pulling 1000 N of force at 20 m/s = 20,000 W = 20 kW (assuming constant velocity, all motor power overcomes drag and friction). Electrical power: P = IV = I²R = V²/R, where I = current, V = voltage, R = resistance. Energy = Power × Time. Common energy units: 1 kWh = 1000 W × 3600 s = 3,600,000 J = 3.6 MJ; electric bill quoted in kWh. A 100 W bulb running 10 hours uses 1 kWh of energy. Efficiency: useful power output / total power input; expressed as percentage; always less than 100% due to losses (heat, sound, friction). Engines typically 20-40% efficient; electric motors 70-95%; LEDs 30-50%. ASVAB MC may ask for power in basic scenarios (Work/Time), efficiency calculations, or unit conversions. The motor in the question above outputs 500 W (about 2/3 of one horsepower) to lift 100 kg vertically 5 meters in 10 seconds.