Power

Average power is the energy moved divided by the time taken, P_avg = W/Δt. It is the right number for an electricity bill and the wrong one for choosing a motor. Take two conveyor drives that each deliver 1.2 MJ in a 60 s cycle. One runs at a steady 20 kW. The other idles, then delivers two 10 s bursts at 60 kW. Both average 20 kW, but only a 60 kW motor can run the second, because a machine is limited by the rate it must sustain at each moment. That moment-by-moment rate is the instantaneous power P = dW/dt. Run the definition backwards and the work over an interval is W = ∫ P dt, the area under a power–time graph: 20 kW × 60 s and 2 × 60 kW × 10 s enclose the same 1.2 MJ. The kilowatt-hour is this area in practical units: 1 kWh = 1000 W × 3600 s = 3.6 MJ.

In a short time dt, a force F⃗ acting on an object that moves through dr⃗ does work dW = F⃗⋅dr⃗. Divide by dt and, since dr⃗/dt = v⃗, P = F⃗⋅v⃗ = F v cos φ. No finite displacement appears, only the velocity at this instant, so the formula holds for forces and speeds that change continuously. A rope pulling a sled with 150 N at 35° to its direction of motion, while the sled moves at 1.2 m s⁻¹, delivers P = 150 × 1.2 × cos 35° = 147 W. The sign carries physics. A 4.0 kN brake force opposing a car moving at 20 m s⁻¹ has P = −80 kW: energy is leaving the car and heating the discs. A force perpendicular to v⃗ has P = 0 however large it is, which is why the normal force from a fixed curved track and the tension in a pendulum string can steer an object without ever changing its speed.

Sum F⃗⋅v⃗ over every force on a particle and Newton's second law turns the sum into ΣF⃗⋅v⃗ = m a⃗⋅v⃗ = d(½mv²)/dt, because d(v⃗⋅v⃗)/dt = 2a⃗⋅v⃗. The net power is the rate of change of kinetic energy: the work–kinetic energy theorem, written per second. That makes it a way to find a driving power you cannot measure directly. A 1200 kg car at 25 m s⁻¹, gaining speed at 1.5 m s⁻², needs dK/dt = m a v = 1200 × 1.5 × 25 = 45 kW of net power. If drag and rolling resistance are removing 12 kW at that speed, the wheels must be supplying 57 kW. It also explains uniform circular motion at a glance: a⃗ is perpendicular to v⃗, so dK/dt = 0 at every instant, however hard the centripetal force pulls.

A motor's rated power is the largest rate at which it can deliver energy continuously without overheating. Because P = F v, a fixed rating trades force against speed: the drive force available is at most Pₘₐₓ/v. Put 75 kW into a van's driven wheels and it can push with 7.5 kN at 10 m s⁻¹ but only 1.5 kN at 50 m s⁻¹. That hyperbola cannot hold near v = 0, where it would demand an unlimited force, so something else takes over there: tyres can push on the road only up to the static-friction limit. With a grip limit of 5.0 kN the van accelerates at constant force until v = P/F = 75 000 W ÷ 5000 N = 15 m s⁻¹, and only above that speed does the rating become the constraint. Ratings also come with conditions. An engine's quoted maximum applies at one shaft speed, and an electric motor's short-term peak can be well above what it can sustain. Horsepower survives as a unit: 1 hp ≈ 746 W.

At steady speed the drive force equals the total resistance, so the power needed is P = Fᵣₑₛ v. On a road Fᵣₑₛ = mg sin θ + Cᵣᵣ mg cos θ + kv², a grade term, a rolling term and air drag with k = ½ρC_d A. The power needed therefore climbs like v³ once drag dominates, and top speed is where it meets the power actually delivered to the wheels. Take the car in the figure: m = 1200 kg, Cᵣᵣ = 0.012, k = 0.40 kg m⁻¹, and a 60 kW engine behind an 85% efficient drivetrain, so 51 kW reaches the wheels. On the level, 0.40v³ + 141v = 51 000 W gives v ≈ 48 m s⁻¹. Tilt the road to 4° and the grade term adds mg sin θ ≈ 820 N, which pulls the top speed down to about 35 m s⁻¹, even though the engine and its rating are unchanged.

No real machine delivers all of its input. Efficiency is the fraction that survives, η = Pₒᵤₜ/Pᵢₙ, and the rest, (1 − η)Pᵢₙ, leaves as heat in windings, bearings, gear teeth and tyres. Stages in series multiply rather than add. An electric car whose inverter is 97% efficient, motor 92% and gearbox 97% passes on 0.97 × 0.92 × 0.97 = 0.866 of the battery's output, so 80 kW drawn becomes 69 kW at the wheels and 11 kW of heat the cooling system must carry away. A petrol engine turns only about 30% of its fuel's chemical energy into shaft work, so running at a 60 kW rating means about 200 kW of fuel power and 140 kW rejected through the radiator and exhaust. Efficiency is not a fixed label either: it depends on the operating point, and most motors do worst far below their rated load. Always ask which power a quoted figure means: fuel in, shaft out, or delivered to the load.