Coaching Centre
Power
Power as the rate of energy transfer (P = ΔE/t), power and velocity (P = Fv), efficiency, the power needed to overcome air resistance and rolling resistance at different speeds, and a practical to measure personal power output.
An elite track cyclist can hold about 400 W for an hour and burst past 2,000 W in a sprint start. A small car cruising on the M1 needs only about 14 kW, the output of roughly 35 of those cyclists. But the car's engine is rated at 100 kW or more. Where does the rest go? Mostly into overtaking, climbing hills and fighting air resistance, which grows brutally with speed.
Work tells you how much energy is transferred. Power tells you how fast. Two students of the same mass climbing the same stairs do the same work, but the one who runs develops more power.
Power as rate of energy transfer
$$P = \frac{\Delta E}{t} = \frac{W}{t}$$Power is measured in watts: 1 W = 1 J s⁻¹. Useful conversions: 1 kW = 1000 W; one horsepower is about 746 W; and the kilowatt-hour on a power bill is an amount of energy, $1000 \text{ W} \times 3600 \text{ s} = 3.6 \times 10^6$ J.
A 60 kg student runs up 18 steps, each 17.0 cm high, in 2.5 s. Find the average useful power.
Power and velocity: P = Fv
When a constant force $F$ moves an object along its direction at speed $v$, the work done in a short time $t$ is $W = Fs$, so
$$P = \frac{Fs}{t} = Fv$$This is the power the force delivers at that instant. If the force is at angle $\theta$ to the velocity, $P = Fv\cos\theta$.
$P = Fv$ is especially useful for vehicles at constant speed. With no acceleration, the driving force equals the total resistance force, so the power needed is just resistance times speed.
A 1500 kg car travels at a constant 100 km/h on a level road. The rolling resistance coefficient is 0.012, and air resistance is $\tfrac12\rho C_dAv^2$ with $\rho = 1.2$ kg m⁻³ and $C_dA = 0.70$ m². Find the driving power needed.
At low speeds, rolling resistance dominates. At highway speeds, air resistance takes over. Since $F_{air} \propto v^2$, the power to overcome it grows with $v^3$. Going from 100 km/h to 110 km/h, only 10% faster, needs 25% more power. This is why fuel use rises sharply at high speed, and why cyclists ride in tight bunches: tucking in behind another rider cuts air resistance by roughly a third.
Efficiency
No machine converts all its input energy into useful output. The rest becomes thermal energy (and some sound).
$$\text{efficiency} = \frac{\text{useful output power}}{\text{input power}} \times 100\%$$An electric winch lifts a 200 kg load at a steady 0.50 m s⁻¹. The motor is 80% efficient. Find the useful output power and the electrical input power.
Measuring your own power
- In Stair climb, a 60 kg student taking 2.5 s gains 1,799 J and develops 720 W. Slow the time to 5.0 s: same energy, half the power.
- In Car at highway speed, 100 km/h needs 500 N of driving force and 13.9 kW. At 110 km/h it is 17.4 kW. At 50 km/h it is only 3.6 kW.
- Watch the green air-resistance curve. The two curves cross near 75 km/h: below that, rolling resistance takes more power; above it, air resistance dominates.
- In Motor lifting a load, 200 kg at 0.50 m/s needs 980 W of useful power, and with 80% efficiency the motor draws 1,225 W.
Test yourself
A 75 kg climber gains 12 m of height in 40 s. The average useful power is
$P = \dfrac{75 \times 9.8 \times 12}{40} = 220$ W.
A cyclist rides at a constant 10 m s⁻¹ against total resistance forces of 25 N. The cyclist's power output to the wheels is
At constant speed the driving force equals the resistance: $P = Fv = 25 \times 10 = 250$ W.
If air resistance is proportional to $v^2$, doubling a car's speed multiplies the power needed to overcome it by
$P = Fv \propto v^2 \times v = v^3$, and $2^3 = 8$.
A 2.0 kW kettle runs for 3.0 minutes. The energy it uses is
$E = Pt = 2000 \times 180 = 3.6 \times 10^5$ J.
Label the measurements needed for the power practical.
Order these by the power needed (smallest first).
Power is the rate at which is transferred, measured in . For a force acting along the motion, power equals force times . At highway speed most of a car's driving power goes into overcoming resistance. The ratio of useful output power to input power is the .
A 1200 kg electric car drives at a constant 90 km/h up a straight road that rises 1.0 m for every 20 m travelled along the road. Total air and rolling resistance is 450 N. The motor system is 90% efficient. Calculate the electrical power drawn from the battery.
$v = 90/3.6 = 25$ m s⁻¹. The road rises 1.0 m per 20 m, so $\sin\theta = 0.050$.
Component of weight down the slope: $mg\sin\theta = 1200 \times 9.8 \times 0.050 = 588$ N.
At constant speed the driving force balances this and the resistance: $F = 588 + 450 = 1038$ N.
Output power: $P = Fv = 1038 \times 25 = 2.6 \times 10^4$ W.
Electrical input: $P_{in} = \dfrac{2.6 \times 10^4}{0.90} = 2.9 \times 10^4$ W, about 29 kW.