Power
When Tesla launched the Supercharger V3 network in 2019, each stall delivered 250,000 W — enough to add 120 km of range in 5 minutes. A 75 kWh battery pack charged from 20% to 80% receives $0.60 \times 75{,}000 = 45{,}000\text{ Wh} = 162\text{ MJ}$ of energy; at 250 kW that takes just $162{,}000{,}000 \div 250{,}000 = 648\text{ s} \approx 10.8$ minutes — a direct application of $P = \Delta E/\Delta t$.
Practise this lesson
Power worksheets — from formula recall to real-world analysis.
A 70 kg athlete sprints up a 5 m staircase in 4 seconds. A 70 kg person walks up the same staircase in 40 seconds. Who does more work? Who is more powerful?
A motor lifts a 200 kg load 10 m in 5 s. Its average power is ($g = 9.8$ m/s²):
Know
- $P = \Delta E/\Delta t$ (energy transferred per unit time)
- $P = Fv\cos\theta$ (force times velocity)
- $P = mgh/\Delta t$ (for lifting at constant speed)
- $1\text{ W} = 1\text{ J/s}$
Understand
- Why the athlete and walker do the same work but different power
- Why $P = Fv$ makes sense: faster means more energy per second
- Why a car at constant speed needs power even though it doesn't accelerate
Can Do
- Calculate average power from $\Delta E/\Delta t$
- Calculate power from force and velocity
- Find time given power and energy
- Design a power measurement investigation
Core Content
Two builders carry identical bricks up a ladder to the same floor. The first takes 2 minutes; the second rushes up in 30 seconds. They both moved the same mass the same height — yet anyone watching knows the second worker worked far harder per second. That difference in rate is power: same total energy transferred, four times faster, four times the power.
The SI unit of power is the watt (W): $1\text{ W} = 1\text{ J/s}$. Common multiples: kW (10³ W), MW (10⁶ W).
Staircase answer: Both athlete and walker do the same work ($W = mgh = 70 \times 9.8 \times 5 = 3430\text{ J}$). Athlete power: $3430/4 = 857.5\text{ W}$. Walker power: $3430/40 = 85.75\text{ W}$. Athlete is 10× more powerful.
An electric motor lifts a 300 kg object 12 m in 8 s at constant speed. Find the average power. ($g = 9.8$ m/s²)
Power is the rate of energy transfer: $P = \Delta E / \Delta t$ (W); 1 W = 1 J s$^{-1}$; for lifting at constant speed: $P = mgh/\Delta t$; power describes how quickly work is done, not the total work done.
Pause — copy the highlighted definition and formula into your book before moving on.
True or false: Two people who climb the same staircase always do the same amount of work, regardless of how fast they move.
We just saw that power is the rate of energy transfer ($P = \Delta E/\Delta t$). That raises a question: when a constant force acts on a moving object, is there a more direct formula using force and speed? This card answers it → $P = Fv$ (force and velocity in the same direction); at constant speed, engine force equals drag so $P = F_{drag} \times v$.
When a constant force acts on an object moving at constant velocity, power can be calculated directly from force times velocity.
For force and velocity in the same direction ($\theta = 0°$): $P = Fv$.
This makes intuitive sense: if the same force acts but the object moves twice as fast, the object covers twice the distance in the same time, so twice the work is done — double the power.
Tesla at cruise: At 110 km/h = 30.6 m/s, drag force ≈ 589 N. $P = Fv = 589 \times 30.6 \approx 18\text{ kW}$ — matching the stated cruise power.
A car engine exerts a forward driving force of 2400 N while travelling at constant 20 m/s. Find the power output of the engine.
Instantaneous power: $P = Fv$ when force and velocity are parallel (W); general: $P = Fv\cos\theta$; at constant speed $F_{engine} = F_{drag}$, so engine power $P = F_{drag} \times v$ — doubling speed with same drag force doubles the power required.
Pause — copy the highlighted formula and constant-speed principle into your book before moving on.
A crane lifts a 500 kg beam at constant speed of 0.4 m/s. The power required is ($g = 9.8$ m/s²):
Power is the ______ of energy transfer; it is measured in ______ which equals ______ per second.
Activities
Design or conduct a stair-climb power experiment. Measure: height of staircase, your mass, and time to climb. Calculate your average power output. Compare walking vs running.
A 1500 kg car travels at constant 100 km/h. Air resistance + rolling friction totals 900 N. (a) Calculate engine power at this speed. (b) The driver doubles speed to 200 km/h. If drag force scales with $v^2$ (now 3600 N), calculate new power required. (c) What does this imply for fuel efficiency at high speed?
UnderstandBand 3(3 marks) 1. Explain the difference between work and power using the staircase scenario as an example. Your answer should include both formulas.
ApplyBand 4(3 marks) 2. A 65 kg student runs up stairs of total height 6 m in 3.5 seconds. Calculate: (a) the work done, (b) the average power output. ($g = 9.8$ m/s²)
AnalyseBand 5(4 marks) 3. A car engine produces 60 kW of power. The car travels at constant 25 m/s on a level road. (a) Calculate the driving force. (b) At this speed, what are the total resistive forces? (c) If the car slows to 15 m/s at the same power output, what driving force is available, and what happens to acceleration?
Show all answers
Short Answer — Model Answers
Q1 (3 marks): Work ($W = Fs\cos\theta$, joules) measures the total energy transferred — it depends only on force and displacement, not time. Power ($P = \Delta E/\Delta t$, watts) measures the rate of energy transfer — how quickly work is done. In the staircase scenario: both athlete and walker do identical work $W = mgh$, because they have the same mass and climb the same height. However, the athlete does this work in less time, so the athlete's power is much higher.
Q2 (3 marks): (a) $W = mgh = 65 \times 9.8 \times 6 = 3822\text{ J}$. (b) $P = W/\Delta t = 3822/3.5 = 1092\text{ W} \approx 1.09\text{ kW}$.
Q3 (4 marks): (a) $F = P/v = 60\,000/25 = 2400\text{ N}$. (b) At constant speed, net force = 0, so resistive forces = driving force = 2400 N. (c) At 15 m/s: $F = P/v = 60\,000/15 = 4000\text{ N}$. Net force = $4000 - 2400 = 1600\text{ N}$ forward. The car accelerates forward since driving force now exceeds resistive forces. This is why cars accelerate more easily at lower speeds with the same engine power.
Same work, different power. The athlete and walker do identical work on the stairs ($W = mgh = 70 \times 9.8 \times 5 = 3{,}430\text{ J}$) — but the athlete is 10× more powerful. The Tesla Supercharger V3 (2019) delivers 250,000 W — about 290 times a typical human's sustained output — which is why it can fill a 75 kWh battery in under 11 minutes using $P = \Delta E/\Delta t$.