D. Energy and Society · Work, Energy and PowerPower: How Fast Energy Is Transferred
D2.5 D3.2 D2.1About 70 minutes
Learning goals- I can define power as the rate of doing work or transferring energy, P = W ÷ Δt = ΔE ÷ Δt.
- I can solve problems relating power, energy and time, in watts, kilowatts, joules and kilowatt-hours.
- I can derive and use P = Fv for an object moving at constant velocity.
- I can explain how energy, work and power are related, and their units.
Words to know
power (P)
The rate at which work is done or energy is transferred: P = W ÷ Δt = ΔE ÷ Δt.
watt (W)
The SI unit of power. 1 W = 1 J/s. Do not confuse the unit W (watt) with the symbol W (work).
kilowatt (kW), megawatt (MW)
1 kW = 10³ W; 1 MW = 10⁶ W; 1 GW = 10⁹ W.
kilowatt-hour (kWh)
A unit of energy: the energy transferred in 1 h at a power of 1 kW. 1 kWh = 3.6 × 10⁶ J.
horsepower (hp)
An older unit of power still used for engines: 1 hp ≈ 746 W.
Same stairs, different story
Two students of the same mass climb the same staircase. One strolls up; the other dashes up. They gain exactly the same gravitational potential energy, so they do the same work. But the dash is clearly more demanding. The difference is how fast the work is done. Physicists call that power.
Defining power
Power is the rate at which work is done, or the rate at which energy is transferred or transformed:
P = W ÷ Δt or P = ΔE ÷ Δt
- P in watts (W), W or ΔE in joules (J), Δt in seconds (s)
- 1 W = 1 J/s
The unit is named for James Watt, the Scottish engineer whose improved steam engines helped power the Industrial Revolution.
Rearranged: ΔE = PΔt and Δt = ΔE ÷ P.
How energy, work and power fit together
- Energy (J) is the ability to do work.
- Work (J) is energy transferred by a force acting along a displacement.
- Power (W = J/s) is how quickly that energy is transferred.
A 9 W LED and a 1500 W kettle can both transfer 1 kJ, but the kettle does it about 170 times faster.
Worked exampleExample 1: Elevator motor
An elevator and its passengers have a total mass of 800 kg. The motor lifts them 30 m at constant speed in 20 s. What power output does the motor need (ignoring friction)?
- Work = increase in Eg = mgΔh = (800)(9.8)(30) = 235 200 J.
- P = W ÷ Δt = 235 200 J ÷ 20 s = 11 760 W.
- P ≈ 1.2 × 10⁴ W = 12 kW
About 12 kW (roughly 16 hp).
Worked exampleExample 2: Energy from power and time
A snowblower's engine delivers 4.0 kW of power for 30 minutes. How much energy does it deliver, in joules and in kilowatt-hours?
- Convert time: 30 min = 1800 s = 0.50 h.
- In joules: ΔE = PΔt = (4000 W)(1800 s) = 7.2 × 10⁶ J (7.2 MJ).
- In kWh: ΔE = (4.0 kW)(0.50 h) = 2.0 kWh.
- Check: 2.0 kWh × 3.6 × 10⁶ J/kWh = 7.2 × 10⁶ J ✓
7.2 × 10⁶ J, which is 2.0 kWh.
Worked exampleExample 3: How long will it take?
A 1500 W kettle must transfer 3.0 × 10⁵ J to the water to bring it to a boil. Assuming all of the kettle's power goes into the water, how long does this take?
- Δt = ΔE ÷ P = 3.0 × 10⁵ J ÷ 1500 W
- Δt = 200 s (3 min 20 s)
200 s, a bit over 3 minutes. (In reality slightly longer, because some energy escapes to the kettle and air.)
Power and velocity: P = Fv
When a force F pushes an object at a constant velocity v along the line of the force, the work in time Δt is W = FΔd, so
P = W ÷ Δt = F(Δd ÷ Δt) = Fv
This is useful for vehicles. At highway speed, a car's engine must supply enough force to balance air resistance and rolling friction. Because air resistance grows rapidly with speed, the power needed rises steeply too, which is one reason driving faster uses more fuel per kilometre.
Worked exampleExample 4: Power at highway speed
A car travels at a constant 25 m/s (90 km/h). Air resistance and rolling friction together are 500 N. What power must the engine deliver to the wheels? Give the answer in kW and hp.
- At constant velocity, the driving force equals the resistive force: F = 500 N.
- P = Fv = (500 N)(25 m/s) = 12 500 W = 12.5 kW.
- In horsepower: 12 500 W ÷ 746 W/hp ≈ 17 hp.
12.5 kW (about 17 hp) to the wheels.
On an energy–time graph, the slope is the power. Motor A transfers 1000 J every 5 s (200 W); motor B, with half the slope, has half the power (100 W).Units to watch
- W in italics in a formula means work; W after a number means watts. Read carefully.
- kW is power; kWh is energy. Your electricity bill charges for kWh.
- 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J.
- Minutes must become seconds (× 60) for joules, or hours (÷ 60) for kWh.
Just for funJames Watt promoted the unit "horsepower" to help sell his steam engines to people who used horses for work. Today the watt is named after him, so every light bulb label is a tiny advertisement for a man who died in 1819.
Hands-on activityPower around the house
You need: notebook, pencil, a 2 L bottle of water (about 2.0 kg), a tape measure and a stopwatch. A grown-up should help you find appliance labels.
Grown-up note: read the power rating only from labels on appliances that are unplugged and cool; do not open or move heavy appliances.
- Find the power rating (in W) on at least five appliances: a kettle, a microwave, a hair dryer, a phone charger, a lamp. (It is printed on a label or the base.)
- Estimate how long each is used per day. Calculate the daily energy in kWh: E = P (kW) × t (h).
- Lift the 2 L bottle from your waist to above your head (measure the height) 10 times slowly, timing it. Repeat 10 times quickly.
- Calculate the work for 10 lifts and your power each time.
- Compare your lifting power with the appliances. Which appliance has about the same power as you?
What to look for: sample: a 1500 W kettle used 10 min a day uses 1.5 kW × (1/6) h = 0.25 kWh. A lift of 0.60 m: W = (2.0)(9.8)(0.60) ≈ 11.8 J per lift, 118 J for 10. Done in 20 s: P ≈ 5.9 W; done in 8.0 s: P ≈ 15 W. Your lifting power here is similar to a small LED bulb, though your body uses much more power than this because muscles are not very efficient.
Your turn
1A motor does 6000 J of work in 15 s. Find its power.
2A 75 W TV is on for 3.0 h. How much energy does it use, in kWh and in joules?
E = 0.075 kW × 3.0 h = 0.225 kWh. In joules: 75 W × 10 800 s = 8.1 × 10⁵ J.
3A 50 kg student climbs 4.0 m of stairs in 5.0 s. Find her power.
W = (50)(9.8)(4.0) = 1960 J. P = 1960 ÷ 5.0 = 392 W ≈ 3.9 × 10² W.
4How long does a 1200 W microwave take to transfer 9.0 × 10⁴ J?
Δt = 90 000 ÷ 1200 = 75 s.
5A boat moves at a constant 8.0 m/s while its engine pushes with 2500 N. What is the engine's useful power output?
P = Fv = (2500)(8.0) = 2.0 × 10⁴ W (20 kW).
6Convert a 150 hp snowmobile engine to kilowatts.
150 × 746 = 111 900 W ≈ 112 kW.
7Explain the difference between a kilowatt and a kilowatt-hour.
A kilowatt is a unit of power (how fast energy is used: 1000 J each second). A kilowatt-hour is a unit of energy (how much is used: 1 kW for 1 h = 3.6 × 10⁶ J).
More practice: warm-up
1Write two formulas for power.
P = W ÷ Δt and P = ΔE ÷ Δt (also P = Fv at constant velocity).
2What is 1 W in terms of joules and seconds?
3A device transfers 500 J in 10 s. Find its power.
4How many watts are in 2.5 kW?
5How many joules are in 1 kWh?
More practice: on your own
1A crane lifts a 1500 kg load 20 m in 25 s at constant speed. Find its useful power output.
W = (1500)(9.8)(20) = 294 000 J. P = 294 000 ÷ 25 = 11 760 W ≈ 12 kW.
2An Ontario home uses about 750 kWh of electrical energy in a month. What is the home's average power in kW? (Use 30 days.)
Hours: 30 × 24 = 720 h. P = 750 ÷ 720 ≈ 1.0 kW.
3A 2.0 kW electric motor runs for 45 min. How much energy does it use? If electricity costs $0.12 per kWh (a made-up simple price), what is the cost?
E = 2.0 × 0.75 = 1.5 kWh. Cost = 1.5 × $0.12 = $0.18.
4A cyclist pedals at a constant 6.0 m/s against total resistive forces of 30 N. What is her power output to the bike?
P = Fv = (30)(6.0) = 180 W.
5Two motors each lift a 100 kg crate 5.0 m. Motor X takes 10 s and motor Y takes 25 s. Compare the work done and the power.
Work: both (100)(9.8)(5.0) = 4900 J. Power: X = 490 W; Y = 196 W. X has 2.5 times the power of Y.
Challenge
1A ski lift carries 40 skiers (average mass 70 kg) up a vertical rise of 200 m every 5.0 minutes. What is the minimum power the lift motor needs to raise the skiers?
W = (40 × 70)(9.8)(200) = 2800 × 9.8 × 200 = 5 488 000 J. P = 5.488 × 10⁶ ÷ 300 s ≈ 1.8 × 10⁴ W (18 kW). The real motor must be larger, to lift the chairs and overcome friction.
2A 1200 kg car accelerates from rest to 27 m/s in 9.0 s. What is the average power delivered to the car's kinetic energy? Give it in hp.
ΔEk = ½(1200)(27)² = 437 400 J. P = 437 400 ÷ 9.0 = 48 600 W ≈ 49 kW ≈ 65 hp.
3A pump must lift 500 L of water (500 kg) each minute from a well 12 m deep. What is the minimum power of the pump? How much energy, in kWh, does it deliver in 8.0 h?
Per minute: W = (500)(9.8)(12) = 58 800 J; P = 58 800 ÷ 60 = 980 W. In 8.0 h: E = 0.98 kW × 8.0 h ≈ 7.8 kWh.
Quiz
2. A 60 W lamp is on for 2.0 minutes. The energy it uses is:
3. Which unit measures energy, not power?
4. A motor lifts a 50 kg mass 2.0 m in 4.0 s. Its useful power is about:
5. A truck moves at a constant 20 m/s with a driving force of 3000 N. Its power output is:
6. On an energy–time graph, the slope represents:
Big idea: Power is the rate of energy transfer (D2.5), and this lesson ties energy, work and power together with their units (D3.2). Grade 9 introduced P = E ÷ t and kWh for electricity; Grade 11 extends this to mechanical work and P = Fv.
How to help: Look at appliance labels together and at your electricity bill (which charges for kWh). Ask "Is this a power or an energy number?"
Common difficulties:
- Confusing kW (power) and kWh (energy).
- Forgetting to convert minutes or hours to seconds when working in joules.
- Mixing the symbol W for work with the unit W for watts.
- Using P = Fv when the object is accelerating (it gives the instantaneous power only).
Safety: read appliance labels only on unplugged, cool appliances; keep the water-bottle lifts gentle.