Grade by Grade
C. Energy and Momentum · Work, Energy and Simple Harmonic Motion

Power and Efficiency in Mechanical Systems

C2.2 C2.1About 70 minutes
Learning goals
  • I can calculate power as the rate of doing work or transferring energy.
  • I can derive and use P = Fv for objects moving at constant velocity, including on slopes.
  • I can calculate efficiency and draw or interpret a Sankey diagram.
  • I can solve multi-step problems involving work, energy, power and efficiency.

Words to know

power
The rate at which work is done or energy is transferred: P = ΔE ÷ Δt, measured in watts (W).
watt (W)
The SI unit of power; 1 W = 1 J/s.
horsepower (hp)
A non-SI unit of power still used for engines; 1 hp ≈ 746 W.
efficiency
The percentage of input energy that becomes useful output energy.
Sankey diagram
An energy-flow diagram in which the width of each arrow is proportional to the energy it represents.

Same job, different speeds

Two elevators lift the same people to the same floor. One takes 16 seconds and the other takes 32 seconds. Both do the same work against gravity, but the fast one needs a motor twice as powerful. Power measures how quickly energy is transferred, and efficiency measures how much of the input energy does the job you want. Engineers use both whenever they design anything that moves.

Remember from Grade 11: SPH3U defined power as P = ΔE ÷ Δt (in watts, W = J/s), measured your own power on the stairs, and defined efficiency = useful output energy ÷ input energy × 100%, with Sankey diagrams. SPH3U also met P = Fv; Grade 12 extends it to forces at an angle (P = Fv cos θ) and to slopes, and applies power and efficiency to vehicles, lifts and multi-step problems.

Power

P = W ÷ Δt = ΔE ÷ Δt

  • Unit: the watt (W). 1 W = 1 J/s. Large powers use kilowatts (kW) and megawatts (MW).
  • Car engines are often rated in horsepower (hp). 1 hp ≈ 746 W.
  • Energy can also be measured in kilowatt-hours: 1 kWh = (1000 W)(3600 s) = 3.6 × 10⁶ J.

Power and velocity

If a constant force F moves an object at constant velocity v along the force's direction, the work in time Δt is W = FΔd, so

P = FΔd ÷ Δt = Fv

If the force is at angle θ to the velocity, P = Fv cos θ. This shows why a car needs much more power at high speed: the air-resistance force grows with speed, and the power is that force times the speed.

Efficiency

Efficiency = useful output energy ÷ input energy × 100% = useful output power ÷ input power × 100%

No real machine is 100% efficient: some energy always becomes thermal energy (friction, electrical resistance) and sound.

time (s)energy transferred (kJ)481216501001502009.8 kW4.9 kWPower = slope= energy ÷ timeSame 157 kJ lift:16 s needs 9.8 kW;32 s needs only 4.9 kW.
Energy transferred against time for two elevator motors doing the same 157 kJ lift. Power is the slope: the solid line (16 s) is twice as steep as the dashed line (32 s).
Worked example

Example 1: An elevator motor

An 800 kg elevator car is lifted 20 m at constant speed in 16 s. (a) What useful power is delivered? (b) The motor system is 75% efficient. What input power does it draw?

  1. (a) Useful energy = ΔEg = mgΔh = (800)(9.8)(20) = 156 800 J. P = 156 800 ÷ 16 = 9.8 × 10³ W (9.8 kW).
  2. Check with P = Fv: F = mg = 7840 N, v = 20 ÷ 16 = 1.25 m/s, P = (7840)(1.25) = 9800 W. ✓
  3. (b) Input power = useful power ÷ efficiency = 9800 ÷ 0.75 = 1.3 × 10⁴ W (13 kW).

(a) 9.8 kW; (b) about 13 kW. The other 3.3 kW becomes thermal energy and sound.

electrical input 100%useful: Eg of the elevator 75%wasted: thermal energy and sound 25%
A Sankey diagram for the elevator motor in Example 1. The width of each arrow shows its share of the input energy: 75% becomes useful gravitational energy and 25% is transformed to thermal energy and sound.
Worked example

Example 2: Highway driving

A car cruises at a constant 30 m/s (108 km/h) on a level highway. Air resistance and rolling resistance total 600 N. What power must the wheels deliver?

  1. At constant velocity, the driving force equals the resistance: F = 600 N.
  2. P = Fv = (600)(30) = 1.8 × 10⁴ W (18 kW, about 24 hp).

18 kW at the wheels. The engine must produce more, because the drivetrain and engine are far from 100% efficient.

Worked example

Example 3: Cycling up a hill

A cyclist and bike (80 kg in total) ride up a 5.0° hill at a steady 4.0 m/s. Ignoring friction and air resistance, what power must the cyclist supply?

  1. Along the slope at constant speed, the cyclist's driving force balances the component of gravity: F = mg sin 5.0° = (80)(9.8)(0.08716) = 68.3 N.
  2. P = Fv = (68.3)(4.0) = 2.7 × 10² W (273 W).
  3. Another way: the cyclist rises at a vertical speed of 4.0 sin 5.0° = 0.349 m/s, and P = mg × (vertical speed) = (784)(0.349) = 273 W. ✓

About 270 W, a hard but sustainable effort for a fit cyclist.

Worked example

Example 4: A ski lift

A chairlift carries 250 skiers per hour, each about 70 kg, up a vertical rise of 300 m. (a) What average useful power does it deliver? (b) If it is 60% efficient, what input power does it need?

  1. (a) Energy per hour: (250)(70)(9.8)(300) = 5.145 × 10⁷ J. P = 5.145 × 10⁷ ÷ 3600 = 1.4 × 10⁴ W (14 kW).
  2. (b) Input = 14 292 ÷ 0.60 = 2.4 × 10⁴ W (24 kW).

(a) About 14 kW; (b) about 24 kW (the chairs themselves balance out: the empty ones going down help pull the full ones up).

Just for fun

James Watt invented the horsepower in the 1700s so customers could compare his steam engines with the horses they were replacing. A real horse can briefly produce well over one horsepower, so the unit is a little unfair to horses.

Hands-on activity

How efficient is your electric kettle?

You need: an electric kettle with its power rating on the label (for example 1500 W), a measuring cup, a kitchen thermometer, and a phone stopwatch. Safety: a grown-up should supervise; boiling water and steam cause serious burns. Do not touch the kettle body, and keep the cord away from the counter edge.

  1. Put exactly 1.00 L (1.00 kg) of cold tap water in the kettle and measure its temperature.
  2. Switch the kettle on and start timing. Switch it off as soon as it reaches about 80 °C (check with the thermometer, held by the grown-up, without touching the element) or let it switch itself off at boiling (use 100 °C as the final temperature).
  3. Useful energy: Q = mcΔT, with c = 4180 J/(kg·°C) for water (from SPH3U, Thermal and Nuclear Energy).
  4. Input energy: E = P × Δt, with P from the label in watts and Δt in seconds.
  5. Efficiency = Q ÷ E × 100%.

What to look for: kettles are usually quite efficient, often somewhere around 80 to 90%, because the element sits right in the water. Where does the rest go? (Warming the kettle itself, the air, and steam escaping.) Compare: heating the same water in a pot on a stove is usually less efficient, because much of the heat goes around the pot.

Your turn

1

A machine does 500 J of work in 4.0 s. Find its power.

2

A student lifts a 50 kg box from the floor onto a loft 3.0 m higher in 6.0 s at constant speed. Find the power.

3

A car's engine drives it at a constant 25 m/s against resistive forces of 800 N. Find the power at the wheels.

4

A motor draws 2000 W and delivers 1500 W of mechanical power. Find its efficiency.

5

Convert 150 hp to kilowatts.

6

Two students climb the same stairs. One is twice as fast. Compare the work they do and their power (same mass).

7

A 1.5 kW space heater runs for 2.0 h. How much energy does it use, in kWh and in J?

More practice: warm-up

1

Define power.

2

What is 1 W in J/s?

3

Write the formula for power in terms of force and velocity.

4

Can a real machine be 100% efficient?

5

How many joules are in 1 kWh?

More practice: on your own

1

A 60 kg hiker climbs 500 m vertically in 2.0 h. Find her average useful power.

2

A crane lifts a 2000 kg steel beam 30 m at a steady 0.50 m/s. Find the power, the time taken and the energy transferred.

3

The cyclist in Example 3 delivers 273 W to the pedals. If the human body is about 25% efficient at turning food energy into mechanical work, at what rate is she using food energy?

4

A car needs 18 kW at its wheels on the highway. If the engine and drivetrain together are 30% efficient, at what rate does it use fuel energy?

5

A pump lifts 500 L of water (500 kg) up 12 m in 2.0 min. It is 60% efficient. Find its useful output power and its input power.

Challenge

1

A 1200 kg car accelerates from rest to 27 m/s in 10 s on a level road. Ignoring friction, what is the average power delivered? Why does this underestimate the real engine power?

2

Derive P = Fv cos θ from the definition of work, for a constant force and constant velocity.

3

A 15 000 kg truck climbs a 3.0° hill at a steady 20 m/s against 3000 N of air and rolling resistance. Find the power needed at the wheels.

Check your understanding

Quiz

1.

Power is defined as:

2.

A motor lifts a 100 kg load 10 m in 20 s at constant speed. The useful power is:

3.

A truck moves at a constant 20 m/s against a total resistive force of 5000 N. The power needed is:

4.

A machine has an input of 800 J and a useful output of 600 J. Its efficiency is:

5.

In a Sankey diagram, the width of each arrow represents:

6.

Two cranes lift identical loads to the same height. Crane X takes half the time of crane Y. Crane X:

For parents and teachers

Big idea (C2.2, C2.1): power is the rate of energy transfer, P = ΔE ÷ Δt = Fv, and efficiency compares useful output with input. These ideas connect the physics of work and energy to motors, vehicles and appliances.

How to help: look at power ratings on appliance labels together and estimate energy use in kWh. Do the kettle activity together (with care around hot water) and talk about where the wasted energy goes.

Common difficulties:

  • Confusing power (rate) with energy (amount).
  • Forgetting to convert minutes or hours to seconds.
  • Dividing the wrong way in efficiency problems (input is always larger than useful output).
  • Using P = Fv when the speed is changing (then it gives the power only at that instant).