A.3.12—Power

Power is rate

Power is the rate of work or energy transfer:

$P=\frac{\Delta W}{\Delta t}=\frac{\Delta E}{\Delta t}$

Mechanical shortcut

For a constant force parallel to velocity,

$P=Fv$

Keep energy and power distinct

Energy is measured in joules; power is measured in watts, or joules per second. Multiply power by time to recover transferred energy.

Worked example from local Question Bank row 29322

A student of weight $600\,\mathrm{N}$ climbs $6.0\,\mathrm{m}$ vertically in $8.0\,\mathrm{s}$ .

$\Delta W=(600)(6.0)=3.6\times10^3\,\mathrm{J}$
$P=\frac{3.6\times10^3}{8.0}=4.5\times10^2\,\mathrm{W}=450\,\mathrm{W}$

The result is the average rate of energy transfer against gravity.

How it is assessed

The evidence asks for the average power supplied while running upstairs and for the energy delivered by a cell over a discharge time.

Command terms

Calculate

What earns marks

Use P=ΔE/Δt or P=Fv with the correct force component and speed. Convert hours to seconds when energy is in joules, and distinguish average power from instantaneous power.

Watch for

Using total energy as power or forgetting to convert the time interval into seconds.

Representative question

[Maximum number: 1]

A student of mass m initially at rest takes t seconds to run up stairs of height h. At the top of the stairs the student has a velocity v.

What is the average power supplied by the student during the climb?

$\frac{m g h}{t}$

$\frac{m\left(g h+\frac{1}{2} v^{2}\right)}{t}$

$\frac{m\left(g h-\frac{1}{2} v^{2}\right)}{t}$

m g v

A.3.12—Power

Calculate Power as a Transfer Rate

A.3.12 Exam Analysis

Assessment in practice

Question 1