Power and Efficiency

Branch role: This page distinguishes average from instantaneous power and develops efficiency as consistent input–output energy accounting.

Overview

This branch note treats power as the rate of energy transfer and efficiency as a matched input-output comparison over the same system and interval.

Core Ideas

• Average power uses a finite energy transfer divided by a finite time.
• Instantaneous mechanical power can be written as $P=\vec{F}\cdot \vec{v}$ for the force doing the work.
• Efficiency compares useful output with total input and is always tied to a clearly defined device or process.

Exam Relevance

Use this note for vehicle, lift, slope, motor and device-efficiency questions where the key decision is which energy or power is useful and which is input.

Power is a rate

Power measures how rapidly work is done or energy is transferred. The SI unit is the watt:

$$1\mathrm{W}=1\mathrm{J}{\mathrm{s}}^{-1}.$$

A more powerful machine can perform the same work in less time. Power alone does not state the total energy transferred.

Average power

Over a finite interval,

$$P_{\mathrm{avg}}=\frac{W}{\Delta t}=\frac{\Delta E}{\Delta t}.$$

Average power depends on the total transfer and elapsed time, not necessarily on the force or velocity at either endpoint.

Instantaneous mechanical power

For a force $\vec{F}$ whose point of application moves with instantaneous velocity $\vec{v}$,

$$P=\frac{\mathrm{d}W}{\mathrm{d}t}.$$

Since an infinitesimal work is

$$\mathrm{d}W=\vec{F}\cdot \mathrm{d}\vec{r},$$

and $\vec{v}=\mathrm{d}\vec{r}/\mathrm{d}t$,

$$P=\vec{F}\cdot \vec{v}=Fv\cos \theta .$$

Figure: Average power uses a finite energy transfer divided by a finite time. Instantaneous mechanical power uses the current force and velocity of the point of application. The parallel component $F\cos \theta$ transfers energy; a perpendicular force gives zero power even when it changes the velocity direction.

Which force belongs in $P=Fv$?

Use the force whose power is asked for:

• engine power: use engine/driving force;
• power dissipated by drag: use drag force and obtain a negative signed power or quote the positive dissipation rate;
• rate of change of kinetic energy: use resultant-force power, ${\vec{F}}_{\mathrm{res}}\cdot \vec{v}$.

These are not interchangeable.

Constant-speed motion

At constant velocity,

$${\vec{F}}_{\mathrm{res}}=0,$$

but individual forces and their powers need not be zero. On a level road, if a car moves at constant speed against resistance $R$,

$$F_{\mathrm{drive}}=R.$$

The engine’s useful mechanical power is

$$P_{\mathrm{drive}}=Rv.$$

The resistive force removes mechanical energy at the same rate, so the kinetic energy stays constant.

Lifts and slopes

For a lift of total mass $m$ rising at constant speed $v$ against a constant frictional force $f$,

$$T=mg+f,$$

so

$$P_{\mathrm{motor}}=Tv=(mg+f)v.$$

For a vehicle moving uphill at constant speed along angle $\alpha$ against resistance $R$,

$$F_{\mathrm{drive}}=R+mg\sin \alpha ,$$

and

$$P_{\mathrm{drive}}=(R+mg\sin \alpha )v.$$

If the vehicle accelerates, include $ma$ in the force equation before using $P=Fv$.

Efficiency

Efficiency is the fraction of total input converted to the intended useful output:

$$\eta =\frac{E_{\mathrm{useful}}}{E_{\mathrm{input}}}.$$

For steady operation or matched time intervals,

$$\eta =\frac{P_{\mathrm{useful}}}{P_{\mathrm{input}}}.$$

Percentage efficiency is

$$100\eta \%.$$

Figure: The total input divides into useful and dissipated outputs over the same interval. Band widths represent energy amounts, so the useful-output width divided by the input width is $\eta$. The dissipated output remains energy, commonly increasing internal energy of the device and surroundings.

For a passive energy-conversion device,

$$0\le \eta \le 1.$$

An answer above $100\%$ signals mismatched quantities, an inverted ratio or inconsistent units.

Worked examples

7.1 Average versus final instantaneous power

A car of mass $1500\mathrm{kg}$ accelerates uniformly from rest to $18.0\mathrm{m}{\mathrm{s}}^{-1}$ in $12.0\mathrm{s}$ against constant resistance $400\mathrm{N}$.

The acceleration is

$$a=\frac{18.0}{12.0}=1.50\mathrm{m}{\mathrm{s}}^{-2}.$$

Hence the constant engine force is

$$F=ma+400=(1500)(1.50)+400=2650\mathrm{N}.$$

For uniform acceleration, the average speed is $9.00\mathrm{m}{\mathrm{s}}^{-1}$, so

$$P_{\mathrm{avg}}=F{v}_{\mathrm{avg}}=(2650)(9.00)=2.39\times 10^4\mathrm{W}.$$

At $12.0\mathrm{s}$,

$$P=Fv=(2650)(18.0)=4.77\times 10^4\mathrm{W}.$$

The final instantaneous power is twice the average because the force is constant while speed rises linearly from zero.

7.2 Lift efficiency

A lift gains GPE at $24.0\mathrm{kW}$ while the motor receives $30.0\mathrm{kW}$ electrical power:

$$\eta =\frac{24.0}{30.0}=0.800=80.0\%.$$

The remaining

$$30.0-24.0=6.0\mathrm{kW}$$

is dissipated, for example through heating and sound.

7.3 Falling water

Water falls through height $h$ at mass flow rate $\dot{m}$. The rate of GPE decrease is

$$P_g=\dot{m}gh.$$

If a turbine–generator has efficiency $\eta$, its electrical output is

$$P_{\mathrm{electrical}}=\eta \dot{m}gh.$$

This illustrates why a mass flow rate, not total mass alone, is needed for continuous power.

Common errors

• Confusing energy in joules with power in watts.
• Using $P=Fs$ instead of $P=Fv$.
• Using resultant force when the question asks for engine power.
• Assuming constant speed means individual forces and powers are zero.
• Using average speed in an instantaneous-power calculation without justification.
• Dividing input by useful output when calculating efficiency.
• Mixing energies measured over different time intervals in an efficiency ratio.
• Describing dissipated energy as destroyed.

Summary

Quantity	Relationship	Condition
average power	$P_{\mathrm{avg}}=\Delta E/\Delta t$	finite interval
instantaneous power by a force	$P=\vec{F}\cdot \vec{v}$	velocity of its point of application
energy efficiency	$\eta =E_{\mathrm{useful}}/E_{\mathrm{input}}$	same device/process
power efficiency	$\eta =P_{\mathrm{useful}}/P_{\mathrm{input}}$	steady or matched interval

Links

• Work, Energy and Power
• Energy Forms and Conservation
• Dynamics
• Current Electricity Fundamentals