**Power** is defined as the rate of work done or energy converted with respect to time.

- $P = \frac{W}{t}$ OR $P = \frac{E}{t}$, where W = work, t = time, E = Energy
- SI Unit for power is watt (W), scalar quantity
- One watt (W) is defined as the rate of work done or energy conversion of one joule per second.
- $\text{One watt} = \frac{\text{one joule}}{\text{one second}}$ $\rightarrow$ $1 \, \text{W} = 1 \, \text{J s}^{-1}$
- Power tells us how fast work is being done or how fast energy is being converted from one form to another.
- Another useful equation for power: P = Fv, where F = force, v = velocity (Simple derivation below)

#### Worked Example 1: Power of engine

An engine does 60 000 J of work in ten minutes. What is the power of the engine?

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$$\begin{aligned} P &= \frac{W}{t} \\ &= \frac{60 000 \, \text{J}}{10 \times 60 \, \text{s}} \\ &= 100 \, \text{W} \end{aligned}$$

#### Self-Test Question 1: Fat Man & Thin Man

A fat man and a thin man (with half of the mass of the fat man) ran to the top of a hill in the same time. Whose power output is higher? Why?

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Recall that power is given by $P = \frac{W}{t}$.

Notice that the time, t, is the same for both fat man and thin man. Hence, we will have to look at the work done by both men.

The work done by the fat man will be larger than the work done by the thin man.

Hence, the power output by the fat man is higher.

### Efficiency

From the Principle of Conservation of Energy, we know that the total energy output of a machine must be equal to its energy input. However, it is found that the energy output of a machine is always less than the energy input. This phenomena can be attributed to the work done against frictional forces, which is considered as wasted energy output.

Hence, we have:

Energy input = useful energy output + wasted energy output

**Efficiency** of a system is given by

- $\text{Efficiency} = \frac{\text{Useful energy output}}{\text{Total energy input}} \times 100 \%$

#### Worked Example 2: Efficiency of Electric Motor

An electric motor is rated at 1.0 kW. If 60% of the input energy is lost as heat and sound, find the amount of useful energy produced in half hour. Hence or otherwise, find the efficiency of the electric motor.

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Let’s begin by calculating the total energy output of the electric motor:

$$\begin{aligned} P &= \frac{W}{t} \\ W &= P \times t \\ &= 1000 \, \text{W} \times 30 \times 60 \, \text{s} \\ &= 1.8 \times 10^{6} \, \text{J} \end{aligned}$$

Since 60% of the energy from the electric motor is lost as heat and sound, only 40% is useful energy output. Hence,

$$\begin{aligned} \text{Useful Energy Output} &= \frac{40}{100} \times 1.8 \times 10^{6} \\ &= 7.2 \times 10^{5} \, \text{J} \end{aligned}$$

As given in the question, the efficiency of the electric motor is 40%.

#### Worked Example 3: Electric Motor

An electric motor is used to lift a 10 N load through 5 m. The total amount of electrical energy input is 65 J. Calculate the amount of energy wasted by the motor. Hence or otherwise, calculate the efficiency of the motor.

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Since efficiency is given by: $\text{Efficiency} = \frac{\text{Useful energy output}}{\text{Total energy input}} \times 100 \%$, we have to find the useful energy output. In this case, the useful energy output is the lifting of the 10 N load through 5 m.

Hence, useful energy output:

$$\begin{aligned} W_{\text{useful}} &= F \times d \\ &= 10 \times 5 \\ &= 50 \text{ J} \end{aligned}$$

Since the total amount of electrical energy input is given to be 65 J, the energy wasted will be:

$$\begin{aligned} E_{\text{wasted}} &= 65-50 \\ &= 15 \text{ J} \end{aligned}$$

Hence, the efficiency of the electric motor will be:

$$\begin{aligned} \text{Efficiency} &= \frac{\text{Useful energy output}}{\text{Total energy input}} \times 100 \% \\ &= \frac{50}{65} \times 100 \% \\ &= 76.9 \% \end{aligned}$$

**Simple derivation of P = Fv**

In some questions, the formulation of power in terms of force and velocity will be useful in the problem-solving. The formula $P = Fv$ can be simply derived as seen below:

$$\begin{aligned} P &= \frac{W}{t} \\ &= \frac{F \times d}{t} \\ &= Fv \text{ where } v = \frac{d}{t} \end{aligned}$$

Ali scribbled

hello! I have a question

the input power to a lamp is 6.0W. The lamp wasted 2.7J of energy in 3.0s.

What is the efficiency of the lamp?

Uzair scribbled

Convert Power into Total input Energy using formulae. Subtract your answer from Wasted Energy , you will get you Useful Energy.Then use Efficiency formulae to solve your answer by dividing useful energy by total energy multiply by 100.

Solution:-

Wasted Energy=2.7 J

Total Energy input=Power x Time

=6 x 3

=18 J

Useful Energy= Total Energy – Wasted Energy

=18 – 2.7

=15.3 J

Efficiency=Useful/Total x 100

=15.3/18 x 100

=85% Answer

Joshua scribbled

In the solution to the worked example 2, the efficiency of the electric motor is stated to be 60%. However, shouldn’t it be 40% efficiency? When the useful energy (7.2 x 10^5) is divided by the total energy (1.8 x 10^6), it amounts to 0.4.

Mini Physics scribbled

Hello. Yes, it should be 40% efficiency. It was a typo. Thanks for the heads-up. 🙂

Joshua scribbled

Thank you for the quick reply and amendment. Your guides are very helpful. Keep it up! 🙂