Any word has multiple meanings which are also quite ambiguous sometimes. However, in science, we see that there are only precise meanings for certain things. Similarly, efficiency is one such word with various meanings, but a specific one in science. We will study about what it actually is and what formula is used to measure the efficiency of anything. In other words, anything which takes input and produces output can be measured in order to see how well it is using the work put into it. It does not matter whether it is an electrical generator or even a pulley system. The efficiency formula will help us in quantifying it and judging the efficiency of any machine.

**Definition**

Efficiency is basically measuring how much work or energy we can conserve in a process. In other words, it is like comparing the output of the energy to the input of the energy in any given system. For instance, we see that in a number of procedures, we lose work or energy like waste heat or vibration.

Thus, the energy efficiency is the output of the energy which we divide by the energy input and then express it in a percentage form. Similarly, any perfect process is said to have an efficiency of 100%.

Moreover, people often confuse efficiency with effectiveness. However, they are very different from each other. In other words, efficiency is when things are being done right. Effectiveness is doing the right thing. For instance, a car is quite an effective form of transportation. Similarly, the way it uses fuel to transport determines its efficiency.

## Efficiency Formula

We can calculate the efficiency of anything by dividing the energy input and the energy output by 100%. We use this equation generally to represent the energy in the form of heat or power.

Therefore, the formula comes as:

\(\frac{energy output}{energy input}\) × 100%

\(\eta = \frac{W_{out}}{W_{in}}\) × 100%

**Derivation**

To understand the formula better, let’s have a look at what the formula stands for.

\(\eta\) refers to the efficiency (Greek letter “eta”)

\(W_{out}\) is the work or energy which produces in the process. Unit- Joules (J)

\(W_{in}\) refers to the work or energy to put in a process. Unit- Joules (J)

**Solved Examples on Efficiency Formula**

**Question-** A laborer puts in around 20 J of energy in one strike of his hammer on the nail’s head. The energy which the laborer inputs to drive the nail in the wood is 8.0 J. Calculate the efficiency of the laborer’s hammering?

**Answer-** We see over here that we have got our energy output as 8.0 J. Further, the energy input of the laborer is 20 J. Thus, we can calculate the efficiency by using the formula as below:

\(\eta = \frac{W_{out}}{W_{in}}\) × 100%

\(\eta\) = \(\frac{8.0 J}{20J}\) × 100%

\(\eta\) = 0.40 × 100%

\(\eta\) = 40%

Therefore, we see that the efficiency of the hammer strike was 40%. Vibrations and heating of the nail are two potential causes for the loss of energy.

**Question-** A certain process of the chemical has an energy efficiency of just 3.00%. In order to complete this chemical process on a large-scale, 140,000 J of energy is put in. Calculate the energy output of this process.

**Answer-** We see that we have energy input which is 140,000 J. We also know the efficiency as 3.00%. In order to calculate the energy output, we need to rearrange the formula for efficiency.

\(\eta = \frac{W_{out}}{W_{in}}\) × 100% \(\rightarrow W_{out}\) = \(W_{in}\frac{\eta }{100%}\)

\(W_{out} = W_{in}\frac{\eta }{100%}\)

\(W_{out}\) = (140,000 J) (0.03)

\(W_{out}\) = 4,200 J

Therefore, the energy output of this chemical process comes out as 4,200 J.

Typo Error>

Speed of Light, C = 299,792,458 m/s in vacuum

So U s/b C = 3 x 10^8 m/s

Not that C = 3 x 108 m/s

to imply C = 324 m/s

A bullet is faster than 324m/s

I have realy intrested to to this topic

m=f/a correct this

M=f/g

Interesting studies

It is already correct f= ma by second newton formula…