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# Kilogram|Definition & Meaning

## Definition

The kilogram is the unit of **mass** defined by the international system of units. It is** denoted by kg.**

The kilogram **has** **been** defined in **relation** **to** the second and the meter, both of which are based on **fundamental** physical **constants.** This enables a metrology lab with the appropriate equipment to calibrate a mass measuring device, such as a **Kibble balance,** to calculate a precise kilogram mass.

One liter of water’s mass was the initial definition of the kilogram in 1795. To within 30 parts per million, the modern definition of a kilogram matches the original. In 1795, the French spelling was chosen in **Great Britain**, while the spelling kilogram was adopted in the United States.

Both spellings are used in the **United Kingdom. **However, “kilogram” has become significantly more prevalent. The use of either spelling is not prohibited by **UK legislation**, which governs the units to be used when trading in weights or measures. Figure 1 illustrates different weights.

The French word kilo, a **contraction of the kilogram**, was introduced into the English language in the 19th century and is now used to denote both a kilogram and a kilometer.

**For** **example,** **according** **to** **The** **Economist,** the kilo is a **good** substitute, **but** the **Government** **of** **Canada’s** Premium Plus system **prohibits** **its** **use** “SI (International System of Units) in scientific and technical writing” and Russ Rowlett’s **“** **A** **dictionary** of **units** of **measurement** **called** **‘common** **informal** **names’.**

The term “kilo” could be used as an alternative to the word “**kilogram**” when the metric system was given legal standing by the **US Congress** in 1866, but that status was abolished in 1990.

**Kilogram As a Base Unit Due to Electromagnetism**

The kilograms rather than the grams were finally chosen as the foundation unit of mass in the SI, partly due to units for **electromagnetism.** The pertinent discussion and decision-making process began around the 1850s and was effectively finished in 1946.

**“Practical** units” for **electrical** and magnetic quantities, **such** **as** **amperes** and **volts,** were **established** in everyday **life** **at** the end of the 19th century. Unfortunately, they did not **coincide** with the centimeter and gram, the **primary** **basic** units for measuring length and **mass** **at** **the** **time.**

In particular, the watt, a purely **mechanical unit** of power, is produced by multiplying the ampere and volt. It was discovered that under a system where the meter and the kilogram served as the basic units of **length and mass,** respectively, **Purely** mechanical **utility** units **like** **Watts** **are** coherent.

**The** watt is a **consistent** unit of power, the base **unit** of length and time **is** **the** **integer** to the **10th** **power** **of** **the** **ratio** **of** **meters** and **kilograms** **(to** **keep** the system “metric”), and the base **unit** of length and mass **is** **the** **magnitude** **Convenient.** for practical use. **Since** no one wanted to replace the second as the base unit of time, they kept the base units of length and **mass** **as** **meters** **and** **kilograms.**

The essentially mechanical practical units, like the watt, are coherent in the **meter-kilogram-second** system, but the explicitly **electrical and magnetic units**, like the volt, the ampere, etc., are not.This **eliminates** purely electrical and magnetic units.

Only by changing the meter-kilogram-second system differently by adding a fourth basic dimension to the existing three (length, mass, and time) can those units be made consistent with the **meter-kilogram-second system** (the previous three, plus one **purely electrical one).**

**Difference Between Kilogram and Pounds**

**One** **kilogram** **of** mass weighs **approximately** 2.20 pounds **(pounds)** **on** **the** surface **of** **the** **earth.** On the other hand, an **object** **weighing** 1 **pound** **on** the **ground** has a mass of **about** 0.454 kg.

Between kilograms and pounds, there are both **qualitative and quantitative differences**. Pounds, not kilograms, are used to describe weight, which is the force a mass applies to a barrier when there is an **acceleration field** operating perpendicular to the barrier. Figure 2 illustrates the weights of 2 kilograms, 5 kilograms, and 10 kilograms.

A mass of 1 kg weighs 0.814 lb on **the surface of Mars**, where the gravitational acceleration is 37% lower than on the surface of the Earth. A mass of 1 kg has no weight when in **orbit **or while coasting in space, where there is zero net acceleration.

## Formula To Calculate Weight

There is no doubt that the mass of an item and the force of gravity on it affect its weight. This is why mass and weight are different. Whether an object was on the Moon or the Earth, its mass would be the same. On Earth compared to the Moon, an object would weigh differently as a result of gravity. So the formula will be

Weight = mass $\times$ gravity

## Some Examples of Kilogram

### Example 1

Find the weight of the person if the mass of the person is 60 kilograms and the gravitational force is 9.8 meters per square second.

### Solution

As we know the formula for weight is equal to mass and gravitational force so from the statement mass of the person is equal to 60 kg and the gravitational force is equal to 9.8 meters per square second.

Weight = 60 $\times$ 9.8

Weight = 588 N

### Example 2

The weight of a person on the moon is equal to 637 Newton, find the mass of the person if the gravitational force acting on the moon is 1.6 meters per square second.

### Solution

According to the formula

Weight = mass $\times$ gravity

As the gravitational force acting on the moon is 1.6 meters per square second and the weight of the person on the moon is 637 N so

Mass=weight/gravity

Mass = 637/1.6

Mass = 398 kg

Hence the mass of the person is equal to 398 kilograms.

### Example 3

Find the mass of the box illustrated in figure 3

### Solution

Ass illustrated in the figure force applied on the object is 100 N and acceleration is equal to 0.05 m/s

Force = mass $\times$ acceleration

Mass = force / acceleration

Mass = 100/0.05

Mass = 2 kg

*All images/ mathematical drawings were created using GeoGebra.*