It is simply the energy that a body possesses due to its position above the surface of the earth. There is a second form of potential energy called elastic potential energy that we are going to see later.

The equation to calculate the gravitational energy is

Gravitational potential energy Ep = mgh

Where m is the mass of the object g is the acceleration due to gravity h is the height of the object above the ground

Let us have a look at example where the gravitational potential energy of an object.

Example 1

A high jumper has a mass of 62 kg. He jumps to a maximum height of 2.3 m. If the acceleration due to gravity is 9.81 ms^{-2}, calculate the gravitational potential energy of the object.

Gravitational potential energy Ep = mgh = 62*9.81*2.3 = 1398 kg = 1400 kg

You should also know how to obtain the mass or the height of an object if the gravitational potential energy is known.

Let us see an example.

Example 2

A person has a potential energy of 11000 J. If the mass of the person is 85 kg and the acceleration due to gravity is 9.81ms^{-2}, determine his height above the ground.

Gravitational potential energy Ep = mgh 11000 = 85*9.81*h h = 11000/ (85*9.81) ` = 13.2 m = 13 m

I hope you have understood. If you have any questions, leave them in the comment section below. See you later my students.

If you want to test your knowledge of gravitational potential energy, do the following questions leave the answer in the comment section?

Question

A bird of mass 0.12 kg is flying at a height of 12 m above the ground. If the acceleration due to gravity is 9.81 m s^{-2}, what would be the gravitational potential energy of the bird?

If the bird has a gravitational potential energy of 110 J, at what height would it be flying?

As we have seen in the last post there are 7 fundamental base units. From these 7 fundamental base units all the other units can be derived.

In fact if a quantity is not a basic quantity then it must certainly be a derived quantity and its unit a derived unit. The derived unit is based on one or more of the 8 fundamental base units. You must keep in mind that some quantities have units of their own based on famous scientists such as Joule (J) for energy or Newton (N) for force. For such quantities the unit can be used or the derived unit can be used.

We are now going to see several derived quantities and their derived units and how these derived units are obtained.

We are now going to see how these derived units are obtained. Once you know how to determine the derived units of common quantities like volume, force, etc then later on you will be able to determine the derived quantity for any other quantity.

So what is the method to obtain the derived unit of a quantity?

Example 1

We are going to start with a simple one: the derived unit for volume.

In order to deduce the derived unit of a quantity you need a formula to calculate the quantity. The formula must contain that you know the units in term of based unit.

Volume = length * width * height

Unit of [volume] = unit of [length * width * height]
= unit of [length] * unit of [width] * unit of [height]
= m * m *m
= m^{3}

Example 2

What is the derived unit for acceleration?

In order to deduce the derived unit for acceleration, you must know a formula to calculate acceleration such as

Acceleration = velocity / time

The formula must contain quantities that you know the units in term of base units.

If in the new formula, there is quantity that you do not know the derived unit then you will also have to know the formula to calculate that quantity.

Velocity = displacement / time

You can now rewrite the formula for acceleration to

Acceleration = velocity / time
= (displacement / time) /time

So

the unit of acceleration = unit of [velocity / time]
= unit of [(displacement / time) /time]
= unit of [(displacement / time)] / unit of time
= (m/s) /s
= m *s^{-1} * s^{-1} = m s^{-2}

Was it easy?

Now let us have a look at another example to confirm your newly learned skills.

Example 3

What is the derived unit for pressure?

Have you worked it out?

Now you would remember that in order to find the derived unit you need to know a formula for the quantity in term of quantities that you already know the units.

Pressure = Force / surface are

Hence unit of pressure = unit of [force] / unit of [surface area]
= unit of [mass * acceleration] / unit of surface area]
= (kg * m/ s^{2})/m^{2} = (kg m * s^{-2} * m ^{-2} = Kg m^{-1} s^{-2}

Do the following questions and post the answers in the comment section. I will give the correct answers later on after a few of you have submitted your answers.

Find the derived units for the following quantities.

1. Work done
2. Kinetic energy
3. Volume
4. Density

As you progress through the course you will meet more quantities that you would be able to derive the units.

Kinetic energy is one if the eight forms of energy. It is simply the energy that a body possesses due to its motion.

The kinetic energy of a body is simply calculated using the equation

Kinetic energy Ek = ½ mv^{2}

Where m is the mass of the mass of the object in kg v is the velocity of the object in ms^{-1}

As you can see the kinetic energy of the object depends on both the velocity and the mass of the object. So it is possible for an elephant with a large mass to have the same kinetic energy as a small object moving at a high velocity.

Now let us see an example where the kinetic energy of an object is calculated.

Example 1

An elephant of mass _{3.00 x 10}^{3} kg is moving at a velocity of 3.00 ms^{-1}. What is the kinetic energy of the elephant?

So let us recall that the kinetic energy of the elephant is calculated using the equation

Ek = ½ mv^{2}

Hence the kinetic energy Ek = ½ mv^{2} = ½(_{3.00*}_{10}^{3})*(3.00)^{2} = 13500 J = _{1.35 *10}^{4} J

Now you can be given the kinetic energy and be asked to calculate the mass or the velocity. Let us look at a second example. Example 2

An object has a mass of 4.0 kg and a kinetic energy of 16 J. Determine the velocity of the object.

Ek = ½ mv^{2} 16 = ½ (4.0)*v^{2}

Making v the subject of formula

v^{2} = (16 *2)/4.0 v = (16*2/4.0)^{½} v = 2.8 ms^{-1}

Now to test your newly acquired you can do the following questions

1. Calculate the kinetic energy possessed by an aeroplane of mass _{3.4 * 10}^{5} kg flying at 110 ms. 2. If a body has _{3.4 x 10}^{5} J of kinetic energy and mass _{3.2* 10}^{2} kg, what is its velocity?

Units are the third aspect that is important when measuring in Physics. It can be the mistake that will ruin the whole measurement process.

Let us recap.

You need to identify the quantity that is to be measured eg length of the book.

you need to use the correct instrument to measure the quantity. In this case the 30 cm ruler.

Lastly you need to measure the length and write it with appropriate unit.

Imagine that you write the length of a 20 cm long book as 20 inches. Then it ruins the whole measurement process.

SI units

To avoid such mistakes scientist around the world have decided to use what is called the SI system of units. It is a standard that is used by all scientists such that when a particular quantity is being measures everyone uses the same units although the Americans and the British continue to use the old systems. This has resulted in costly mistakes and loss of human lives in the space industry.

Base units

The SI system is based on fundamental base units on which all other units can be derived.

The 7 fundamental base units are given in the table below.

Basic Quantity

Base unit

Symbol of base unit

Length

Metre

m

Time

Second

s

Mass

Kilogram

kg

Thermodynamic temperature

Kelvin

K

Luminous intensity

Candela

Cd

Electric current

Ampere

A

Amount of matter

Mole

Mol

These units are the 7 fundamental base units that are used in the SI system of units. Hence even though the quantities in the above table can be measured using other units, it is recommended that only these units are used.

The next part this series on derived units can be found here.

The physics students are often faced with calculations involving logarithm either to base 10 or the natural logarithm. Just like addition, subtraction, multiplication and division, the answer to a calculation involving either lg or ln has to be given in a certain form.

Let us have a look at an example and then later on we are going to see the law.

lg 12.3 = 1.0899

Now is 1.0899 the correct answer?

Let us see how the answer should have been written and after that we will discuss the reasoning behind.

lg 12.3 = 1.0899 = 1.090

Can you guess the reasoning behind? The number being used has 3 significant figures and as a result the answer is given to 3 decimal places.

Let us have a look at a second example.

ln 1.98 = 0.98503 = 0.985

Since the number used has three significant figures then the answer is given to three decimal places.

Did you get it?

The rule when calculating with lg and ln is simply

The answer to a calculation involving lg and ln is given to a certain number of decimal places equal to the number of significant figures of the number used.

Now some questions for you.

lg 0.67

ln 6.7

lg 122

ln 6.777

Here you go my students. I will give you the answers to these questions later on.

The young physicist is faced with a lot of calculations during his studies. Sadly however very few of you are able to perform the properly and this will be a problem for you both in theoretical and practical classes. In this five part series we are going to teach our young physicists, you, how to perform calculations the way it should be and how to give the answer in the appropriate form.

I have divided this series into the following parts;

5. Other mathematical operations (under construction)

At the end of this series, you would be well equipped for the physics course so sit tight and hop in. Remember if you have any questions leave them at the end of the blog in the comment section.

Division and multiplication is often encountered in physics and it is very important that you know how to give the answer in the appropriate form.

Let us have a look at how division and multiplication is performed.

Multiplication

Let us see an example of a multiplication and see how to give the answer in the appropriate form.

23.2 * 3.1 = 71.92 = 72

Can you guess why the answer is 72?

Firstly you have to look at the number of significant figures in the two number used.

23.2 3 significant figures

3.1 2 significant figures

One of the numbers (3.1) has the smallest number of significant figures (2) as a result the answer must be given to 2 significant figures. Hence the answer is given as 72.

Let us have a look at a second example.

22.34 * 123 = 2747.82 = 2750

123 has 3 significant figures and 22.34 has 4 significant figures so the answer is given to 3 significant figures.

Have you guessed the law yet? If yes write it down. If not keep thinking. We will see it at the end of the post.

Division

In a way division is somewhat related to multiplication. Let us have a look at an example and see how the answer is given.

14.2 / 2.333 = 6.08658 = 6.09

As you can see it is similar to multiplication. The two number used has 3 and 4 significant figures respectively. And as a result the answer is given to 3 significant figures.

Let us have a look at a second example.

1.33 / 3.0 = 0.44333

Can you guess how to write the answer in the correct form?

Yes you have guessed right. It is 0.44. Since the two numbers used are at 3 and 2 significant figures respectively, then the answer is given to 2 significant figures.

So have you guessed the rule that govern the multiplication and division? Here it is!!

When a multiplication or a division is performed, the answer is given to the same number of significant figures as the number used with the smallest number of significant figures.

It is now time to do some exercises.

1.2 * 343

2.3 / 3.55

3.4 * 0.2222

3.555 * 434 / 0.32

I will give the correct answer after a few of you have submitted your answers in the comment section.