## What is Kinetic Energy?

For us to perfectly understand how to calculate kinetic energy, we will need to first define and and then derive its formula.

**Definition of Kinetic Energy:** *The kinetic energy of a body can be defined as the energy of the body due to motion.*

The symbol for kinetic energy is KE or K.E.

We measure kinetic energy in Joules (J).

Kinetic energy is a scalar quantity.

The formula for calculating kinetic is (1/2)mv^{2}

## Derivation of Kinetic Energy (K.E) Formula

Assuming an object of mass m in kilogram has an initial velocity of u in meters per second. The object attains a final velocity of v in meters per second through a distance s in meters. We can derive the kinetic energy formula by applying the **third equation of motion**, the **Second newton’s law of motion**, and the **formula for work done**.

### Step 1

Let’s get to work, start by making acceleration a subject of the formula from the equation v^{2} = u^{2} + 2as

We have an equation that says v^{2} = u^{2} + 2as

When we take the initial velocity u to the left-hand side, the above equation becomes

v^{2} – u^{2} = 2as

We can now divide both sides by 2s to make a subject of the formula [ (v^{2} – u^{2})/2s= 2as/2s ] to obtain

a = (v^{2} – u^{2})/2s

Now, remember that second Newton’s law says the force, f = ma [where m = mass and a = acceleration]

Hence, plugin a = (v^{2} – u^{2})/2s into f = ma

Which implies that

f = m x [ (v^{2} – u^{2})/2s ]

Therefore, we can break the right-hand side into

f = (1/2s) x m x (v^{2} – u^{2})

Remember that the formula for work done is w = f x s [ where w = work done, f = force, and s = distance]

### Step 2

Hence, we can substitute f in the equation w = f x s into f = (1/2s) x m x (v^{2} – u^{2}) to obtain

K.E = (1/2s) x m x (v^{2} – u^{2}) x s

and s will cancel each other to leave us with

K.E = (1/2) x m x (v^{2} – u^{2})

If the initial velocity is zero (0) we will end up with

K.E = (1/2) x m x (v^{2} – 0)

Which is

K.E = (1/2) mv^{2} or 0.5mv^{2}

## Units of Kinetic Energy (K.E)

The units of kinetic energy are as follows:

- Joules
- Kilojoules
- Calories
- We also use Foot-pound as a unit of kinetic energy
- Other units for kinetic energy are watt-hour and electron volt

## Examples: Solved Problems on How to Calculate Kinetic Energy

Here are solved problems on how to calculate kinetic energy to help improve your understanding of the topic:

### Example 1

A body of mass 5 kilograms is acted upon by a constant force of 10 newtons for 4 seconds. Calculate the kinetic energy gained by the body.

**Solution**

**Data**

mass, m = 5 kg

force, f = 10 N

Time, t = 4 s

K.E=?

The formula for calculating K.E is (1/2)mv^{2}

But we don’t have v and we need to apply the above data to find it

When we read the question again, we will see that

Initial velocity, u = 0

acceleration, a =f/m [Because f = ma]

Thus, a = 10/5 = 2 ms^{-2}

Now, plugin t= 4 s, u=0, and a=2 ms^{-2} into the equation v = u+at to obtain

v = 0 + 2 x 4 = 8 m/s

Hence,

K.E = (1/2) mv^{2} = 0.5 x 5 x 8^{2} = 160 J

**Therefore, the kinetic energy of the body is 160 Joules.**

### Example 2

An express train of mass 500 kilograms is moving with a velocity of 180 kilometers per hour. What is the kinetic energy of the train?

**Solution **

**Data**

Mass of the train, m = 500 kilograms

Velocity of the train, v = 180 km/h = (180 x1000m)/ 60 x 60s = 180,000m/3,600s = 50 m/s

And the formula for calculating kinetic energy is K.E = (1/2) mv^{2}

Hence, we substitute our data into the above formula

K.E = 0.5 x 500 x 50^{2} = 625,000 Joules = 625 kJ

**Therefore, the kinetic energy of the train is 625 kilojoules**

### Example 3

A hydrogen molecule of mass 3.3 x 10^{-27} kilograms is moving with a velocity of 1.7 x10^{3} m/s. Calculate its kinetic energy.

**Solution**

**Data**

Mass of the hydrogen, m = 3.3 x 10^{-27} kg

Velocity of hydrogen, v = 1.7 x 10^{3} m/s

We can now apply the formula for K.E to solve the problem

K.E = (1/2) mv^{2}

We can insert our data into the above formula to find the kinetic energy

K.E = 0.5 x 3.3 x 10^{-27} x (1.7 x 10^{3})^{2}

**Which implies that K.E = 4.7685 x 10 ^{-21} Joules **

### Example 4

A stone of mass 5 kilograms is thrown vertically upward with a velocity of 10m/s. Find the kinetic energy on reaching the ground.

**Solution**

**Data**

Extract your data from the question, it will make it easy for you to solve the problem.

Mass of the stone, m = 5 kg

velocity, v = 10 m/s

K.E =?

The formula for calculating kinetic energy is K.E = 0.5 mv^{2}

Thus,

K.E = 0.5 x 5 x 10^{2} = 2.5 x 100 = 250 J

**Therefore, the kinetic energy is 250 Joules **

### Example 5

A ball of mass 1000 grams is dropped from a height of 5 meters and rebounded to a height of 25 meters. Calculate the kinetic energy of the ball before impact.

**Solution**

**Data**

Mass of the ball, m = 1000 grams = 1 kg

height before impact = 5 meters

To find the velocity before impact, we need to understand that the ball is falling on the ground. Thus, we can apply the formula for potential energy

Potential energy (P.E) = mgh [ where m = mass, g = gravitational acceleration, and h = height]

We also know that K.E = P.E

Hence,

(1/2) mv^{2} = mgh

and m will cancel each other to give

(1/2) v^{2} = gh

But g is constant and is 9.8ms^{-2}

Therefore,

v^{2} = 2gh = 2 x 9.8 x 5 = 98

Now, we can apply a square root to each side

**v = √98 = 9.9 or 10 m/s which is the velocity before impact**

Let’s insert our v above into the formula for kinetic energy

**The kinetic energy is, K.E = 0.5 mv ^{2} = 0.5 x 1 x 10^{2} = 50 Joules**

### Example 6

A body of mass 20 kilograms, initially at rest is subjected to a force of 40 newtons for one second. Calculate the change in kinetic energy.

**Solution**

**Data**

The change in kinetic energy is the same as the work done on the body.

Hence

Mass = 20 kg

The force f = 40 Newtons

Time, t = 1 Second

K.E = (1/2)mv^{2}

and v = u + at [where a = acceleration and u = initial velocity]

u = 0 because the body was at rest

a = f/m = 40/20 = 2ms^{-2}

Hence

v = 0 + 2 x 1 = 2 m/s

Therefore,

K.E = 0.5 x 20 x 2^{2} = 10 x 4 = 40 J

**Therefore, the kinetic energy of the body is 40 Joules**

### Example 7

A body of mass 0.6 kg is thrown vertically upward from the ground with a speed of 20 m/s. Calculate its kinetic energy just before it hits the ground.

**Solution**

**Data**

Mass of the body, m = 0.6 kg

Velocity, v = 20 m/s

K.E =?

We can now apply the formula for kinetic energy which is K.E=(1/2)mv^{2} to solve the problem

After inserting our data into the above formula. We will get

K.E = 0.5 x 0.6 x 20^{2} = 0.3 x 400 = 120 J

**Therefore the kinetic energy before hitting the ground is 120 Joules**

### Example 8

A body of mass 5 kilograms falls from a height of 10 meters above the ground. What is the kinetic energy of the body just before it strikes the ground?

[Neglect energy losses and take g = 10 m/s^{2}]

**Solution**

**Data**

Mass of the body, m = 5 kg

height is the distance, s = 10 m

K.E = (1/2)mv^{2}

But we need to find v

Let us apply one of the equations of motion (v^{2} = u^{2} + 2as) to find v

After inserting our data into the above equation of motion, we will have

v^{2} = 0^{2} + 2 x 9.8 x 10 [ Where a=g=10m/s^{2} and u=initial velocity= 0 because the body starts from rest]

v = **√** 196 =14 m/s

Now we can apply the formula for kinetic energy

K.E = (1/2)mv^{2} = 0.5 x 5 x 14^{2} = 490 Joules

### Example 9

A bus of mass 800 kilograms is traveling at a speed of 5 m/s. Determine the kinetic energy of the bus.

**Solution**

**Data**

Mass of the bus, m = 800 kg

Speed of the bus, v = 5 m/s

K.E =?

Apply the formula below to solve the problem

K.E = (1/2)mv^{2}

Substitute the equations above with our data

K.E = 0.5 x 800 x 5^{2} = 400 x 25 = 10,000 J = 10 kJ

**Therefore, the kinetic energy of the bus is 10 kilojoules**

### Example 10

A ball of mass 0.7 kilograms is thrown at a speed of 10 m/s. Calculate the kinetic energy of the ball

**Solution**

**Data**

Mass of the ball, m = 0.7 kg

Speed of the ball, v = 10 m/s

K.E =?

Solve the problem by applying K.E=(1/2)mv^{2}

K.E = 0.5 x 0.7 x 10^{2} = 0.35 x 100 = 35 J

**Therefore, the kinetic energy of the ball is 35 Joules **

## Other Forms of Energy that Involves Kinetic Energy?

Kinetic energy can be:

- Mechanical
- Thermal
- In form of Sound
- Electrical or
- Radiation

## Applications of Kinetic Energy

- To generate electricity
- Sailor working on the boat
- Firing a bullet
- Sending a rocket from one planet to another

## Important Points on Kinetic Energy

- The energy is due to motion
- The energy is a horizontal energy
- Kinetic energy does not act against gravity
- The symbol of Kinetic energy is K.E or KE

*You may also like to read:*

How to Calculate Kinetic Energy Without Velocity

Mechanical Energy: Definition and Types

How to Calculate Work Done in Physics

**Sources:**