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## Kinetic energy

Definition 1: Kinetic energy

Kinetic energy is the energy an object has due to its motion.

Quantity: Kinetic energy $E K$ Unit name: Joule         Unit symbol: J

Kinetic energy is the energy an object has because of its motion. This means that any moving object has kinetic energy. Kinetic energy is defined as:

$E K =1 2mv 2$(1)

where $E K$ is the kinetic energy (measured in joules, J)

m = mass of the the object (measured in kg)

v = velocity of the object (measured in m·s−1).

Therefore the kinetic energy $E K$ depends on the mass and velocity of an object. The faster it moves, and the more massive it is, the more kinetic energy it has. A truck of 2000 kg, moving at 100 km·hr−1 will have more kinetic energy than a car of 500 kg, also moving at 100 km·hr−1.

## Tip:

You may sometimes see kinetic energy written as KE. This is simply another way to write kinetic energy. We will not use this form in this book, but you may see it written like this in other books.

Consider the 1 kg suitcase on the cupboard that was discussed earlier. When it is on the top of the cupboard, it will not have any kinetic energy because it is not moving:

$E K =1 2mv 2 =1 2(1kg)(0m·s -1 ) 2 =0J$(2)
.

When the suitcase falls, its velocity increases (falls faster), until it reaches the ground with a maximum velocity. As its velocity velocity increases, it will gain kinetic energy. Its kinetic energy will increase until it is a maximum when the suitcase reaches the ground. If it has a velocity of 6,26 m·s−1 when it reaches the ground, its kinetic energy will be:

$E K =1 2mv 2 =1 2(1kg)(6,26m·s -1 ) 2 =19,6J.$(3)

## Example 1: Calculation of kinetic energy

### Question

A 1 kg brick falls off a 4 m high roof. It reaches the ground with a velocity of 8,85 m·s−1. What is the kinetic energy of the brick when it starts to fall and when it reaches the ground?

#### Analyse the question to determine what information is provided

• The mass of the brick m = 1 kg

• The velocity of the brick at the bottom v = 8,85 m·s−1

These are both in the correct units so we do not have to worry about unit conversions.

#### Analyse the question to determine what is being asked

We are asked to find the kinetic energy of the brick at the top and the bottom. From the definition we know that to work out $E K$, we need to know the mass and the velocity of the object and we are given both of these values.

#### Calculate the kinetic energy at the top

Since the brick is not moving at the top, its kinetic energy is zero.

#### Substitute and calculate the kinetic energy

$E K =1 2mv 2 =1 2(1kg)(8,85m·s -1 ) 2 =39,2J$(4)

## Example 2: Kinetic energy of 2 moving objects

### Question

A herder is herding his sheep into the kraal. A mother sheep and its lamb are both running at 2,7 m·s−1 towards the kraal. The sheep has a mass of 80 kg and the lamb has a mass of 25 kg. Calculate the kinetic energy for each of the sheep and the lamb.

#### Analyse the question to determine what information is provided

• the mass of the mother sheep is 80 kg

• the mass of the lamb is 25 kg

• both the sheep and the lamb have a velocities of 2,7 m·s−1

#### Analyse the question to determine what is being asked

We need to find the kinetic energy of the sheep and the kinetic energy of its lamb

#### Use the definition to calculate the sheep's kinetic energy

$E K =1 2mv 2 =1 2(80kg)(2,7m·s -1 ) 2 =291,6J$(5)

#### Use the definition to calculate the lamb's kinetic energy

$E K =1 2mv 2 =1 2(25kg)(2,7m·s -1 ) 2 =91,13J$(6)

Note: Even though the sheep and the lamb are running at the same velocity, due to their different masses, they have different amounts of kinetic energy. The sheep has more than the lamb because it has a higher mass.

## Checking units

According to the equation for kinetic energy, the unit should be kg·m·s−2. We can prove that this unit is equal to the joule, the unit for energy.

$kgm·s -1 2 =kg·m·s -2 ·m=N·mbecause ForceN=masskg×accelerationm·s -2 =JWorkJ=ForceN×distancem$(7)

We can do the same to prove that the unit for potential energy is equal to the joule:

$kgm·s -2 m=N·m=J$(8)

### Example 3: Mixing units & energy calculations

#### Question

A bullet, having a mass of 150 g, is shot with a muzzle velocity of 960 m·s−1. Calculate its kinetic energy.

##### Analyse the question to determine what information is provided
• We are given the mass of the bullet m = 150 g. This is not the unit we want mass to be in. We need to convert to kg.

$Mass in grams÷1000=Mass in kg150g÷1000=0,150kg$(9)
• We are given the initial velocity with which the bullet leaves the barrel, called the muzzle velocity, and it is v = 960 m·s−1.

##### Analyse the question to determine what is being asked
• We are asked to find the kinetic energy.

##### Substitute and calculate

We just substitute the mass and velocity (which are known) into the equation for kinetic energy:

$E K =1 2mv 2 =1 2(0,150kg)(960m·s -1 ) 2 =69120J$(10)

### Exercise 1: Kinetic energy

Describe the relationship between an object's kinetic energy and its:

1. mass and

2. velocity

a) The kinetic energy is directly proportional to the mass. As the mass increases, so does the kinetic energy.

b) The kinetic energy is directly proportional to the square of the velocity, i.e. $KE\propto {v}^{2}$.

A stone with a mass of 100 g is thrown up into the air. It has an initial velocity of 3 m·s−1. Calculate its kinetic energy:

1. as it leaves the thrower's hand.

2. when it reaches its turning point.

We draw a sketch of the problem:

a) We convert the mass to kilograms: $\frac{100}{1000}=0,1kg$.

We are given the initial velocity and since the question asks us to find the kinetic energy at the start (i.e. when it leaves the thrower's hand) we use the initial velocity.

 $KE=\frac{1}{2}m{v}^{2}$ $KE=\frac{1}{2}\left(0,1\right){\left(3\right)}^{2}$ $KE=0,45J$

b) At the turning point the velocity of the stone is 0. So the kinetic energy is:

$KE=\frac{1}{2}\left(0,1\right){\left(0\right)}^{2}=0J$

A car with a mass of 700 kg is travelling at a constant velocity of 100 km·hr−1. Calculate the kinetic energy of the car.

We first convert the velocity to the correct units:

$\frac{100×1000}{3600}=27,78m\cdot {s}^{-1}$ . (To get m we multiply by 1 000 and to get s we divide by $60×60$ ).

Now we can work out the kinetic energy:

 $KE=\frac{1}{2}m{v}^{2}$ $KE=\frac{1}{2}\left(700\right)\left(27,78\right)$ $KE=270104,94J$