Physics
Cambridge IGCSE

TOPIC 1B: ENERGY and PRESSURE

# 1.7c Kinetic and gravitational energy

How high can you throw a ball? 3 metres? 10 metres? What makes it slow down as it goes upwards? Do you think you could throw it higher if you were on the Moon?
The reason all objects slow down when projected upwards is because of the force of gravity. Astronauts found that playing golf was really easy on the Moon due to the lower strength of gravity - the golf ball could be hit over huge distances.

When we project any object upwards - it could be a ball, it could be a firework rocket - it is gaining gravitational energy. This increase in gravitational energy has to come from some other source of energy. When a ball is thrown upwards and released it has kinetic energy, and at the ball's highest point, this has been transferred to gravitational energy. When the ball falls again, this process is reversed.

Have a look at this video to show that this change of energy stores is a fully reversible process:

Youtube 1. The reversible transfer of kinetic and gravitational energy

A simple ball throw like this is often used in exam questions to check you understand that energy is being transferred from kinetic energy to gravitational, and then back again. It is a really important principle that is often misunderstood.

In this section, we will look more closely at these two types of energy, and also find out how to calculate the energy in both gravitational and kinetic energy stores.

## Kinetic energy (K.E. or Ek)

Scientists are worried about asteroid impacts on Earth because asteroids are extremely massive, and have a very high speed compared to Earth. If they collide with Earth this huge quantity of kinetic energy will be released in the impact, and transferred to thermal energy and many other forms. The kinetic energy stored in a moving object depends on two factors - how fast the object is moving, and the mass of the object.

There is a standard maths formula for calculating the kinetic energy of any object:

kinetic energy = ½ x mass x (speed)2

K.E. = ½ m v2

[joules] = [kilograms] x [m/s]2

Learn this formula!

Notes:

• In this formula, the letter 'v' is used for the speed, instead of velocity, as the object does not need to be travelling in a straight line. See section 1.3 for more on the difference between speed and velocity.
• The speed is squared, but not the mass. A common mistake is to get this mixed up and square everything on the right-hand side of the formula.
• This formula is very similar to the formula for momentum (section 1.8) - another source of confusion. Learn them both and learn the difference between them!

Using this formula, scientists are able to show that the impact of an asteroid could be devastating - many times the energy released from a nuclear bomb. Evidence from a huge ancient crater in Mexico indicates this impact caused the mass extinction event and end of the dinosaurs, 66 million years ago.

Figure 1. The Chicxulub Crater, Mexico
Mapped using gravity anomalies - dated 66 million years ago, 150 km wide

You should be able to use this kinetic energy formula to calculate kinetic energy, and also use your algebra skills to work backwards and find the velocity or mass of an object with a known quantity of kinetic energy.

Example:
Calculate the kinetic energy stored in a 50 g tennis ball moving at 23 m/s.

The mass of the ball needs to be in kilograms, and 50 g = 0.05 kg (divide by 1000)
The formula for k.e. is K.E.= ½mv2 = 0.5 x 0.05 x 232
This gives a K.E.= 13.2 J (to 3 sig. figs.)

Questions:

1. A 2 kg cat is running across a room at 4 m/s. What is the kinetic energy of the cat?

We know that:
K.E. = ½ m v2
Substituting in the values from the question gives:
K.E. = ½ x 2 x 42
So K.E. = ½ x 2 x 16
So the cat has a K.E. = 16 J

2. A dog is chasing the cat from question 1. It has a mass of 14 kg and a kinetic energy of 63 J. How fast is the dog running?

K.E. = ½ m v2
Substituting in the values from the question gives:
63 = ½ x 14 x v2
63 = 7 x v2
rearranging the equation gives:

 v2 = 63 7

v2 = 9
giving v = 3 m/s

## Gravitational energy (G.P.E.)

Gravitational energy is often called 'gravitational potential energy' (the word 'potential' meaning 'stored'), and is therefore shortened to G.P.E. The energy stored depends on the height the object has been lifted (h), the mass of the object (m), and also the gravitational field strength (g). See section 1.3 on mass and weight for more about 'g'. On Earth, it is always taken as 9.8 N/kg as explained in 1.3.

There is a standard maths formula for calculating the G.P.E. of any object:

gravitational energy = mass x gravitational field strength x height

G.P.E. = m g h

[joules] = [kilograms] x [N/kg] x [m]

Learn this formula!

Note: The height in this formula is always the vertical height - you can ignore the distance travelled sideways.

Example:
Calculate the gravitational energy stored in a 200 g book on a shelf 1.2 m above the ground.

The mass of the book needs to be in kilograms, and 200 g = 0.2 kg (divide by 1000)
As stated above, 'g' on the Earth is always taken as 9.8 N/kg.
The formula for G.P.E. is G.P.E.= mgh = 0.2 x 9.8 x 1.2
This gives G.P.E. = 2.35 J

Have a go at the following examples to check you understand how to use this formula:

Questions:

3. A student of mass 60 kg climbs up a long flight of stairs of height 30 m. Calculate the gain in gravitational energy.

We know that:
G.P.E. = mgh
'g' = 9.8 N/kg as we can assume the student is on Earth!
Substituting in the values from the question gives:
G.P.E. = 60 x 9.8 x 30
So the gain in G.P.E. = 17 640 J (17.6 kJ to 3 sig.figs.)

4. A farmer has a small hydroelectric scheme that stores water in a large tank high on a hill. When needed, the water flows down hill and through a generator, transferring the stored gravitational energy to electrical energy. The tank is 50 m above the nearby river.
What mass of water needs to be stored at this height so that the tank stores 2 MJ of gravitational energy?

We know that h = 50 m. g = 9.8 N/kg and the total G.P.E. must be 2 MJ (2 million joules)
Substituting in the values from the question gives:
G.P.E. = mgh
2 000 000 = m x 9.8 x 50
2 000 000 = m x 490

rearranging the equation gives:

 m = 2 000 000 490

So m = 4082 kg (4.08 tonnes to 3 sig. figs.)

## Going up or down?

The diver in figure 2 below is going to dive into the water 8 m below.

Figure 2. A diver on a high board above the pool

At the top of the board, the diver has gravitational potential energy. When the diver jumps, this G.P.E. will convert to K.E. However, what is clever about this is that - using the formulas above - we can find the speed of impact of the diver.

This is how it works:

We know the G.P.E. of the diver from the formula G.P.E. = mgh.
Substituting in the values gives:
G.P.E. = 50 x 9.8 x 8
So the G.P.E. = 3920 J

We know that this will convert into kinetic energy. For this problem (and all like it that you will meet) we are going to ignore the friction caused by air resistance, and assume that this transfer is 100% efficient.
That means the diver will have 3920 J of K.E. at the bottom of the dive, just as they hit the water. If we know the kinetic energy and the mass of the diver, we can work backwards and find the speed of impact.

The formula for K.E. is:
K.E. = ½ m v2
Substituting in the values from the question gives:
3920 = ½ x 50 x v2
3920 = 25 x v2
rearranging the equation gives:

 v2 = 3920 25

v2 = 156.8
so v = √156.8
giving v = 12.5 m/s (to 3 sig. figs.)

This method is, again, reversible and we can do the same to find out how high an object goes when projected upwards, using the same method. Have a go at this question to check you understand how to do this:

Questions:

5. A 100 g tennis ball is thrown upwards at 20 m/s. If we ignore air resistance, how high will it go?

The mass of the ball is 0.1 kg (don't forget to convert 100g to kg!)
If the velocity is 20 m/s we can calculate the K.E. of the ball:
K.E. = ½ m v2, therefore
K.E. = ½ x 0.1 x (20)2
K.E. = 20 J

This is converted to G.P.E., so G.P.E. = 20 J.
G.P.E. = m g h
20 = 0.1 x 9.8 x h
so h = 20.4 m

Extension:

Can you work out a general formula for finding the velocity of impact of an object that is dropped from stationary, regardless of it's mass? if you want a challenge, have a go first before checking the solution here. This is tough, and not needed for the exam!

We know that initially, the object has stored gravitational energy, and G.P.E. = m g h.
On falling down, this converts to kinetic energy, and K.E. = ½ m v2.
So we can equate these two formulas:
½ m v2 = m g h
The mass 'm' on each side of the equation cancels leaving:
½ v2 = g h
So v2 = 2 g h,
and finally:
v = √2 g h

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