These concepts not only form the basis for numerous mathematical problems but also have practical applications in fields like engineering and physics.

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## Gravitational Potential Energy (PE)

Gravitational potential energy is the energy an object possesses due to its position in a gravitational field, typically relative to Earth.

**Formula:**$PE = mgh$**m:**Mass in kg**g:**Gravity ($9.81 m/s²$ on Earth)**h:**Height in meters

### Example:

Calculate the gravitational potential energy (PE) of a 10 kg object that is 5 meters above the ground.

**Solution:**

**Given Data:**

- Mass $(m)$ = 10 kg
- Acceleration due to gravity $(g)$ = 9.81 m/s²
- Height $(h)$ = 5 m

**Calculating PE:**

- Formula: $PE = mgh$
- Substitution: $PE = 10 \times 9.81 \times 5$
- Calculation: $PE = 490.5 \text{ Joules}$

**Conclusion:** The gravitational potential energy of the object, when it is 5 meters above the ground, is 490.5 Joules, demonstrating the influence of height and mass on gravitational potential energy.

## Kinetic Energy (KE)

Kinetic energy represents the energy of motion - any moving object possesses kinetic energy.

**Formula:**$KE = ½mv²$**m:**Mass in kg**v:**Velocity in m/s

### Example:

Calculate the kinetic energy (KE) of a car with a mass of 1500 kg, traveling at a speed of 20 m/s.

**Solution:**

**Given Data:**

- Mass $(m)$ = 1500 kg
- Velocity $(v)$ = 20 m/s

**Calculating KE:**

- Formula: $KE = \frac{1}{2}mv^2$
- Substitution: $KE = \frac{1}{2} \times 1500 \times (20)^2$
- Calculation: $KE = \frac{1}{2} \times 1500 \times 400 = 300,000 \text{ Joules}$

**Conclusion:** The kinetic energy of the car, traveling at 20 m/s, is 300,000 Joules, illustrating the significant impact of velocity on kinetic energy.

## Energy Transformation: PE to KE

- Conservation of Energy: Energy changes form but doesn't disappear.

### Example:

Determine the kinetic energy of a 2 kg ball just before it hits the ground, if it is dropped from a height of 10 meters.

**Solution:**

**Calculating Initial Potential Energy (PE):**

- Formula: $PE = mgh$
- Where $m = 2 \, \text{kg}), (g = 9.81 \, \text{m/s}^2), and (h = 10 \, \text{m}$

- Calculation:$PE = 2 \times 9.81 \times 10 = 196.2 \, \text{Joules}$

**Kinetic Energy at the Ground:**

- Since energy is conserved, the kinetic energy (KE) of the ball just before hitting the ground equals its initial potential energy.
- KE at the ground = Initial PE = 196.2 Joules

**Conclusion:** The kinetic energy of the ball just before impact is 196.2 Joules, demonstrating energy conservation where gravitational potential energy converts into kinetic energy during free fall.

**Energy and Motion Problems**

**Example: Maximum Height of a Thrown Stone**

Calculate the maximum height reached by a 1 kg stone thrown vertically upwards with an initial velocity of 15 m/s.

**Solution:**

**Calculating Initial Kinetic Energy (KE):**

- Formula: $KE = \frac{1}{2}mv^2$
- Where $m = 1 \, \text{kg}$ and $v = 15 \, \text{m/s}$

- Calculation: $KE = \frac{1}{2} \times 1 \times (15)^2 = 112.5 \, \text{Joules}$

**Energy Transformation at Maximum Height:**

- At maximum height, all kinetic energy is converted into potential energy (PE).
- Formula for PE: $PE = mgh$
- Where $g = 9.81 \, \text{m/s}^2$ and $h$ is the height.

- Setting $PE = KE:$ $112.5 = 1 \times 9.81 \times h$

**Solving for Height ((h)):**

- Rearrange and calculate: $h = \frac{112.5}{9.81} \approx 11.47 \, \text{meters}$

**Conclusion:** The stone reaches a maximum height of approximately 11.47 meters, illustrating the conversion of kinetic energy into potential energy during its ascent.

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.