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.