IB MYP 4-5 Physics- Gravitational potential energy - Study Notes - New Syllabus
IB MYP 4-5 Physics-Gravitational potential energy – Study Notes
Key Concepts
- Gravitational potential energy
Gravitational Potential Energy (GPE)
Gravitational Potential Energy (GPE)
Gravitational potential energy is the stored energy an object possesses due to its position in a gravitational field. It depends on the object’s mass, the gravitational field strength, and its height above a chosen reference point.
The work done to move an object of mass \( m \) from a reference level to a height \( h \) against the gravitational field.
Formula:
\( E_p = m g h \)
where:
- \( E_p \) = gravitational potential energy (J)
- \( m \) = mass of the object (kg)
- \( g \) = gravitational field strength (\( 9.8 \, \text{m/s}^2 \) on Earth)
- \( h \) = vertical height above the reference point (m)
Key points:
- The reference level can be any convenient position, often the ground.
- GPE increases as height increases and decreases as the object moves down.
- It is a form of mechanical energy that can convert to kinetic energy during motion.
Relation to Work Done:
Work done in lifting an object = increase in its gravitational potential energy.
Energy Conservation Connection:
When an object falls freely (neglecting air resistance):
Loss of GPE = Gain in kinetic energy
\( m g h = \dfrac{1}{2} m v^2 \)
Example:
A 5 kg object is lifted to a height of 4 m. Calculate its gravitational potential energy.
▶️ Answer/Explanation
Given: \( m = 5 \, \text{kg}, \, h = 4 \, \text{m}, \, g = 9.8 \, \text{m/s}^2 \)
Using \( E_p = m g h \):
\( E_p = 5 \times 9.8 \times 4 = 196 \, \text{J} \)
Final Answer: \(\boxed{196 \, \text{J}}\)
Example:
A ball of mass 0.6 kg is dropped from a height of 10 m. Assuming no air resistance, calculate its speed just before hitting the ground using GPE–KE conversion.
▶️ Answer/Explanation
Loss of GPE = Gain in KE
\( m g h = \dfrac{1}{2} m v^2 \)
Cancelling \( m \): \( g h = \dfrac{1}{2} v^2 \)
\( 9.8 \times 10 = \dfrac{1}{2} v^2 \)
\( 98 = \dfrac{1}{2} v^2 \)
\( v^2 = 196 \)
\( v = 14 \, \text{m/s} \)
Final Answer: \(\boxed{14 \, \text{m/s}}\)