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IB MYP 4-5 Physics- Gravitational potential energy – Study Notes

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 

IB MYP 4-5 Physics Study Notes – All topics

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: \(\boxed{14 \, \text{m/s}}\)

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