Gravitational Potential Energy

What you’ll learn

Calculate a change in gravitational energy near Earth.

គណនាបម្រែបម្រួលថាមពលប៉ូតង់ស្យែលទំនាញជិតផែនដី។

Meaning and use

Near Earth’s surface, lifting a mass through a vertical height increases gravitational potential energy by approximately mgh. The height is relative to a chosen reference level. Only the vertical change enters this model, not the length of a sloping path. Use the supplied local g.

ជិតផ្ទៃផែនដី បម្រែបម្រួលថាមពលប៉ូតង់ស្យែលស្មើ mgΔh។ កម្ពស់ជាបម្រែបម្រួលបញ្ឈរ មិនមែនប្រវែងផ្លូវទ្រេត។ ត្រូវមានកម្រិតយោង។

Formulas

Change near a uniform field

\[\Delta E_p=mg\Delta h\]

Reference-level version

\[E_p=mgh\]

Symbols and units

Symbol	Meaning	Unit or restriction
Ep, ΔEp	Potential energy relative to a reference, or its change	J
m	Mass	kg
g	Gravitational field strength	N/kg
h, Δh	Vertical height above reference, or height change	m

Conditions and limits

• Assume g is approximately constant over the height change.
• Examples use g = 10 N/kg unless otherwise supplied.
• ថាមពលដែលបាត់បង់ក្នុងការលើកអាចធ្វើឱ្យថាមពលបញ្ចូលលើស mgh។

Worked examples

Raising a mass

Lift 2 kg vertically through 3 m; g = 10 N/kg.

1. Use \(\Delta E_p=mg\Delta h\).
2. Calculate \(2\times10\times3=60\) J.

60 J increase

កម្ពស់គិតបញ្ឈរ ហើយ g បានកំណត់។

Lowering a mass

A 4 kg mass drops 2 m vertically; g = 10 N/kg.

1. Choose upward positive: \(\Delta h=-2\) m.
2. Calculate \(\Delta E_p=4\times10\times(-2)=-80\) J.

−80 J, an 80 J decrease

ការថយកម្ពស់ធ្វើឱ្យថាមពលប៉ូតង់ស្យែលថយ។

Common mistakes

Instead of	Use	Why
Use the slope length in mgh.	Use the vertical height change.	The energy depends on gravitational height difference.
Potential energy has one universal zero level.	Its zero is a chosen reference in this model.	Energy changes do not depend on that reference choice.

Practice

Solve these before revealing the worked solutions. Include units where relevant.

សាកល្បងដោះស្រាយមុនបើកមើលចម្លើយ។ សូមបង្ហាញជំហាន និងឯកតាដែលពាក់ព័ន្ធ។

1. Find energy increase for 5 kg raised 2 m with g = 10 N/kg.
2. A 3 kg object rises 4 m. Use g = 10 N/kg.
3. If height doubles with mass and g fixed, what happens to potential energy above the same reference?

Answers and steps

Show solutions / បង្ហាញដំណោះស្រាយ

1. 100 J

   5 × 10 × 2 = 100.

2. 120 J increase

   3 × 10 × 4 = 120.

3. It doubles.

   Ep is proportional to h.

Quick recap

• Use vertical height.
• State g and the reference level.
• Distinguish energy gain from total input if losses occur.