Gravitational potential energy (PE or $U_g$) is energy an object has due to its position in a gravitational field.

Formula for Gravitational PE

Near Earth's surface (constant $g$):

$PE_g = mgh$

where:

• $PE_g$ (or $U_g$) = gravitational potential energy (J)
• $m$ = mass (kg)
• $g$ = gravitational field strength = 9.8 m/s² (on Earth)
• $h$ = height above reference point (m)

Key Concepts

1. Reference Point (Zero Level)

• You must choose where $h = 0$ (the reference point)
• Common choices: ground, floor, table top, lowest point in motion
• PE is relative to this choice
• Different reference points give different PE values, but changes in PE ($\Delta PE$) are the same!

2. Only Changes in PE Matter

For energy problems, what matters is:

$\Delta PE = mg\Delta h = mg(h_f - h_i)$

The actual value of PE depends on your reference choice, but $\Delta PE$ does not.

3. Sign of PE Change

• Increase height → $\Delta PE > 0$ (gain PE)
• Decrease height → $\Delta PE < 0$ (lose PE)

Relationship to Work

Work done by gravity:

$W_g = -\Delta PE = -mg\Delta h$

The negative sign appears because:

• If object goes up: gravity does negative work, PE increases
• If object goes down: gravity does positive work, PE decreases

Alternatively, work done against gravity:

$W_{against\,g} = \Delta PE = mg\Delta h$

This is the work you must do to lift an object.

Conservative Forces

Gravity is a conservative force because:

1. Work is path-independent - Only start and end heights matter, not the path taken
2. PE can be defined - Conservative forces have associated potential energies
3. Energy is conserved - Can convert between KE and PE

Path Independence

Whether you:

• Lift straight up
• Take a ramp
• Take a complicated winding path

The work done against gravity is the same: $W = mgh$ (for same height change).

Examples with Different Reference Points

Scenario: A 2 kg book on a table 1 m high.

Reference 1: Floor is $h = 0$

$PE = mgh = (2)(9.8)(1) = 19.6 \text{ J}$

Reference 2: Table is $h = 0$

$PE = mgh = (2)(9.8)(0) = 0 \text{ J}$

Reference 3: Floor is $h = -1$ m (table at $h = 0$)

$PE = mg(-1) = -19.6 \text{ J}$

All valid! But if the book falls to the floor:

$\Delta PE = 0 - 19.6 = -19.6 \text{ J (all three methods)}$

When Can We Use $PE = mgh$?

This formula is valid when:

• Near Earth's surface ($g \approx$ constant)
• Height change is small compared to Earth's radius

For satellites or large heights, use:

$PE = -\frac{GMm}{r}$

(This gives PE = 0 at $r = \infty$)

⚠️ Common Mistakes

Mistake 1: Forgetting to Choose Reference

❌ Wrong: "The PE is 100 J" (without stating reference) ✅ Right: "The PE is 100 J above the ground" or "relative to the floor"

Mistake 2: Thinking PE is Always Positive

PE can be negative if you're below your reference point. This is fine!

Mistake 3: Confusing PE with Work

• PE is energy an object has (due to position)
• Work is energy transferred (by a force through displacement)
• Related by: $W_g = -\Delta PE$

Mistake 4: Using Wrong Height

Use vertical height change, not distance along a ramp!

If sliding 5 m down a 30° ramp:

• Distance along ramp: $d = 5$ m
• Vertical drop: $h = 5\sin(30°) = 2.5$ m
• Use $h = 2.5$ m for PE!

Problem-Solving Strategy

1. Choose a reference point where $h = 0$
2. Identify initial and final heights above reference
3. Calculate initial PE: $PE_i = mgh_i$
4. Calculate final PE: $PE_f = mgh_f$
5. Find change: $\Delta PE = PE_f - PE_i = mg(h_f - h_i)$

Or directly: $\Delta PE = mg\Delta h$

Applications

Hydroelectric Dams

Water at height $h$ has PE. When it falls, PE converts to KE, which turns turbines.

Roller Coasters

PE at the top of hills converts to KE on descents. Highest hill must have most PE to complete the circuit.

Pendulums

Continuous conversion between PE (at extremes) and KE (at bottom).

Falling Objects

As object falls, PE decreases and KE increases by the same amount (if no air resistance).

Key Formulas Summary

(a) Find change in gravitational PE

Step 1: Calculate height change

$\Delta h = h_f - h_i = 1200 - 500 = 700 \text{ m}$

Step 2: Calculate change in PE

$\Delta PE = mg\Delta h$

$\Delta PE = (60)(9.8)(700)$

$\Delta PE = 411,600 \text{ J} = 411.6 \text{ kJ}$

(b) Work done against gravity

Work done against gravity equals the change in PE:

$W_{against\,g} = \Delta PE = 411,600 \text{ J}$

Answers:

• (a) Change in PE: 411,600 J or 411.6 kJ (positive because height increased)
• (b) Work against gravity: 411,600 J

Note: The hiker did 411.6 kJ of work to increase their PE by 411.6 kJ. If they slide back down, gravity would do +411.6 kJ of work on them.

Find: Final speed $v_f$ just before hitting ground

Strategy: Use conservation of energy (we'll learn this more in the next topic, but we can preview it here!)

Step 1: Calculate initial energy

At the top:

• $KE_i = \frac{1}{2}mv_i^2 = 0$ (at rest)
• $PE_i = mgh_i = (3)(9.8)(10) = 294$ J

Total initial energy: $E_i = 0 + 294 = 294$ J

Step 2: Calculate final energy

At the ground:

• $KE_f = \frac{1}{2}mv_f^2$ (unknown)
• $PE_f = mgh_f = (3)(9.8)(0) = 0$ J

Total final energy: $E_f = \frac{1}{2}mv_f^2 + 0$

Step 3: Apply conservation of energy

$E_i = E_f$

$294 = \frac{1}{2}(3)v_f^2$

$294 = 1.5v_f^2$

$v_f^2 = 196$

$v_f = 14 \text{ m/s}$

Alternative Method: Using work-energy theorem

Work done by gravity: $W_g = mgh = (3)(9.8)(10) = 294 \text{ J}$

This equals change in KE: $294 = \frac{1}{2}(3)v_f^2 - 0$

Same result: $v_f = 14$ m/s

Answer: The speed just before hitting the ground is 14 m/s (about 31 mph).

Check: Notice the speed doesn't depend on mass! All objects (ignoring air resistance) fall at the same rate.