Gravitational Potential Energy

Energy stored due to position in a gravitational field

๐Ÿ”๏ธ Gravitational Potential Energy

What is Potential Energy?

Potential energy is stored energy due to position or configuration.

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


Formula for Gravitational PE

Near Earth's surface (constant gg):

PEg=mghPE_g = mgh

where:

  • PEgPE_g (or UgU_g) = gravitational potential energy (J)
  • mm = mass (kg)
  • gg = gravitational field strength = 9.8 m/sยฒ (on Earth)
  • hh = height above reference point (m)

Key Concepts

1. Reference Point (Zero Level)

  • You must choose where h=0h = 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 (ฮ”PE\Delta PE) are the same!

2. Only Changes in PE Matter

For energy problems, what matters is:

ฮ”PE=mgฮ”h=mg(hfโˆ’hi)\Delta PE = mg\Delta h = mg(h_f - h_i)

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

3. Sign of PE Change

  • Increase height โ†’ ฮ”PE>0\Delta PE > 0 (gain PE)
  • Decrease height โ†’ ฮ”PE<0\Delta PE < 0 (lose PE)

Relationship to Work

Work done by gravity:

Wg=โˆ’ฮ”PE=โˆ’mgฮ”hW_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:

Wagainstโ€‰g=ฮ”PE=mgฮ”hW_{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=mghW = 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=0h = 0

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

Reference 2: Table is h=0h = 0

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

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

PE=mg(โˆ’1)=โˆ’19.6ย JPE = mg(-1) = -19.6 \text{ J}

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

ฮ”PE=0โˆ’19.6=โˆ’19.6ย Jย (allย threeย methods)\Delta PE = 0 - 19.6 = -19.6 \text{ J (all three methods)}


When Can We Use PE=mghPE = mgh?

This formula is valid when:

  • Near Earth's surface (gโ‰ˆg \approx constant)
  • Height change is small compared to Earth's radius

For satellites or large heights, use:

PE=โˆ’GMmrPE = -\frac{GMm}{r}

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


โš ๏ธ Common Mistakes

Mistake 1: Forgetting to Choose Reference

โŒ Wrong: "The PE is 100 J" (without stating reference) โœ… 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: Wg=โˆ’ฮ”PEW_g = -\Delta PE

Mistake 4: Using Wrong Height

Use vertical height change, not distance along a ramp!

If sliding 5 m down a 30ยฐ ramp:

  • Distance along ramp: d=5d = 5 m
  • Vertical drop: h=5sinโก(30ยฐ)=2.5h = 5\sin(30ยฐ) = 2.5 m
  • Use h=2.5h = 2.5 m for PE!

Problem-Solving Strategy

  1. Choose a reference point where h=0h = 0
  2. Identify initial and final heights above reference
  3. Calculate initial PE: PEi=mghiPE_i = mgh_i
  4. Calculate final PE: PEf=mghfPE_f = mgh_f
  5. Find change: ฮ”PE=PEfโˆ’PEi=mg(hfโˆ’hi)\Delta PE = PE_f - PE_i = mg(h_f - h_i)

Or directly: ฮ”PE=mgฮ”h\Delta PE = mg\Delta h


Applications

Hydroelectric Dams

Water at height hh 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

| Concept | Formula | Notes | |---------|---------|-------| | Gravitational PE | PEg=mghPE_g = mgh | hh relative to chosen reference | | Change in PE | ฮ”PE=mgฮ”h\Delta PE = mg\Delta h | Same regardless of reference choice | | Work by gravity | Wg=โˆ’ฮ”PEW_g = -\Delta PE | Negative of PE change | | Work against gravity | W=ฮ”PE=mghW = \Delta PE = mgh | Work needed to lift object |

๐Ÿ“š Practice Problems

1Problem 1easy

โ“ Question:

A 0.5 kg ball is held 2 m above the ground. What is its gravitational potential energy relative to the ground?

๐Ÿ’ก Show Solution

Given Information:

  • Mass: m=0.5m = 0.5 kg
  • Height above ground: h=2h = 2 m
  • Reference point: ground (h=0h = 0)

Find: Gravitational potential energy


Solution:

Use the gravitational PE formula:

PEg=mghPE_g = mgh

PEg=(0.5)(9.8)(2)PE_g = (0.5)(9.8)(2)

PEg=9.8ย JPE_g = 9.8 \text{ J}


Answer: The gravitational potential energy is 9.8 J relative to the ground.

Note: If we chose a different reference (say, the 2 m height as h=0h = 0), the PE would be 0 J at that point. But changes in PE would be the same!

2Problem 2medium

โ“ Question:

A 60 kg hiker climbs from an elevation of 500 m to 1200 m. (a) What is the change in gravitational potential energy? (b) How much work did the hiker do against gravity?

๐Ÿ’ก Show Solution

Given Information:

  • Mass: m=60m = 60 kg
  • Initial height: hi=500h_i = 500 m
  • Final height: hf=1200h_f = 1200 m
  • g=9.8g = 9.8 m/sยฒ

(a) Find change in gravitational PE


Step 1: Calculate height change

ฮ”h=hfโˆ’hi=1200โˆ’500=700ย m\Delta h = h_f - h_i = 1200 - 500 = 700 \text{ m}


Step 2: Calculate change in PE

ฮ”PE=mgฮ”h\Delta PE = mg\Delta h

ฮ”PE=(60)(9.8)(700)\Delta PE = (60)(9.8)(700)

ฮ”PE=411,600ย J=411.6ย kJ\Delta PE = 411,600 \text{ J} = 411.6 \text{ kJ}


(b) Work done against gravity


Work done against gravity equals the change in PE:

Wagainstโ€‰g=ฮ”PE=411,600ย JW_{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.

3Problem 3hard

โ“ Question:

A 3 kg object is released from rest at a height of 10 m above the ground. What is its speed just before it hits the ground? (Use energy methods and ignore air resistance.)

๐Ÿ’ก Show Solution

Given Information:

  • Mass: m=3m = 3 kg
  • Initial height: hi=10h_i = 10 m
  • Final height: hf=0h_f = 0 m (ground)
  • Initial velocity: vi=0v_i = 0 m/s (released from rest)
  • No air resistance

Find: Final speed vfv_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:

  • KEi=12mvi2=0KE_i = \frac{1}{2}mv_i^2 = 0 (at rest)
  • PEi=mghi=(3)(9.8)(10)=294PE_i = mgh_i = (3)(9.8)(10) = 294 J

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


Step 2: Calculate final energy

At the ground:

  • KEf=12mvf2KE_f = \frac{1}{2}mv_f^2 (unknown)
  • PEf=mghf=(3)(9.8)(0)=0PE_f = mgh_f = (3)(9.8)(0) = 0 J

Total final energy: Ef=12mvf2+0E_f = \frac{1}{2}mv_f^2 + 0


Step 3: Apply conservation of energy

Ei=EfE_i = E_f

294=12(3)vf2294 = \frac{1}{2}(3)v_f^2

294=1.5vf2294 = 1.5v_f^2

vf2=196v_f^2 = 196

vf=14ย m/sv_f = 14 \text{ m/s}


Alternative Method: Using work-energy theorem

Work done by gravity: Wg=mgh=(3)(9.8)(10)=294ย JW_g = mgh = (3)(9.8)(10) = 294 \text{ J}

This equals change in KE: 294=12(3)vf2โˆ’0294 = \frac{1}{2}(3)v_f^2 - 0

Same result: vf=14v_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.