Energy stored due to position in a gravitational field
How can I study Gravitational Potential Energy effectively?โพ
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 3 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
Is this Gravitational Potential Energy study guide free?โพ
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What course covers Gravitational Potential Energy?โพ
Gravitational Potential Energy is part of the AP Physics 1 course on Study Mondo, specifically in the Energy section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Gravitational Potential Energy?
PEgโ=mgh
where:
PEgโ (or Ugโ) = gravitational potential energy (J)
m = mass (kg)
g = gravitational field strength = 9.8 m/sยฒ (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 (ฮPE) are the same!
2. Only Changes in PE Matter
For energy problems, what matters is:
ฮPE=mgฮh=mg(hfโโhiโ)
The actual value of PE depends on your reference choice, but ฮPE does not.
3. Sign of PE Change
Increase height โ ฮPE>0 (gain PE)
Decrease height โ ฮPE<0 (lose PE)
Relationship to Work
Work done by gravity:
Wgโ=โฮPE=โmgฮh
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:
Wagainstgโ=ฮPE=mgฮh
This is the work you must do to lift an object.
Conservative Forces
Gravity is a conservative force because:
Work is path-independent - Only start and end heights matter, not the path taken
PE can be defined - Conservative forces have associated potential energies
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)=19.6ย J
Reference 2: Table is h=0
PE=mgh=(2)(9.8)(0)=0ย J
Reference 3: Floor is h=โ1 m (table at h=0)
PE=mg(โ1)=โ19.6ย J
All valid! But if the book falls to the floor:
ฮPE=0โ19.6=โ19.6ย Jย (allย threeย methods)
When Can We Use PE=mgh?
This formula is valid when:
Near Earth's surface (gโ constant)
Height change is small compared to Earth's radius
For satellites or large heights, use:
PE=โrGMmโ
(This gives PE = 0 at r=โ)
โ ๏ธ Common Mistakes
Mistake 1: Forgetting to Choose Reference
โ Wrong: "The PE is 100 J" (without stating reference)
โ 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: Wgโ=โฮPE
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=5 m
Vertical drop: h=5sin(30ยฐ)=2.5 m
Use h=2.5 m for PE!
Problem-Solving Strategy
Choose a reference point where h=0
Identify initial and final heights above reference
Calculate initial PE: PEiโ=mghiโ
Calculate final PE: PEfโ=mghfโ
Find change: ฮPE=PEfโโPEiโ
Or directly: ฮPE=mgฮh
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
Concept
Formula
Notes
Gravitational PE
PEgโ=mgh
h relative to chosen reference
Change in PE
ฮPE=mgฮh
Same regardless of reference choice
Work by gravity
Wgโ=โฮPE
Negative of PE change
Work against gravity
W=ฮPE=mgh
Work needed to lift object
h=2
Reference point: ground (h=0)
Find: Gravitational potential energy
Solution:
Use the gravitational PE formula:
PEgโ=mgh
PEgโ=(0.5)(9.8)(2)
PEgโ=9.8ย 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=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?
๐ก Show Solution
Given Information:
Mass: m=60 kg
Initial height: hiโ=500 m
Final height: hfโ=1200 m
g=9.8 m/sยฒ
(a) Find change in gravitational PE
Step 1: Calculate height change
ฮh=hfโโhiโ=1200โ
Step 2: Calculate change in PE
ฮPE=mgฮh
ฮPE=(60)(9.8)(700)
ฮPE=411,600ย J=411.6ย kJ
(b) Work done against gravity
Work done against gravity equals the change in PE:
Wagainstgโ=ฮPE=411,600ย 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.)
๐ก Show Solution
Given Information:
Mass: m=3 kg
Initial height: hiโ=10 m
Final height: hfโ=0 m (ground)
Initial velocity: viโ=0 m/s (released from rest)
No air resistance
Find: Final speed vfโ 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:
KEiโ=21โmv (at rest)
Total initial energy: Eiโ=0+294=294 J
Step 2: Calculate final energy
At the ground:
KEfโ=21โmv (unknown)
Total final energy: Efโ=21โmv
Step 3: Apply conservation of energy
Eiโ=Efโ
294=21โ(3)vf2โ
294=1.5vf2โ
vf2โ=196
vfโ=14ย m/s
Alternative Method: Using work-energy theorem
Work done by gravity:
Wgโ=mgh=(3)(9.8)(10)=294ย J
This equals change in KE:
294=21โ(3)vf2โ
Same result: vfโ=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.
โพ
Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.