Potential Energy
Why this matters
Picture this: you're at Six Flags, strapped into the seat of a roller coaster. You hear the clack-clack-clack of the chain lift pulling you up the first massive hill. As you get higher, the ground gets smaller, and the tension builds. You're not moving fast, but you can feel the energy in the air. You're holding your breath, waiting for that moment at the very peak before the plunge.
That feeling of stored, ready-to-be-unleashed energy is exactly what potential energy is. It’s the energy of position and possibility. In this lesson, we'll break down exactly what potential energy is, how to calculate it for different systems, and how a simple choice you make can change all your numbers—but still lead to the right answer.
Concept overview
flowchart TD
A[System has stored energy?] -->|Yes| B{Source of energy?}
B -->|Gravity| C{Location?}
B -->|Spring/Elastic| D[U_s = (1/2)k(Δx)²]
C -->|Near Planet Surface| E[ΔU_g = mgΔy]
C -->|Astronomical Scale| F[U_g = -G(m₁m₂)/r]
Core explanation
Hello everyone! Let's dive into one of the most fundamental ideas in physics: potential energy. It’s the "stored" part of the energy story.
It's All About the System
First, let's get a big idea straight. Potential energy doesn't belong to a single object. It belongs to a system.
Think about a basketball. If you just say "the basketball has 50 Joules of potential energy," that statement is incomplete. What you really mean is "the system of the basketball and the Earth has 50 Joules of potential energy." The energy is stored in the gravitational interaction between them. If the Earth vanished, the idea of the ball "falling" would be meaningless, and so would its gravitational potential energy.
This stored energy only exists when objects interact through what we call conservative forces. For AP Physics 1, the two you need to know are gravity and the elastic force from an ideal spring. These are forces that "give back" the energy you put in. If you lift a book (doing work against gravity), gravity gives that work back as kinetic energy when the book falls.
You Decide Where Zero Is
This is the part that can feel a little strange, but it's incredibly powerful. The "zero" level for potential energy is completely arbitrary. You get to choose where you set y = 0 or U = 0.
Imagine you're measuring the height of a plant on a table in a Boston apartment. Do you measure its height from the tabletop? The floor? The sidewalk outside? Sea level?
Any of these is a valid choice for a "zero" reference point. What truly matters in physics problems is the change in potential energy (ΔU). As long as you are consistent with your zero point throughout the problem, you will always calculate the correct change.
Gravitational Potential Energy (Near Earth)
When we're talking about objects near the Earth's surface (like a person jumping, a car on a hill, or a falling apple), the gravitational field is basically constant. In this familiar territory, the change in gravitational potential energy is simple:
ΔU_g = mgΔyΔU_gis the change in gravitational potential energy, in Joules (J).mis the mass of the object (kg).gis the acceleration due to gravity (approx. 9.8 m/s²).Δyis the change in vertical height (y_final - y_initial).
If you lift an object up, Δy is positive, and its potential energy increases. If it moves down, Δy is negative, and its potential energy decreases.
Elastic (Spring) Potential Energy
What about the energy stored in a stretched or compressed spring? That's called elastic potential energy. Think of a pinball launcher or the springs in your car's suspension. The formula is:
U_s = (1/2)k(Δx)²U_sis the elastic potential energy stored in the spring (J).kis the spring constant in Newtons per meter (N/m). This tells you how stiff the spring is. A higherkmeans a stiffer spring.Δxis the distance the spring is stretched or compressed from its natural, equilibrium length (m).
Notice the (Δx)² term. This is important! It means two things:
- The energy increases with the square of the displacement. Compressing a spring by 2 cm stores four times the energy as compressing it by 1 cm.
- It doesn't matter if you stretch (
+Δx) or compress (-Δx). The energy stored is always positive because the displacement is squared.
A Quick Look at Universal Gravity
The mgΔy formula is an approximation that works great near a planet's surface. But what about a satellite orbiting Earth, or the Moon orbiting Earth? For these vast distances where g is not constant, we use the universal form of gravitational potential energy:
U_g = -G(m₁m₂)/rHere, r is the distance between the centers of the two masses (m₁ and m₂), and G is the universal gravitational constant. The AP exam wants you to recognize this formula. The key takeaway is that it's negative, which signifies a "bound" system—you'd have to add energy to pull the two objects infinitely far apart. For most of your work in this course, you'll be using mgΔy.
Total Potential Energy
Finally, what if a system has multiple sources of potential energy? For example, a mass hanging from a spring. The total potential energy of the system is simply the sum of the potential energies of each type:
U_total = U_g + U_s
You just calculate each one and add them up. Because energy is a scalar (a number without direction), you can just add them like regular numbers.
Worked examples
Let's walk through a few examples to make these ideas concrete.
Lifting a Box
Problem: Priya works at a warehouse in Dallas. She lifts a 15 kg box of inventory from the floor and places it on a shelf that is 2.0 meters high. What is the change in the gravitational potential energy of the box-Earth system?
Solution:
- 1Identify the goalWe need to find the change in gravitational potential energy,
ΔU_g. - 2Choose the formulaSince this is happening near the Earth's surface, we'll use
ΔU_g = mgΔy. - 3Define your zero pointLet's make this easy and define the floor as our
y = 0level. This is a smart choice because it's the starting position.y_initial = 0 my_final = 2.0 m
- 4Calculate the change in height,
Δy:Δy = y_final - y_initial = 2.0 m - 0 m = 2.0 m
- 5Plug in the values and solve
m = 15 kgg ≈ 9.8 m/s²Δy = 2.0 mΔU_g = (15 kg)(9.8 m/s²)(2.0 m) = 294 J
So, the potential energy of the system increased by 294 Joules.
Compressing a Spring
Problem: Carlos is playing with a toy dart gun. The spring inside has a spring constant of k = 400 N/m. He compresses the spring by 5.0 cm to load a dart. How much elastic potential energy is stored in the spring?
Solution:
- 1Identify the goalWe need to find the elastic potential energy,
U_s. - 2Choose the formulaThe correct equation is
U_s = (1/2)k(Δx)². - 3Check your units!This is a critical step. The spring constant is in N/m, so our displacement must be in meters, not centimeters.
Δx = 5.0 cm = 0.05 m- Why this step is crucial: This is the single most common place to make a mistake on spring problems. The formula will give a wildly incorrect answer if you use 5 instead of 0.05.
- 4Plug in the values and solve
k = 400 N/mΔx = 0.05 mU_s = (1/2)(400 N/m)(0.05 m)²U_s = (200 N/m)(0.0025 m²)U_s = 0.5 J
The compressed spring stores 0.5 Joules of energy, ready to launch the dart.
Try it yourself
Ready to try a couple on your own? Remember to lay out your steps clearly.
- 1Basketball JumpLiam, a basketball player with a mass of 80 kg, jumps straight up. At the peak of his jump, his center of mass is 1.2 meters higher than when he started on the ground. Relative to the floor, how much gravitational potential energy does the Liam-Earth system have at the peak of his jump?
- Hint: What's a convenient choice for your
y=0level?
- Hint: What's a convenient choice for your
- 2Archery PracticeSofia is practicing archery. She draws her bowstring back 0.70 meters, and in doing so, stores 120 J of elastic potential energy in the bow's limbs. Assuming the bow acts like an ideal spring, what is its effective spring constant,
k?- Hint: You have
U_sandΔx. Rearrange the formula to solve for the unknown,k.
- Hint: You have
Practice — 8 questions
In simple terms, potential energy is stored energy a system has because of the position or arrangement of its parts, like a book held high or a compressed spring.
ΔU_g = mgΔy- 3.3.A: Describe the potential energy of a system.
- 3.3.A.1
- A system composed of two or more objects has potential energy if the objects within that system only interact with each other through conservative forces.
- 3.3.A.2
- Potential energy is a scalar quantity associated with the position of objects within a system.
- 3.3.A.3
- The definition of zero potential energy for a given system is a decision made by the observer considering the situation to simplify or otherwise assist in analysis.
- 3.3.A.4
- The potential energy of common physical systems can be described using the physical properties of that system.
- 3.3.A.4.i
- The elastic potential energy of an ideal spring is given by the following equation, where Δx is the distance the spring has been stretched or compressed from its equilibrium length. Relevant equation: U_s = (1/2)k(Δx)²
- 3.3.A.4.ii
- The general form for the gravitational potential energy of a system consisting of two approximately spherical distributions of mass (e.g., moons, planets or stars) is given by the equation U_g = -G(m₁m₂)/r
- 3.3.A.4.iii
- Because the gravitational field near the surface of a planet is nearly constant, the change in gravitational potential energy in a system consisting of an object with mass m and a planet with gravitational field of magnitude g when the object is near the surface of the planet may be approximated by the equation ΔU_g = mgΔy.
- 3.3.A.5
- The total potential energy of a system containing more than two objects is the sum of the potential energy of each pair of objects within the system.
flowchart TD
A[System has stored energy?] -->|Yes| B{Source of energy?}
B -->|Gravity| C{Location?}
B -->|Spring/Elastic| D[U_s = (1/2)k(Δx)²]
C -->|Near Planet Surface| E[ΔU_g = mgΔy]
C -->|Astronomical Scale| F[U_g = -G(m₁m₂)/r]
Read what Saavi narrates
(Sound of a pencil writing, then fades)
Hello everyone, I'm Saavi, and welcome to Shrutam.
Have you ever been on a roller coaster? Think about that slow ride up the first big hill... that *clack-clack-clack* of the chain. You're not moving fast, but you can feel the energy building up, right? You're at the very top, holding your breath, just waiting for that drop. That feeling... that stored energy of being up so high... that's potential energy. It's the energy of possibility.
In physics, potential energy isn't something an object has by itself. It's a property of a system. It's not just the roller coaster car; it's the car *and* the Earth, interacting through gravity. Today, we're going to learn how to calculate this stored energy.
Let's take a simple, real-world example. Imagine Priya is working in a warehouse. She lifts a 15-kilogram box from the floor onto a shelf that's 2 meters high. How much did the potential energy of that box-Earth system change?
First, we need our formula for gravitational potential energy near the Earth's surface. That's... change in potential energy equals m g delta y.
Now, we need to define our zero point. This is a step students often find tricky, but it's actually your choice! Let's make it easy and say the floor is our zero height. So, the initial height is zero, and the final height is two meters.
The change in height, delta y, is just final minus initial... so, two minus zero, which is two meters.
Now we just plug in the numbers. The mass, m, is 15 kilograms. The acceleration due to gravity, g, is about 9.8 meters per second squared. And the change in height, delta y, is 2 meters.
So, 15... times 9.8... times 2... gives us 294 Joules. The potential energy of the system increased by 294 Joules.
One of the most common mistakes I see every year is students getting stuck trying to find the "correct" place to call zero. They'll ask, "But shouldn't it be sea level?" The truth is, it doesn't matter. You could have set the shelf as zero, and the floor as negative two meters. The *change* in energy would still be the same. The key is to pick a zero point that makes the problem easy for you, and then stick with it.
Potential energy is a foundational concept for understanding how energy transforms from one type to another. Keep practicing with it, and you'll see how it connects to everything else we study. You've got this.
The spring constant `k` is in Newtons per *meter*. If you plug in `Δx` in centimeters, your units don't cancel and your answer will be off by a factor of 10,000 (since `Δx` is squared).
Before plugging any numbers into the spring potential energy formula, `U_s = (1/2)k(Δx)²`, always convert your displacement `Δx` to meters.
You're calculating `(1/2)k(Δx)` instead of `(1/2)k(Δx)²`. This fundamentally misunderstands the relationship between displacement and stored energy, which is quadratic, not linear.
When you write the formula, say it out loud in your head: "one-half k delta x *squared*." Double-check your calculator entry to make sure you hit the square button.
Gravitational potential energy only depends on the change in vertical height, not the path taken. An object pushed 5 meters up a gentle ramp has the same `ΔU_g` as an object lifted straight up by the same vertical height.
Always identify the initial and final *vertical* heights (`y_initial` and `y_final`). `Δy` is always `y_final - y_initial`.
This leads to hesitation and confusion. The physics works for any chosen zero point. The *change* in energy, `ΔU`, will be the same regardless of your choice.
Confidently choose a `y=0` that simplifies your problem. Often, this is the lowest point of interest. State your choice clearly in your work: "Let `y=0` be the ground."
You're double-counting the direction. The formula `ΔU_g = mgΔy` already accounts for everything. `Δy` itself will be positive or negative based on whether the object went up or down.
Trust the formula. Use `g = +9.8 m/s²` and let the sign of `Δy = y_final - y_initial` determine whether the potential energy increases or decreases.