Potential Energy
Author: Matthew Lewine (mlewine3)
Afua Kessie (akessie) Fall 2021 Matthew Jacobs (mjacobs44) Fall 2022
Potential energy (referred as [math]\displaystyle{ U }[/math]) is the stored energy of position possessed by an object and is that some body possesses due to their position relative to other bodies, configuration or stresses within itself, electric charges, and other factors. These factors can include a variety of many things, the main one being the pull of the earth. However, other common sources of potential energy include springs and other planets. Potential Energy can be pictured as a stored energy which has the potential to do some work but hasn't done it yet. Changing the parameters that give a system a potential energy (e.g. distance between objects), [math]\displaystyle{ U }[/math] changes as well. The change in [math]\displaystyle{ U }[/math] can result in a change in [math]\displaystyle{ K }[/math] (Kinetic Energy) For a lot of beginner-level physics the total energy in a system is generally created by the addition of [math]\displaystyle{ U }[/math] and [math]\displaystyle{ K }[/math].
Examples of Potential Energy
Example 1: A rock sitting at the edge of a cliff. At the moment it will have potential energy since it is being stored on the height H from the surface of the Earth. If the rock falls, the potential energy will be converted to kinetic energy because now the object is in motion.
Example 2: Even the simplest example would be even holding a yo-yo in ones hand, for example, the yo-yo has the potential to do some work once you release it from the string it's tied on.
Example 3: When throwing a ball up into the air the speed is constantly changing. However, the total energy in the ball-earth system is still the same. This is because as the kinetic energy increases/decreases the potential energy does the opposite. The ball has the most speed (and thus the most kinetic energy) at the points where the ball is nearest to the ground. However, at the apex of the flight the speed of the ball is 0. This is where the potential energy is at its maximum because the kinetic energy of the ball is 0 (because of the speed being equal to 0.)
When referring to [math]\displaystyle{ U }[/math] being at a minimum or maximum, it is important to note that these can be in absolutes. If a system has all potential energy, then all of its energy is stored and there is absolutely no kinetic energy. An example to help think of this is a stationary ball on top of a cliff. It is not moving whatsoever, but if it were to be pushed off of the cliff, the potential energy would begin to be transformed into kinetic energy as it falls. There is also a moment, just before the ball hits the ground and while its velocity is at its theoretical maximum, that virtually zero potential energy is present. All of the energy would be kinetic energy in this case.
Energy is conserved, so the total amount of energy in a system never changes, but it can change forms, such as from potential to kinetic energy, or the other way around. In non-ideal systems, so most realistic earthly systems, energy can be transformed into other forms such as thermal energy. An example of this would be a toy racecar rolling along a flat surface. It eventually slows down and stops due to friction, which results in a decrease in kinetic energy since that energy is transformed into thermal energy on the tires.
In terms of potential energy, its capacity for doing work is a result of its position in a gravitational field (gravitational potential energy), an electric field (electric potential energy), or a magnetic field (magnetic potential energy). It may have elastic potential energy due to a stretched spring or other elastic deformation.
It is interesting to note, that the universe naturally prefers lower potential configuration of systems. This peculiarity is partly why balancing a pencil on one finger is so hard; the pencil contains potential energy that can be released easily. As taught in many engineering classes, this can be explained by one of the thermodynamic laws, where the universe always increase randomness, or entropy.
The Main Idea
When we want to discuss Energy, the first step is always defining your system. In fact, the Energy Principle states: [math]\displaystyle{ \vartriangle \!\, Esys = Wsurr }[/math]. Up to this point, energy was always studied in mono-particular systems. However, most situations in life are more complex and involve multi-particular systems.
An example would be any object that is displaced, but still has stuff interacting and working in it. We can imagine the earth, on which work is done by the Sun and other planets. Above the surface of the earth, there is a ball which also interacts with it. Surrounding planets also do work on the ball. So how can we analyze the behavior of both the ball and the Earth when they interact with each other and the Sun?
This analysis is infinite and can be very difficult to handle. That's why sometimes you need to consider multi-particular systems for your analysis. We might want to fix earth+ball as a point to analyze the work done by the Sun on this system. But the problem is, we'll be neglecting the interaction happening within the system.
If the ball is the system, the Earth represents the surroundings (image 1). The Kinetic energy of the system (ball only) increases gradually due to the work done by the Earth. Imagine the ball is 1 kg and is released from rest 10 m above the surface. When the ball has fallen to 5 m, what is [math]\displaystyle{ K }[/math]?
[math]\displaystyle{ \vartriangle \!\, Eball = WEarth }[/math]
[math]\displaystyle{ \vartriangle \!\, Kball = WEarth }[/math]
[math]\displaystyle{ \vartriangle \!\, Kball = Fy\vartriangle \!\,y = -mg\vartriangle \!\,y }[/math]
[math]\displaystyle{ Kf = -(1 kg)(9.8 N/Kg)(-5 m) = 5 J }[/math]
What if we decide the choose the ball and the Earth as the system?
In this case, we consider there is nothing significant in the surroundings. As a result
[math]\displaystyle{ \vartriangle \!\, Eball = 0 }[/math]
[math]\displaystyle{ \vartriangle \!\, Kball + \vartriangle \!\, Kearth = 0 }[/math]
However, we know that the kinetic energy of the ball increased, and the kinetic energy of the Earth also increased. We also know the surroundings did zero work. But the change in kinetic energy of the system is bigger than zero and so different from the work done by the surroundings. This is happening since we are overlooking a kind of energy in the system related to the interaction between both bodies: it is potential energy.
In any systems containing two or more interacting bodies (stars interacting in a galaxy), there is energy associated with the interactions between pairs of particles inside the system.
So, since the ball+earth system contains interacting objects, its energy is:
[math]\displaystyle{ Esys = EEarth + Eball + Uball/earth }[/math]
Since the interaction force is gravity (dependent of distance), a change in potential energy should be associated with a change of the separation between objects. This can also mean a change of shape of the multi-particle system, such as a spring stretching or compressing.
Getting back to our example above, as the ball and Earth get closer, the kinetic energy of the system increases but it is compensated by a decrease in potential energy (interaction energy). In fact, since there are no surroundings:
[math]\displaystyle{ \vartriangle \!\, Eball = 0 }[/math]
[math]\displaystyle{ \vartriangle \!\, Kball + \vartriangle \!\, Kearth + \vartriangle \!\, Uball/earth = 0 }[/math]
A Mathematical Model
Taking ball+earth as the system makes the force exerted on the ball interior to the system: it is an internal force which makes internal work and so changes [math]\displaystyle{ U }[/math].
[math]\displaystyle{ \vartriangle \!\, Kball = Wearth }[/math]
[math]\displaystyle{ \vartriangle \!\, Kball + (- Wearth) = 0 }[/math]
Definition: for a multi-particle system we define the change of potential energy [math]\displaystyle{ \vartriangle \!\, U }[/math] to be the negative of the internal work:
[math]\displaystyle{ \vartriangle \!\, U = -Wearth }[/math]
For the system of the ball + Earth:
[math]\displaystyle{ \vartriangle \!\, U = -(Fy \vartriangle \!\, y) = -(-mg)\vartriangle \!\, y = \vartriangle \!\,(mgy) }[/math]
To conclude, the difference between a single-particle and a multi-particle one is that the second has pairwise potential energy on top of the particular energy.
Revised energy principle for a multiparticle system
[math]\displaystyle{ \vartriangle \!\, (E1 + E2 + E3 + ...) + \vartriangle \!\, (U12 + U13 + U23 + ...) = W }[/math]
with [math]\displaystyle{ E1 = E1rest + K1 }[/math]
An algebraic process can help us derive equations for all kinds of Potential energies with respect to the parameters that affect it.
A force is considered conservative if it is acting on an object as a function of position only.
We can relate work to potential energy using the equation
[math]\displaystyle{ U = -\int \vec{F}\cdot\vec{dr} = -W }[/math]
This says that the potential energy U is equal to thework you must do to move an object from an arbitrary reference point [math]\displaystyle{ U=0 }[/math] to the position [math]\displaystyle{ r }[/math].
The potential energy of a rock on the top of a cliff is equal to the work you've done to bring the rock up to this point.
If we take the derivative of both sides of this equation and obtain:
[math]\displaystyle{ \frac{-dU}{dx} = F(x) }[/math]
Which means that the force on an object is the negative of the derivative of the potential energy function U. Therefore, the force on an object is the negative of the slope of the potential energy curve. Plots of potential functions are valuable aids to visualizing the change of the force in a given region of space.
Let's apply this relationship. If the potential energy function U is known, the force at any point can be obtained by taking the derivative of the potential. Let's consider gravitational potential and elastic potential.
The potential energy function U of gravitational potential is [math]\displaystyle{ mgh }[/math], where [math]\displaystyle{ m }[/math] is mass, [math]\displaystyle{ g }[/math] is the gravitational constant, and [math]\displaystyle{ h }[/math] is some distance away from the reference point at which U = 0. Then the force is
[math]\displaystyle{ F = \frac{-d}{dh}mgh = -mg }[/math]
We can go the other way as well. We know the force of gravity is [math]\displaystyle{ -mg }[/math], and integrating with respect to h we obtain [math]\displaystyle{ U = mgh }[/math].
This process can be done with elastic potential as well, where the force [math]\displaystyle{ F = -kx }[/math] and the potential energy function is [math]\displaystyle{ U = \frac{1}{2}k^{2} }[/math]
Here are the potential energy functions for all forms:
| Type | Equation | Variables |
| Gravitational Potential | [math]\displaystyle{ U = \frac{GMm}{r} }[/math] [math]\displaystyle{ U = mgh }[/math] close to Earth's surface |
[math]\displaystyle{ G }[/math] is the gravitational constant, [math]\displaystyle{ M }[/math] and [math]\displaystyle{ m }[/math], and [math]\displaystyle{ r }[/math] is distance |
| Elastic Potential | [math]\displaystyle{ U = \frac{1}{2}kx^{2} }[/math] | [math]\displaystyle{ k }[/math] is the spring constant, [math]\displaystyle{ x }[/math] is how much the spring is stretched. |
| Electric Potential | [math]\displaystyle{ U = k\frac{Qq}{r} }[/math] | [math]\displaystyle{ k }[/math] is Coulomb's constant, [math]\displaystyle{ Q }[/math] and [math]\displaystyle{ q }[/math] are point charges, [math]\displaystyle{ r }[/math] is distance |
| Magnetic Potential | [math]\displaystyle{ U = -μ \cdot B }[/math] | [math]\displaystyle{ μ }[/math] is the dipole moment and [math]\displaystyle{ μ = IA }[/math] in a current loop and [math]\displaystyle{ I }[/math] is the current and [math]\displaystyle{ A }[/math] is the area |
A Computational Model
A spring is like a rubber band, as you stretch it, it stores potential energy. Once released, this stored energy is converted to kinetic energy. The simulation at this link shows it https://trinket.io/embed/glowscript/fa7fbeaf3b[1]
It is interesting to see how potential(yellow) and kinetic(cyan) energy alter. When U is at a maximum, K is at a minimum. When K is at a maximum, U is at a minimum. This suggests an energy transfer: U is transformed into kinetic energy which causes the ball to move. Also total Energy is constant because there are no surroundings.
"See "Demo 3" at: https://drive.google.com/drive/folders/0B5m-O1TAXl8GWkxYNGhwQm1xMkU?usp=sharing"
Examples
This is a link to a video displaying an exercise resolved step by step by Dr. Greco [2]
Be sure to show all steps in your solution and include diagrams whenever possible
Simple
Question
An object of mass 3 kg is pressed by a spring with stiffness = 8 and is compressed a length of .02 meters. How much potential energy does this object have?
Solution
Using the equation [math]\displaystyle{ U = \frac{1}{2}kx^{2} }[/math] we can say [math]\displaystyle{ U = 1/2*8*(.02)^2 = .0016 }[/math] J.
Question
An object of mass 5 kg is held 10 meters above the Earth's surface. Relative to the surface, how much potential energy does this object have?
Solution
Using the equation [math]\displaystyle{ U = mgh }[/math] we can say [math]\displaystyle{ U = 5*9.8*10 = 490 }[/math] J.
Middling
Question
If it takes 4J of work to stretch a Hooke's law spring 10 cm from its unstretched length, determine the extra work required to stretch it an additional 10 cm.
Solution
The work done in stretching or compressing a spring is proportional to the square of the displacement. If we double the displacement, we do 4 times as much work. It takes 16 J to stretch the spring 20 cm from its unstretched length, so it takes 12 J to stretch it from 10 cm to 20 cm.
Formally:
[math]\displaystyle{ W = \frac{1}{2}kx^{2}. }[/math] Given W and x we find k.
[math]\displaystyle{ 4 J = \frac{1}{2}k(0.1)^{2} }[/math]
[math]\displaystyle{ k =\frac{8}{0.1^{2}} = 800 }[/math] N/m.
Using [math]\displaystyle{ x = 0.2 }[/math] m, [math]\displaystyle{ W = \frac{1}{2}(800)(0.2)^{2} = 16 }[/math] J
Extra work: 16 J - 4 J = 12 J.
Difficult
Question
We have a point charge A of charge +Q at the origin. Let's say we want to move another charge B of +q, located 10m away from particle A, to a location 5m away from particle B. How much work does it require to move the particle B 5m closer to particle A?
Solution
We have a nonuniform electric field, so we need to integrate the potential energy function to find the amount of work needed. [math]\displaystyle{ W = \int_{10}^{5} \frac{-kQq}{r^2}dr= -kQq\int_{10}^{5}\frac{1}{r^2}dr = \frac{kQq}{10} }[/math]
Connectedness
Potential energy is the driving force behind voltage, or the electric potential difference, expressed in volts. In fact, the circuit has to have a potential in order to light stuff and do some electrical work. The energy use for this work is subtracted from this potential which explains the difference of potential known as V. As a computer engineering major, I recognize the importance of this concept, as without potential difference we would not have transistors or circuits to power our machines.
Nuclear potential energy also exists, and is the potential energy of the particles inside an atomic nucleus. Most nuclear elements like Pd and Ur have unstable nucleus due to a disproportion of nuclear particles like protons and neutrons. Unstable nucleus means a nucleus which is susceptible to change to another configuration: they have a potential. Elements used in nuclear reactions thus have kernels with a high potential energy. As we previously discussed, nature always try to go to the most stable state of energy, the state where it has the least potential energy. That's why most radioactive elements naturally decay and change composition to be in a more stable stage. The difference of energy between previous and next stage is the energy liberated in nuclear reactions as rays.
History
The term "potential energy" was coined by William Rankine, a nineteenth-century Scottish engineer and physicist. It corresponds to energy that is stored within a system. It exists when there is a force that tends to pull an object back towards some original position when the object is displaced.
William Rankine contributed to Civil engineering and was a founding contributor, with Rudolf Clausius and William Thomson (Lord Kelvin), to the science of thermodynamics.
He unfortunately passed away on December 24 1872 in Glasglow, Scotland at the age 52. Cause of death remains unknown as of now
Gravitational Potential Energy
Gravitational energy or gravitational potential energy is the potential energy a massive object has in relation to another massive object due to gravity. It is the potential energy associated with the gravitational field, which is released when the objects fall towards each other.
For two pairwise interacting point particles, the gravitational potential energy [math]\displaystyle{ U }[/math] is given by
- [math]\displaystyle{ U = -\frac{GMm}{R}, }[/math]
Close to the Earth's surface, the gravitational field is approximately constant, and the gravitational potential energy of an object reduces to
- [math]\displaystyle{ U = mgh }[/math]
where [math]\displaystyle{ m }[/math] is the object's mass, [math]\displaystyle{ g = GM_{\oplus}/R_{\oplus}^2 }[/math] is the gravity of Earth, and [math]\displaystyle{ h }[/math] is the height of the object's center of mass above a chosen reference level.[1]
Examples of Gravitational Energy
- A raised weight
- Water that is behind a dam
- A car that is parked at the top of a hill
- A yoyo before it is released
- River water at the top of a waterfall
- A book on a table before it falls
- A child at the top of a slide
- Ripe fruit before it falls
m represents the mass of the object, h represents the height of the object and g represents the gravity of the earth surface which is (9.8 N/kg on Earth).
Elastic Potential Energy
Elastic energy, or elastic potential energy, is the energy that an object has when being compressed by a spring. Found by the equation : [math]\displaystyle{ U = \frac{1}{2}kx^{2} }[/math] Elastic potential energy deals with springs and conservation of energy. When a mass in put next to a compressed spring, the spring has the potential to extend and thus push the mass. The value of how much energy the spring can push with is seen with the elastic potential energy.
See also
Kinetic Energy
Potential Energy for a Magnetic Dipole
Potential Energy of a Multiparticle System
Work
External links
References
“Potential Energy.” The Physics Classroom, https://www.physicsclassroom.com/class/energy/Lesson-1/Potential-Energy.
“What Is Gravitational Potential Energy? (Article).” Khan Academy, Khan Academy, https://www.khanacademy.org/science/physics/work-and-energy/work-and-energy-tutorial/a/what-is-gravitational-potential-energy.
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