Potential Energy

Author: Matthew Lewine (mlewine3)

Added material by:

-Brittney Vidal (bvidal3) *did not create the page but I added information as well as videos and additional reading*

-William Xia (wxia33)

-Joe Zein (Jzein3) 2017

Potential energy (referred as [math]\displaystyle{ U }[/math]) is the energy that some body possesses due to their position relative to other bodies, configuration or stresses within itself, electric charges, and other factors. It can be pictured as a stored energy which has the potential to do some work but haven't done yet. Changing the parameters that gives 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).

Example 1: Imagine a rock on top of cliff. It has somehow been displaced to a height H from the surface of the Earth. Due to this position, we say that this rock has some potential energy: if thrown from up there, it has the potential to do some work as it falls (break the floor). In fact, by throwing the ball, H is being reduced, thus U is being reduced. The change in U is translated to changes in other forms of energy. This difference appear as an increase in kinetic energy: that's why the ball gains speed as it falls.

Example 2: The stretched rubber band (Fig.1) for example, has the potential to do some work once you release it.

*Fig.1* The energy stored in a stretched rubber band is a form of elastic potential energy.

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.

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 who also interact 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 (Fig.2). 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 wok 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

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} }[/math]

This says that the potential energy U is equal to the work you must do to move an object from an arbitrary reference point [math]\displaystyle{ U=0 }[/math] to the position [math]\displaystyle{ r }[/math]. We can 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]

Potential energy can also be seen in the Energy Principle Equation. In words, the Energy Principle equation is "Energy change in a system=external energy input". The mathematical model of this equation is [math]\displaystyle{ deltaK=Wsurr + Wint }[/math]. As stated above, [math]\displaystyle{ -Wint=deltaU }[/math], meaning that negative work internal is equal to the change in potential energy. Putting this all into the Energy Principle equation makes [math]\displaystyle{ deltaK+(-Wint)=Wsurr }[/math] which makes [math]\displaystyle{ deltaK+deltaU=Wsurr }[/math]. Simply stated, this equation means that the total work done on a given system by forces that are external is equal to the change in kinetic energy and the change in potential energy.

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}k^{2} }[/math]	[math]\displaystyle{ k }[/math] is the spring constant
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

Examples

Be sure to show all steps in your solution and include diagrams whenever possible

Simple

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

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

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. As a computer science 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. Nuclear particles like protons and neutrons are not destroyed in fission and fusion processes, but collections of them have less mass than if they were individually free, and this mass difference is liberated as heat and radiation in nuclear reactions.

History

The term potential energy was introduced by the 19th century Scottish engineer and physicist William Rankine, although it has links to Greek philosopher Aristotle's concept of potentiality.

Potential Energy

Contents

The Main Idea

A Mathematical Model

Examples

Simple

Middling

Difficult

Connectedness

History

See also

Further reading

External links

References

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