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		<id>http://www.physicsbook.gatech.edu/index.php?title=Potential_Energy&amp;diff=23417</id>
		<title>Potential Energy</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Potential_Energy&amp;diff=23417"/>
		<updated>2016-10-21T02:11:11Z</updated>

		<summary type="html">&lt;p&gt;Ylee620: /* The Main Idea */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Author: Matthew Lewine (mlewine3)&lt;br /&gt;
&lt;br /&gt;
Added material by: &lt;br /&gt;
&lt;br /&gt;
-Brittney Vidal (bvidal3) *did not create the page but I added information as well as videos and additional reading*&lt;br /&gt;
&lt;br /&gt;
-William Xia (wxia33)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Potential energy is stored energy which results from position or configuration.  It is often contrasted with [[Kinetic Energy|kinetic energy]].&lt;br /&gt;
&lt;br /&gt;
[[File:Rubberband2.jpg|thumb|The energy stored in a stretched rubber band is a form of elastic potential energy.]]&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Potential energy is intimately related to kinetic energy.  Potential energy is defined as the capacity for an object to do work; in terms of gravity, potential energy is related to the distance of an object to the center of mass of an object, and in terms of electric energy, potential energy is related to the distance a charged object has to another charged object.&lt;br /&gt;
&lt;br /&gt;
The change in work is translated to changes in other forms of energy.  For example, if a rock is on top of a cliff, it has zero kinetic energy because it is not moving, but it has plenty of potential energy.  Once the rock is dropped, the potential energy is decreasing and the kinetic energy is increasing.  Mathematically, the sign of the change of the potential energy is negative, while the change in kinetic energy is positive.&lt;br /&gt;
&lt;br /&gt;
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 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.&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
A force is considered conservative if it is acting on an object as a function of position only.&lt;br /&gt;
&lt;br /&gt;
We can relate work to potential energy using the equation &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;U = -\int \vec{F}\cdot\vec{dr}&amp;lt;/math&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
This says that the potential energy U is equal to the [[work]] you must do to move an object from an arbitrary reference point &amp;lt;math&amp;gt;U=0&amp;lt;/math&amp;gt; to the position &amp;lt;math&amp;gt;r&amp;lt;/math&amp;gt;. We can take the derivative of both sides of this equation and obtain: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\frac{-dU}{dx} = F(x)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
Let&#039;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&#039;s consider gravitational potential and elastic potential. &lt;br /&gt;
&lt;br /&gt;
The potential energy function U of gravitational potential is &amp;lt;math&amp;gt;mgh&amp;lt;/math&amp;gt;, where &amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt; is mass, &amp;lt;math&amp;gt;g&amp;lt;/math&amp;gt; is the gravitational constant, and &amp;lt;math&amp;gt;h&amp;lt;/math&amp;gt; is some distance away from the reference point at which U = 0. Then the force is &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;F = \frac{-d}{dh}mgh = -mg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can go the other way as well. We know the force of gravity is &amp;lt;math&amp;gt;-mg&amp;lt;/math&amp;gt;, and integrating with respect to h we obtain &amp;lt;math&amp;gt;U = mgh&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
This process can be done with elastic potential as well, where the force &amp;lt;math&amp;gt;F = -kx&amp;lt;/math&amp;gt; and the potential energy function is &amp;lt;math&amp;gt;U = \frac{1}{2}k^{2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Potential energy can also be seen in the Energy Principle Equation. In words, the Energy Principle equation is &amp;quot;Energy change in a system=external energy input&amp;quot;. The mathematical model of this equation is &amp;lt;math&amp;gt;deltaK=Wsurr + Wint&amp;lt;/math&amp;gt;. As stated above, &amp;lt;math&amp;gt;-Wint=deltaU&amp;lt;/math&amp;gt;, meaning that negative work internal is equal to the change in potential energy. Putting this all into the Energy Principle equation makes &amp;lt;math&amp;gt;deltaK+(-Wint)=Wsurr&amp;lt;/math&amp;gt; which makes &amp;lt;math&amp;gt;deltaK+deltaU=Wsurr&amp;lt;/math&amp;gt;. 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. &lt;br /&gt;
&lt;br /&gt;
Here are the potential energy functions for all forms: &lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot;&lt;br /&gt;
| &#039;&#039;&#039;Type&#039;&#039;&#039; || &#039;&#039;&#039;Equation&#039;&#039;&#039; || &#039;&#039;&#039;Variables&#039;&#039;&#039;&lt;br /&gt;
|-&lt;br /&gt;
| Gravitational Potential || &amp;lt;math&amp;gt;U = \frac{GMm}{r}&amp;lt;/math&amp;gt; &amp;lt;br&amp;gt;&amp;lt;math&amp;gt;U = mgh&amp;lt;/math&amp;gt; close to Earth&#039;s surface     || &amp;lt;math&amp;gt;G&amp;lt;/math&amp;gt; is the gravitational constant, &amp;lt;math&amp;gt;M&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt;, and &amp;lt;math&amp;gt;r&amp;lt;/math&amp;gt; is distance&lt;br /&gt;
|-&lt;br /&gt;
| Elastic Potential || &amp;lt;math&amp;gt;U = \frac{1}{2}k^{2}&amp;lt;/math&amp;gt; || &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt; is the spring constant&lt;br /&gt;
|-&lt;br /&gt;
| Electric Potential || &amp;lt;math&amp;gt;U = k\frac{Qq}{r}&amp;lt;/math&amp;gt; || &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt; is Coulomb&#039;s constant, &amp;lt;math&amp;gt;Q&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;q&amp;lt;/math&amp;gt; are point charges, &amp;lt;math&amp;gt;r&amp;lt;/math&amp;gt; is distance&lt;br /&gt;
|-&lt;br /&gt;
| Magnetic Potential || &amp;lt;math&amp;gt;U = -μ \cdot B&amp;lt;/math&amp;gt; || &amp;lt;math&amp;gt;μ&amp;lt;/math&amp;gt; is the dipole moment and &amp;lt;math&amp;gt;μ = IA&amp;lt;/math&amp;gt; in a current loop and &amp;lt;math&amp;gt;I&amp;lt;/math&amp;gt; is the current and &amp;lt;math&amp;gt;A&amp;lt;/math&amp;gt; is the area&lt;br /&gt;
|-&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Be sure to show all steps in your solution and include diagrams whenever possible&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
An object of mass 5 kg is held 10 meters above the Earth&#039;s surface. Relative to the surface, how much potential energy does this object have?&lt;br /&gt;
&lt;br /&gt;
Solution:&lt;br /&gt;
Using the equation &amp;lt;math&amp;gt;U = mgh&amp;lt;/math&amp;gt; we can say &amp;lt;math&amp;gt;U = 5*9.8*10 = 490&amp;lt;/math&amp;gt; J.&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
If it takes 4J of work to stretch a Hooke&#039;s law spring 10 cm from its unstretched length, determine the extra work required to stretch it an additional 10 cm.&lt;br /&gt;
&lt;br /&gt;
Solution:&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
Formally:&lt;br /&gt;
&amp;lt;math&amp;gt;W = \frac{1}{2}kx^{2}.&amp;lt;/math&amp;gt;  Given W and x we find k.&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt;4 J = \frac{1}{2}k(0.1)^{2}&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt;k =\frac{8}{0.1^{2}} = 800&amp;lt;/math&amp;gt; N/m.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
Using &amp;lt;math&amp;gt;x = 0.2&amp;lt;/math&amp;gt; m, &amp;lt;math&amp;gt;W = \frac{1}{2}(800)(0.2)^{2} = 16&amp;lt;/math&amp;gt; J&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
Extra work:  16 J - 4 J = 12 J.&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
We have a point charge A of charge +Q at the origin. Let&#039;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?&lt;br /&gt;
&lt;br /&gt;
Solution: &lt;br /&gt;
We have a nonuniform electric field, so we need to integrate the potential energy function to find the amount of work needed. &lt;br /&gt;
&amp;lt;math&amp;gt;W = \int_{10}^{5} \frac{-kQq}{r^2}dr= -kQq\int_{10}^{5}\frac{1}{r^2}dr = \frac{kQq}{10} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The term potential energy was introduced by the 19th century Scottish engineer and physicist William Rankine, although it has links to Greek philosopher Aristotle&#039;s concept of potentiality.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[[Kinetic Energy]]&amp;lt;br&amp;gt;&lt;br /&gt;
[[Potential Energy for a Magnetic Dipole]]&amp;lt;br&amp;gt;&lt;br /&gt;
[[Potential Energy of a Multiparticle System]]&amp;lt;br&amp;gt;&lt;br /&gt;
[[Work]]&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/pegrav.html&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elepe.html&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
[http://www.scientificamerican.com/article/bring-science-home-rubber-bands-energy/]&lt;br /&gt;
[https://www.youtube.com/watch?v=elJUghWSVh4]&lt;br /&gt;
[https://www.youtube.com/watch?v=Y3xv-Oz68jQ]&lt;br /&gt;
[https://www.youtube.com/watch?v=zM7Cz1sQj9c]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Potential Energy.&amp;quot; HyperPhysics. Georgia State University, n.d. Web. 04 Dec. 2015.&amp;lt;br&amp;gt;&lt;br /&gt;
Chabay, Ruth W., and Bruce A. Sherwood. Matter &amp;amp; Interactions. Hoboken, NJ: Wiley, 2011. Print.&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Electric Potential Energy.&amp;quot; HyperPhysics. Georgia State University, n.d. Web. 16 Apr. 2016.&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:Which Category did you place this in?]]&lt;/div&gt;</summary>
		<author><name>Ylee620</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Potential_Energy&amp;diff=23416</id>
		<title>Potential Energy</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Potential_Energy&amp;diff=23416"/>
		<updated>2016-10-21T02:09:44Z</updated>

		<summary type="html">&lt;p&gt;Ylee620: /* The Main Idea */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Author: Matthew Lewine (mlewine3)&lt;br /&gt;
&lt;br /&gt;
Added material by: &lt;br /&gt;
&lt;br /&gt;
-Brittney Vidal (bvidal3) *did not create the page but I added information as well as videos and additional reading*&lt;br /&gt;
&lt;br /&gt;
-William Xia (wxia33)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Potential energy is stored energy which results from position or configuration.  It is often contrasted with [[Kinetic Energy|kinetic energy]].&lt;br /&gt;
&lt;br /&gt;
[[File:Rubberband2.jpg|thumb|The energy stored in a stretched rubber band is a form of elastic potential energy.]]&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Potential energy is intimately related to kinetic energy.  Potential energy is defined as the capacity for an object to do work; in terms of gravity, potential energy is related to the distance of an object to the center of mass of an object, and in terms of electric energy, potential energy is related to the distance a charged object has to another charged object.&lt;br /&gt;
&lt;br /&gt;
The change in work is translated to changes in other forms of energy.  For example, if a rock is on top of a cliff, it has zero kinetic energy because it is not moving, but it has plenty of potential energy.  Once the rock is dropped, the potential energy is decreasing and the kinetic energy is increasing.  Mathematically, the sign of the change of the potential energy is negative, while the change in kinetic energy is positive.&lt;br /&gt;
&lt;br /&gt;
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 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.&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
A force is considered conservative if it is acting on an object as a function of position only.&lt;br /&gt;
&lt;br /&gt;
We can relate work to potential energy using the equation &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;U = -\int \vec{F}\cdot\vec{dr}&amp;lt;/math&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
This says that the potential energy U is equal to the [[work]] you must do to move an object from an arbitrary reference point &amp;lt;math&amp;gt;U=0&amp;lt;/math&amp;gt; to the position &amp;lt;math&amp;gt;r&amp;lt;/math&amp;gt;. We can take the derivative of both sides of this equation and obtain: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\frac{-dU}{dx} = F(x)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
Let&#039;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&#039;s consider gravitational potential and elastic potential. &lt;br /&gt;
&lt;br /&gt;
The potential energy function U of gravitational potential is &amp;lt;math&amp;gt;mgh&amp;lt;/math&amp;gt;, where &amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt; is mass, &amp;lt;math&amp;gt;g&amp;lt;/math&amp;gt; is the gravitational constant, and &amp;lt;math&amp;gt;h&amp;lt;/math&amp;gt; is some distance away from the reference point at which U = 0. Then the force is &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;F = \frac{-d}{dh}mgh = -mg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can go the other way as well. We know the force of gravity is &amp;lt;math&amp;gt;-mg&amp;lt;/math&amp;gt;, and integrating with respect to h we obtain &amp;lt;math&amp;gt;U = mgh&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
This process can be done with elastic potential as well, where the force &amp;lt;math&amp;gt;F = -kx&amp;lt;/math&amp;gt; and the potential energy function is &amp;lt;math&amp;gt;U = \frac{1}{2}k^{2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Potential energy can also be seen in the Energy Principle Equation. In words, the Energy Principle equation is &amp;quot;Energy change in a system=external energy input&amp;quot;. The mathematical model of this equation is &amp;lt;math&amp;gt;deltaK=Wsurr + Wint&amp;lt;/math&amp;gt;. As stated above, &amp;lt;math&amp;gt;-Wint=deltaU&amp;lt;/math&amp;gt;, meaning that negative work internal is equal to the change in potential energy. Putting this all into the Energy Principle equation makes &amp;lt;math&amp;gt;deltaK+(-Wint)=Wsurr&amp;lt;/math&amp;gt; which makes &amp;lt;math&amp;gt;deltaK+deltaU=Wsurr&amp;lt;/math&amp;gt;. 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. &lt;br /&gt;
&lt;br /&gt;
Here are the potential energy functions for all forms: &lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot;&lt;br /&gt;
| &#039;&#039;&#039;Type&#039;&#039;&#039; || &#039;&#039;&#039;Equation&#039;&#039;&#039; || &#039;&#039;&#039;Variables&#039;&#039;&#039;&lt;br /&gt;
|-&lt;br /&gt;
| Gravitational Potential || &amp;lt;math&amp;gt;U = \frac{GMm}{r}&amp;lt;/math&amp;gt; &amp;lt;br&amp;gt;&amp;lt;math&amp;gt;U = mgh&amp;lt;/math&amp;gt; close to Earth&#039;s surface     || &amp;lt;math&amp;gt;G&amp;lt;/math&amp;gt; is the gravitational constant, &amp;lt;math&amp;gt;M&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt;, and &amp;lt;math&amp;gt;r&amp;lt;/math&amp;gt; is distance&lt;br /&gt;
|-&lt;br /&gt;
| Elastic Potential || &amp;lt;math&amp;gt;U = \frac{1}{2}k^{2}&amp;lt;/math&amp;gt; || &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt; is the spring constant&lt;br /&gt;
|-&lt;br /&gt;
| Electric Potential || &amp;lt;math&amp;gt;U = k\frac{Qq}{r}&amp;lt;/math&amp;gt; || &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt; is Coulomb&#039;s constant, &amp;lt;math&amp;gt;Q&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;q&amp;lt;/math&amp;gt; are point charges, &amp;lt;math&amp;gt;r&amp;lt;/math&amp;gt; is distance&lt;br /&gt;
|-&lt;br /&gt;
| Magnetic Potential || &amp;lt;math&amp;gt;U = -μ \cdot B&amp;lt;/math&amp;gt; || &amp;lt;math&amp;gt;μ&amp;lt;/math&amp;gt; is the dipole moment and &amp;lt;math&amp;gt;μ = IA&amp;lt;/math&amp;gt; in a current loop and &amp;lt;math&amp;gt;I&amp;lt;/math&amp;gt; is the current and &amp;lt;math&amp;gt;A&amp;lt;/math&amp;gt; is the area&lt;br /&gt;
|-&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/File:Mediaeval_archery_reenactment.jpg&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Be sure to show all steps in your solution and include diagrams whenever possible&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
An object of mass 5 kg is held 10 meters above the Earth&#039;s surface. Relative to the surface, how much potential energy does this object have?&lt;br /&gt;
&lt;br /&gt;
Solution:&lt;br /&gt;
Using the equation &amp;lt;math&amp;gt;U = mgh&amp;lt;/math&amp;gt; we can say &amp;lt;math&amp;gt;U = 5*9.8*10 = 490&amp;lt;/math&amp;gt; J.&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
If it takes 4J of work to stretch a Hooke&#039;s law spring 10 cm from its unstretched length, determine the extra work required to stretch it an additional 10 cm.&lt;br /&gt;
&lt;br /&gt;
Solution:&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
Formally:&lt;br /&gt;
&amp;lt;math&amp;gt;W = \frac{1}{2}kx^{2}.&amp;lt;/math&amp;gt;  Given W and x we find k.&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt;4 J = \frac{1}{2}k(0.1)^{2}&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt;k =\frac{8}{0.1^{2}} = 800&amp;lt;/math&amp;gt; N/m.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
Using &amp;lt;math&amp;gt;x = 0.2&amp;lt;/math&amp;gt; m, &amp;lt;math&amp;gt;W = \frac{1}{2}(800)(0.2)^{2} = 16&amp;lt;/math&amp;gt; J&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
Extra work:  16 J - 4 J = 12 J.&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
We have a point charge A of charge +Q at the origin. Let&#039;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?&lt;br /&gt;
&lt;br /&gt;
Solution: &lt;br /&gt;
We have a nonuniform electric field, so we need to integrate the potential energy function to find the amount of work needed. &lt;br /&gt;
&amp;lt;math&amp;gt;W = \int_{10}^{5} \frac{-kQq}{r^2}dr= -kQq\int_{10}^{5}\frac{1}{r^2}dr = \frac{kQq}{10} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The term potential energy was introduced by the 19th century Scottish engineer and physicist William Rankine, although it has links to Greek philosopher Aristotle&#039;s concept of potentiality.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[[Kinetic Energy]]&amp;lt;br&amp;gt;&lt;br /&gt;
[[Potential Energy for a Magnetic Dipole]]&amp;lt;br&amp;gt;&lt;br /&gt;
[[Potential Energy of a Multiparticle System]]&amp;lt;br&amp;gt;&lt;br /&gt;
[[Work]]&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/pegrav.html&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elepe.html&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
[http://www.scientificamerican.com/article/bring-science-home-rubber-bands-energy/]&lt;br /&gt;
[https://www.youtube.com/watch?v=elJUghWSVh4]&lt;br /&gt;
[https://www.youtube.com/watch?v=Y3xv-Oz68jQ]&lt;br /&gt;
[https://www.youtube.com/watch?v=zM7Cz1sQj9c]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Potential Energy.&amp;quot; HyperPhysics. Georgia State University, n.d. Web. 04 Dec. 2015.&amp;lt;br&amp;gt;&lt;br /&gt;
Chabay, Ruth W., and Bruce A. Sherwood. Matter &amp;amp; Interactions. Hoboken, NJ: Wiley, 2011. Print.&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Electric Potential Energy.&amp;quot; HyperPhysics. Georgia State University, n.d. Web. 16 Apr. 2016.&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:Which Category did you place this in?]]&lt;/div&gt;</summary>
		<author><name>Ylee620</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Potential_Energy&amp;diff=23415</id>
		<title>Potential Energy</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Potential_Energy&amp;diff=23415"/>
		<updated>2016-10-21T02:08:07Z</updated>

		<summary type="html">&lt;p&gt;Ylee620: /* Potential energy for near Earth gravity */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Author: Matthew Lewine (mlewine3)&lt;br /&gt;
&lt;br /&gt;
Added material by: &lt;br /&gt;
&lt;br /&gt;
-Brittney Vidal (bvidal3) *did not create the page but I added information as well as videos and additional reading*&lt;br /&gt;
&lt;br /&gt;
-William Xia (wxia33)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Potential energy is stored energy which results from position or configuration.  It is often contrasted with [[Kinetic Energy|kinetic energy]].&lt;br /&gt;
&lt;br /&gt;
[[File:Rubberband2.jpg|thumb|The energy stored in a stretched rubber band is a form of elastic potential energy.]]&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Potential energy is intimately related to kinetic energy.  Potential energy is defined as the capacity for an object to do work; in terms of gravity, potential energy is related to the distance of an object to the center of mass of an object, and in terms of electric energy, potential energy is related to the distance a charged object has to another charged object.&lt;br /&gt;
&lt;br /&gt;
The change in work is translated to changes in other forms of energy.  For example, if a rock is on top of a cliff, it has zero kinetic energy because it is not moving, but it has plenty of potential energy.  Once the rock is dropped, the potential energy is decreasing and the kinetic energy is increasing.  Mathematically, the sign of the change of the potential energy is negative, while the change in kinetic energy is positive.&lt;br /&gt;
&lt;br /&gt;
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 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.&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
A force is considered conservative if it is acting on an object as a function of position only.&lt;br /&gt;
&lt;br /&gt;
We can relate work to potential energy using the equation &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;U = -\int \vec{F}\cdot\vec{dr}&amp;lt;/math&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
This says that the potential energy U is equal to the [[work]] you must do to move an object from an arbitrary reference point &amp;lt;math&amp;gt;U=0&amp;lt;/math&amp;gt; to the position &amp;lt;math&amp;gt;r&amp;lt;/math&amp;gt;. We can take the derivative of both sides of this equation and obtain: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\frac{-dU}{dx} = F(x)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
Let&#039;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&#039;s consider gravitational potential and elastic potential. &lt;br /&gt;
&lt;br /&gt;
The potential energy function U of gravitational potential is &amp;lt;math&amp;gt;mgh&amp;lt;/math&amp;gt;, where &amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt; is mass, &amp;lt;math&amp;gt;g&amp;lt;/math&amp;gt; is the gravitational constant, and &amp;lt;math&amp;gt;h&amp;lt;/math&amp;gt; is some distance away from the reference point at which U = 0. Then the force is &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;F = \frac{-d}{dh}mgh = -mg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can go the other way as well. We know the force of gravity is &amp;lt;math&amp;gt;-mg&amp;lt;/math&amp;gt;, and integrating with respect to h we obtain &amp;lt;math&amp;gt;U = mgh&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
This process can be done with elastic potential as well, where the force &amp;lt;math&amp;gt;F = -kx&amp;lt;/math&amp;gt; and the potential energy function is &amp;lt;math&amp;gt;U = \frac{1}{2}k^{2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Potential energy can also be seen in the Energy Principle Equation. In words, the Energy Principle equation is &amp;quot;Energy change in a system=external energy input&amp;quot;. The mathematical model of this equation is &amp;lt;math&amp;gt;deltaK=Wsurr + Wint&amp;lt;/math&amp;gt;. As stated above, &amp;lt;math&amp;gt;-Wint=deltaU&amp;lt;/math&amp;gt;, meaning that negative work internal is equal to the change in potential energy. Putting this all into the Energy Principle equation makes &amp;lt;math&amp;gt;deltaK+(-Wint)=Wsurr&amp;lt;/math&amp;gt; which makes &amp;lt;math&amp;gt;deltaK+deltaU=Wsurr&amp;lt;/math&amp;gt;. 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. &lt;br /&gt;
&lt;br /&gt;
Here are the potential energy functions for all forms: &lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot;&lt;br /&gt;
| &#039;&#039;&#039;Type&#039;&#039;&#039; || &#039;&#039;&#039;Equation&#039;&#039;&#039; || &#039;&#039;&#039;Variables&#039;&#039;&#039;&lt;br /&gt;
|-&lt;br /&gt;
| Gravitational Potential || &amp;lt;math&amp;gt;U = \frac{GMm}{r}&amp;lt;/math&amp;gt; &amp;lt;br&amp;gt;&amp;lt;math&amp;gt;U = mgh&amp;lt;/math&amp;gt; close to Earth&#039;s surface     || &amp;lt;math&amp;gt;G&amp;lt;/math&amp;gt; is the gravitational constant, &amp;lt;math&amp;gt;M&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt;, and &amp;lt;math&amp;gt;r&amp;lt;/math&amp;gt; is distance&lt;br /&gt;
|-&lt;br /&gt;
| Elastic Potential || &amp;lt;math&amp;gt;U = \frac{1}{2}k^{2}&amp;lt;/math&amp;gt; || &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt; is the spring constant&lt;br /&gt;
|-&lt;br /&gt;
| Electric Potential || &amp;lt;math&amp;gt;U = k\frac{Qq}{r}&amp;lt;/math&amp;gt; || &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt; is Coulomb&#039;s constant, &amp;lt;math&amp;gt;Q&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;q&amp;lt;/math&amp;gt; are point charges, &amp;lt;math&amp;gt;r&amp;lt;/math&amp;gt; is distance&lt;br /&gt;
|-&lt;br /&gt;
| Magnetic Potential || &amp;lt;math&amp;gt;U = -μ \cdot B&amp;lt;/math&amp;gt; || &amp;lt;math&amp;gt;μ&amp;lt;/math&amp;gt; is the dipole moment and &amp;lt;math&amp;gt;μ = IA&amp;lt;/math&amp;gt; in a current loop and &amp;lt;math&amp;gt;I&amp;lt;/math&amp;gt; is the current and &amp;lt;math&amp;gt;A&amp;lt;/math&amp;gt; is the area&lt;br /&gt;
|-&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Be sure to show all steps in your solution and include diagrams whenever possible&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
An object of mass 5 kg is held 10 meters above the Earth&#039;s surface. Relative to the surface, how much potential energy does this object have?&lt;br /&gt;
&lt;br /&gt;
Solution:&lt;br /&gt;
Using the equation &amp;lt;math&amp;gt;U = mgh&amp;lt;/math&amp;gt; we can say &amp;lt;math&amp;gt;U = 5*9.8*10 = 490&amp;lt;/math&amp;gt; J.&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
If it takes 4J of work to stretch a Hooke&#039;s law spring 10 cm from its unstretched length, determine the extra work required to stretch it an additional 10 cm.&lt;br /&gt;
&lt;br /&gt;
Solution:&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
Formally:&lt;br /&gt;
&amp;lt;math&amp;gt;W = \frac{1}{2}kx^{2}.&amp;lt;/math&amp;gt;  Given W and x we find k.&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt;4 J = \frac{1}{2}k(0.1)^{2}&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt;k =\frac{8}{0.1^{2}} = 800&amp;lt;/math&amp;gt; N/m.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
Using &amp;lt;math&amp;gt;x = 0.2&amp;lt;/math&amp;gt; m, &amp;lt;math&amp;gt;W = \frac{1}{2}(800)(0.2)^{2} = 16&amp;lt;/math&amp;gt; J&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
Extra work:  16 J - 4 J = 12 J.&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
We have a point charge A of charge +Q at the origin. Let&#039;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?&lt;br /&gt;
&lt;br /&gt;
Solution: &lt;br /&gt;
We have a nonuniform electric field, so we need to integrate the potential energy function to find the amount of work needed. &lt;br /&gt;
&amp;lt;math&amp;gt;W = \int_{10}^{5} \frac{-kQq}{r^2}dr= -kQq\int_{10}^{5}\frac{1}{r^2}dr = \frac{kQq}{10} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The term potential energy was introduced by the 19th century Scottish engineer and physicist William Rankine, although it has links to Greek philosopher Aristotle&#039;s concept of potentiality.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[[Kinetic Energy]]&amp;lt;br&amp;gt;&lt;br /&gt;
[[Potential Energy for a Magnetic Dipole]]&amp;lt;br&amp;gt;&lt;br /&gt;
[[Potential Energy of a Multiparticle System]]&amp;lt;br&amp;gt;&lt;br /&gt;
[[Work]]&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/pegrav.html&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elepe.html&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
[http://www.scientificamerican.com/article/bring-science-home-rubber-bands-energy/]&lt;br /&gt;
[https://www.youtube.com/watch?v=elJUghWSVh4]&lt;br /&gt;
[https://www.youtube.com/watch?v=Y3xv-Oz68jQ]&lt;br /&gt;
[https://www.youtube.com/watch?v=zM7Cz1sQj9c]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Potential Energy.&amp;quot; HyperPhysics. Georgia State University, n.d. Web. 04 Dec. 2015.&amp;lt;br&amp;gt;&lt;br /&gt;
Chabay, Ruth W., and Bruce A. Sherwood. Matter &amp;amp; Interactions. Hoboken, NJ: Wiley, 2011. Print.&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Electric Potential Energy.&amp;quot; HyperPhysics. Georgia State University, n.d. Web. 16 Apr. 2016.&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:Which Category did you place this in?]]&lt;/div&gt;</summary>
		<author><name>Ylee620</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Potential_Energy&amp;diff=23414</id>
		<title>Potential Energy</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Potential_Energy&amp;diff=23414"/>
		<updated>2016-10-21T02:07:37Z</updated>

		<summary type="html">&lt;p&gt;Ylee620: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Author: Matthew Lewine (mlewine3)&lt;br /&gt;
&lt;br /&gt;
Added material by: &lt;br /&gt;
&lt;br /&gt;
-Brittney Vidal (bvidal3) *did not create the page but I added information as well as videos and additional reading*&lt;br /&gt;
&lt;br /&gt;
-William Xia (wxia33)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Potential energy is stored energy which results from position or configuration.  It is often contrasted with [[Kinetic Energy|kinetic energy]].&lt;br /&gt;
&lt;br /&gt;
[[File:Rubberband2.jpg|thumb|The energy stored in a stretched rubber band is a form of elastic potential energy.]]&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Potential energy is intimately related to kinetic energy.  Potential energy is defined as the capacity for an object to do work; in terms of gravity, potential energy is related to the distance of an object to the center of mass of an object, and in terms of electric energy, potential energy is related to the distance a charged object has to another charged object.&lt;br /&gt;
&lt;br /&gt;
The change in work is translated to changes in other forms of energy.  For example, if a rock is on top of a cliff, it has zero kinetic energy because it is not moving, but it has plenty of potential energy.  Once the rock is dropped, the potential energy is decreasing and the kinetic energy is increasing.  Mathematically, the sign of the change of the potential energy is negative, while the change in kinetic energy is positive.&lt;br /&gt;
&lt;br /&gt;
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 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.&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
A force is considered conservative if it is acting on an object as a function of position only.&lt;br /&gt;
&lt;br /&gt;
We can relate work to potential energy using the equation &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;U = -\int \vec{F}\cdot\vec{dr}&amp;lt;/math&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
This says that the potential energy U is equal to the [[work]] you must do to move an object from an arbitrary reference point &amp;lt;math&amp;gt;U=0&amp;lt;/math&amp;gt; to the position &amp;lt;math&amp;gt;r&amp;lt;/math&amp;gt;. We can take the derivative of both sides of this equation and obtain: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\frac{-dU}{dx} = F(x)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
Let&#039;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&#039;s consider gravitational potential and elastic potential. &lt;br /&gt;
&lt;br /&gt;
The potential energy function U of gravitational potential is &amp;lt;math&amp;gt;mgh&amp;lt;/math&amp;gt;, where &amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt; is mass, &amp;lt;math&amp;gt;g&amp;lt;/math&amp;gt; is the gravitational constant, and &amp;lt;math&amp;gt;h&amp;lt;/math&amp;gt; is some distance away from the reference point at which U = 0. Then the force is &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;F = \frac{-d}{dh}mgh = -mg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can go the other way as well. We know the force of gravity is &amp;lt;math&amp;gt;-mg&amp;lt;/math&amp;gt;, and integrating with respect to h we obtain &amp;lt;math&amp;gt;U = mgh&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
This process can be done with elastic potential as well, where the force &amp;lt;math&amp;gt;F = -kx&amp;lt;/math&amp;gt; and the potential energy function is &amp;lt;math&amp;gt;U = \frac{1}{2}k^{2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Potential energy can also be seen in the Energy Principle Equation. In words, the Energy Principle equation is &amp;quot;Energy change in a system=external energy input&amp;quot;. The mathematical model of this equation is &amp;lt;math&amp;gt;deltaK=Wsurr + Wint&amp;lt;/math&amp;gt;. As stated above, &amp;lt;math&amp;gt;-Wint=deltaU&amp;lt;/math&amp;gt;, meaning that negative work internal is equal to the change in potential energy. Putting this all into the Energy Principle equation makes &amp;lt;math&amp;gt;deltaK+(-Wint)=Wsurr&amp;lt;/math&amp;gt; which makes &amp;lt;math&amp;gt;deltaK+deltaU=Wsurr&amp;lt;/math&amp;gt;. 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. &lt;br /&gt;
&lt;br /&gt;
Here are the potential energy functions for all forms: &lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot;&lt;br /&gt;
| &#039;&#039;&#039;Type&#039;&#039;&#039; || &#039;&#039;&#039;Equation&#039;&#039;&#039; || &#039;&#039;&#039;Variables&#039;&#039;&#039;&lt;br /&gt;
|-&lt;br /&gt;
| Gravitational Potential || &amp;lt;math&amp;gt;U = \frac{GMm}{r}&amp;lt;/math&amp;gt; &amp;lt;br&amp;gt;&amp;lt;math&amp;gt;U = mgh&amp;lt;/math&amp;gt; close to Earth&#039;s surface     || &amp;lt;math&amp;gt;G&amp;lt;/math&amp;gt; is the gravitational constant, &amp;lt;math&amp;gt;M&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt;, and &amp;lt;math&amp;gt;r&amp;lt;/math&amp;gt; is distance&lt;br /&gt;
|-&lt;br /&gt;
| Elastic Potential || &amp;lt;math&amp;gt;U = \frac{1}{2}k^{2}&amp;lt;/math&amp;gt; || &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt; is the spring constant&lt;br /&gt;
|-&lt;br /&gt;
| Electric Potential || &amp;lt;math&amp;gt;U = k\frac{Qq}{r}&amp;lt;/math&amp;gt; || &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt; is Coulomb&#039;s constant, &amp;lt;math&amp;gt;Q&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;q&amp;lt;/math&amp;gt; are point charges, &amp;lt;math&amp;gt;r&amp;lt;/math&amp;gt; is distance&lt;br /&gt;
|-&lt;br /&gt;
| Magnetic Potential || &amp;lt;math&amp;gt;U = -μ \cdot B&amp;lt;/math&amp;gt; || &amp;lt;math&amp;gt;μ&amp;lt;/math&amp;gt; is the dipole moment and &amp;lt;math&amp;gt;μ = IA&amp;lt;/math&amp;gt; in a current loop and &amp;lt;math&amp;gt;I&amp;lt;/math&amp;gt; is the current and &amp;lt;math&amp;gt;A&amp;lt;/math&amp;gt; is the area&lt;br /&gt;
|-&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Be sure to show all steps in your solution and include diagrams whenever possible&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
An object of mass 5 kg is held 10 meters above the Earth&#039;s surface. Relative to the surface, how much potential energy does this object have?&lt;br /&gt;
&lt;br /&gt;
Solution:&lt;br /&gt;
Using the equation &amp;lt;math&amp;gt;U = mgh&amp;lt;/math&amp;gt; we can say &amp;lt;math&amp;gt;U = 5*9.8*10 = 490&amp;lt;/math&amp;gt; J.&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
If it takes 4J of work to stretch a Hooke&#039;s law spring 10 cm from its unstretched length, determine the extra work required to stretch it an additional 10 cm.&lt;br /&gt;
&lt;br /&gt;
Solution:&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
Formally:&lt;br /&gt;
&amp;lt;math&amp;gt;W = \frac{1}{2}kx^{2}.&amp;lt;/math&amp;gt;  Given W and x we find k.&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt;4 J = \frac{1}{2}k(0.1)^{2}&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt;k =\frac{8}{0.1^{2}} = 800&amp;lt;/math&amp;gt; N/m.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
Using &amp;lt;math&amp;gt;x = 0.2&amp;lt;/math&amp;gt; m, &amp;lt;math&amp;gt;W = \frac{1}{2}(800)(0.2)^{2} = 16&amp;lt;/math&amp;gt; J&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
Extra work:  16 J - 4 J = 12 J.&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
We have a point charge A of charge +Q at the origin. Let&#039;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?&lt;br /&gt;
&lt;br /&gt;
Solution: &lt;br /&gt;
We have a nonuniform electric field, so we need to integrate the potential energy function to find the amount of work needed. &lt;br /&gt;
&amp;lt;math&amp;gt;W = \int_{10}^{5} \frac{-kQq}{r^2}dr= -kQq\int_{10}^{5}\frac{1}{r^2}dr = \frac{kQq}{10} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Potential energy for near Earth gravity===&lt;br /&gt;
The following function is called the potential energy of a near earth gravity field:&lt;br /&gt;
&lt;br /&gt;
{\displaystyle U(\mathbf {r} )=mg\,h(\mathbf {r} ),} U({\mathbf  {r}})=mg\,h({\mathbf  {r}}),&lt;br /&gt;
where m is in kg, g is 9.81 for earth and h is in metres.&lt;br /&gt;
&lt;br /&gt;
In classical physics, gravity exerts a constant downward force F=(0, 0, Fz) on the center of mass of a body moving near the surface of the Earth. The work of gravity on a body moving along a trajectory r(t) = (x(t), y(t), z(t)), such as the track of a roller coaster is calculated using its velocity, v=(vx, vy, vz), to obtain&lt;br /&gt;
&lt;br /&gt;
{\displaystyle W=\int _{t_{1}}^{t_{2}}{\boldsymbol {F}}\cdot {\boldsymbol {v}}\mathrm {d} t=\int _{t_{1}}^{t_{2}}F_{z}v_{z}\mathrm {d} t=F_{z}\Delta z.} W=\int_{t_1}^{t_2}\boldsymbol{F}\cdot\boldsymbol{v}\mathrm{d}t = \int_{t_1}^{t_2}F_z v_z \mathrm{d}t = F_z\Delta z. &lt;br /&gt;
where the integral of the vertical component of velocity is the vertical distance. Notice that the work of gravity depends only on the vertical movement of the curve r(t).&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The term potential energy was introduced by the 19th century Scottish engineer and physicist William Rankine, although it has links to Greek philosopher Aristotle&#039;s concept of potentiality.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[[Kinetic Energy]]&amp;lt;br&amp;gt;&lt;br /&gt;
[[Potential Energy for a Magnetic Dipole]]&amp;lt;br&amp;gt;&lt;br /&gt;
[[Potential Energy of a Multiparticle System]]&amp;lt;br&amp;gt;&lt;br /&gt;
[[Work]]&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/pegrav.html&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elepe.html&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
[http://www.scientificamerican.com/article/bring-science-home-rubber-bands-energy/]&lt;br /&gt;
[https://www.youtube.com/watch?v=elJUghWSVh4]&lt;br /&gt;
[https://www.youtube.com/watch?v=Y3xv-Oz68jQ]&lt;br /&gt;
[https://www.youtube.com/watch?v=zM7Cz1sQj9c]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Potential Energy.&amp;quot; HyperPhysics. Georgia State University, n.d. Web. 04 Dec. 2015.&amp;lt;br&amp;gt;&lt;br /&gt;
Chabay, Ruth W., and Bruce A. Sherwood. Matter &amp;amp; Interactions. Hoboken, NJ: Wiley, 2011. Print.&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Electric Potential Energy.&amp;quot; HyperPhysics. Georgia State University, n.d. Web. 16 Apr. 2016.&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:Which Category did you place this in?]]&lt;/div&gt;</summary>
		<author><name>Ylee620</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Newton%27s_Third_Law_of_Motion&amp;diff=23413</id>
		<title>Newton&#039;s Third Law of Motion</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Newton%27s_Third_Law_of_Motion&amp;diff=23413"/>
		<updated>2016-10-21T02:00:43Z</updated>

		<summary type="html">&lt;p&gt;Ylee620: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;By Karan Shah&lt;br /&gt;
[[File: Newton&#039;s Third Law Explained.png | thumb | right | 400px |Newton&#039;s Third Law Explained]] &lt;br /&gt;
&lt;br /&gt;
==Main Idea==&lt;br /&gt;
&lt;br /&gt;
Newton’s Third Law of Motion describes a push or pull that acts on an object as a result of its interaction with another object. According to this law for every action there is an equal and opposite re-action. This means that for every force  there is a reaction force that is equal in size, but opposite in direction. Meaning that when an object 1 pushes another object 2 then object 1 gets pushed back with equal force but in the opposite direction. [[File: Law3 f1.gif | thumb | left | 250px |If you push an object with 100N it will push back on you with equal but opposite force.]][[File: Snip20151127_7.png| thumb | right | 200px| Mathematically Formula to describe Newton&#039;s Third Law ]]&lt;br /&gt;
&lt;br /&gt;
The third law of motion is also referred to as the action-reaction law because both objects are part of a single interaction and neither force can exist without the other. An important concept to remember about Newton&#039;s Third Law of Motion is that the two forces are of the same type. For example, when you throw a ball in the sky the Earth exerts a gravitational force on the ball and the ball also exerts a [[File: 924.gif | thumb | right | 250px | The canon pushes the canon ball forward and the canon pushes the canon back with equal force.]] gravitational force that is equal in magnitude and opposite in direction on the earth. Another example, that can sum up the concept of Newton&#039;s Third Law is when you walk. When you push down upon the ground and ground pushes with the same force upward. Similarly, the tires of a car push against the road while the road pushes back on the tires.&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Here are some problems regarding Newton&#039;s Third Law.&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
==== Question ====&lt;br /&gt;
&lt;br /&gt;
[[File: Snip20151128_10.png| thumb | left | 250px |Simple Example]]Car B is stopped at a red light. The brakes in Car A have failed and Car A is coming towards Car B at 60 kmh. Car B then runs into the back of Car A, What can be said about the force on Car A on Car B and the force on Car B on Car A?&lt;br /&gt;
&lt;br /&gt;
==== Answer ====&lt;br /&gt;
&lt;br /&gt;
 B exerts the same amount of force on A as A exerts on B. &lt;br /&gt;
 Just the direction of both the forces will be in the opposite direction.&lt;br /&gt;
&lt;br /&gt;
===Middle===&lt;br /&gt;
&lt;br /&gt;
====Question====&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Blocks with masses of 1 kg, 2 kg, and 3 kg are lined up in a row on a frictionless table. All three are pushed forward by a 8 N force applied to the 1 kg block. (a) How much force does the 2 kg block exert on the 3 kg block? (b) How much force does the 2 kg block exert on the 1 kg block?&lt;br /&gt;
&lt;br /&gt;
====Answer====&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 (a) &lt;br /&gt;
 Find the Acceleration of the Whole Object:&lt;br /&gt;
 Total Mass: 6kg&lt;br /&gt;
 8 = (6) a &lt;br /&gt;
 a = 8 / 6 = 1.33 m/s^2&lt;br /&gt;
 Total Acceleration: 1.33 m/s^2 (Acceleration will be the same for all three blocks)&lt;br /&gt;
 F(2 on 3) = m(3) * a  &lt;br /&gt;
 3 * 1.33 = 3.999 N&lt;br /&gt;
&lt;br /&gt;
 (b) &lt;br /&gt;
 Total Acceleration: 1.33 m/s^2&lt;br /&gt;
 Mass to push: 5 kg (Because we are also pushing the 3 kg block)&lt;br /&gt;
 F(1 on 2) = 5 * 1.33&lt;br /&gt;
 F(1 on 2) = 5.33 N&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
====Question====&lt;br /&gt;
A massive steel cable drags a 30 kg block across a horizontal, frictionless surface. A 100 N force applied to the cable causes the block to reach a speed of 5.0 m/s in a distance of 5.0 m. What is the mass of the cable?&lt;br /&gt;
&lt;br /&gt;
====Answer====&lt;br /&gt;
&lt;br /&gt;
 a = V² / (2x) = 2.5 m/s² &lt;br /&gt;
 F = M * a → 100 = (30+m) * 2.5 &lt;br /&gt;
 m = 10 kg&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
&lt;br /&gt;
 Newton&#039;s Third Law is connected to the concept of a spacecraft flying in space. When a spacecraft &lt;br /&gt;
 fires a thruster rocket, the exhaust gas pushes against the thruster and the thruster pushes &lt;br /&gt;
 against the exhaust gas. The gas and rocket move in opposite directions. This is an &lt;br /&gt;
 example of Newton&#039;s Third Law because both forces are equal in magnitude and opposite &lt;br /&gt;
 in direction. This topic is related to space travel a topic that I am interested in and &lt;br /&gt;
 passionate about learning more.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Sir Isaac Newton was a renowned scientist and mathematician who helped create a foundation for modern studies. He was born in England in 1643 and worked his way to earn a bachelor’s and master’s degree from Trinity College Cambridge. He was highly interested in math, physics, and astronomy and wrote many of his ideas in a journal. One of those ideas was about the three laws of motion. [[File: Sir_Isaac_Newton_(1643-1727).jpg| thumb | left | 150px| Sir Isaac Newton (1643 - 1727)]] In 1687 Isaac Newton made his work on his book, Philosophiae Naturalis Principia Mathematic or Principia known to the public.  He discussed the principles of time, force, and motion that helped create modern physical science and helped account for much of the phenomena viewed in the world. Some of the principles he discusses include acceleration, initial movement, fluid dynamics, and motion. Newton’s Laws first appeared in the Principia and discussed the relationship that exists between forces acting on a body and the motion of the body. For the third law, he stated that for every action/force in nature, there will be an equal and opposite reaction.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Books, Articles or other print media on this topic&lt;br /&gt;
&lt;br /&gt;
[http://www.wired.com/2013/10/a-closer-look-at-newtons-third-law/ A Closer Look at Newton’s Third Law]&lt;br /&gt;
&lt;br /&gt;
[http://phys.org/news/2015-05-newton-law-broken.html What happens when Newton&#039;s third law is broken?]&lt;br /&gt;
&lt;br /&gt;
[http://www.livestrong.com/article/423739-newtons-three-laws-motion-used-baseball/ How Are Newton&#039;s Three Laws of Motion Used in Baseball?]&lt;br /&gt;
&lt;br /&gt;
[https://www.newscientist.com/article/dn24411-light-can-break-newtons-third-law-by-cheating/ Light can break Newton’s third law – by cheating]&lt;br /&gt;
&lt;br /&gt;
[http://science360.gov/obj/video/d0e16d27-05d4-4511-9394-2758aa066981/science-nfl-football-newtons-third-law-motion Science of Football]&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
Internet resources on this topic&lt;br /&gt;
&lt;br /&gt;
[http://teachertech.rice.edu/Participants/louviere/Newton/law3.html The Third Law of Motion]&lt;br /&gt;
&lt;br /&gt;
[http://www.physicsclassroom.com/class/newtlaws/Lesson-4/Newton-s-Third-Law Newton&#039;s Third Law of Motion]&lt;br /&gt;
&lt;br /&gt;
[https://www.grc.nasa.gov/www/k-12/WindTunnel/Activities/third_law_motion.html Newton&#039;s Third Law of Motion]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
This section contains the the references you used while writing this page&lt;br /&gt;
&lt;br /&gt;
Knight, R., &amp;amp; Jones, B. (n.d.). College physics: A strategic approach (Third edition, Global ed.).&lt;br /&gt;
&lt;br /&gt;
http://www.physicsclassroom.com/class/newtlaws/Lesson-4/Newton-s-Third-Law&lt;br /&gt;
&lt;br /&gt;
https://www.grc.nasa.gov/www/k-12/airplane/newton3.html&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/newt.html&lt;br /&gt;
&lt;br /&gt;
http://science360.gov/obj/video/d0e16d27-05d4-4511-9394-2758aa066981/science-nfl-football-newtons-third-law-motion&lt;br /&gt;
&lt;br /&gt;
http://www.livescience.com/46561-newton-third-law.html&lt;br /&gt;
&lt;br /&gt;
[[Category:Which Category did you place this in?]]&lt;/div&gt;</summary>
		<author><name>Ylee620</name></author>
	</entry>
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