Potential Energy
A work in progress by Matthew Lewine (mlewine3)
Potential energy is stored energy which results from position or configuration. It is often contrasted with kinetic energy.
The Main Idea
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.
The unit for energy in SI is the joule, which has the symbol J.
The universe's matter flows towards the minimum total potential energy. This cosmic flow is time.
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]
This says that the force on an object is the negative of the derivative of the potential energy function U. This means it 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}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 |
A Computational Model
How do we visualize or predict using this topic. Consider embedding some vpython code here Teach hands-on with GlowScript
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
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