 Potential Energy Functions Printer Friendly Version
In physics, the use of the term mechanical energy usually involves three types of energies: potential gravitational energy, kinetic energy, and elastic potential energy. Although potential energy is often represented by the expression PE, in this lesson we will use the variable U; similarly, kinetic energy will be represented by the variable K.

In the absence of non-conservative, or dissipative forces, these energies obey the law of conservation of energy, or ΔU + ΔK = 0. That is, when a system is only acting under the influence of conservative forces its total energy content never changes, the energy just converts between forms. Let's practice with an example of a problem using conservation of energy and a potential energy graph.

Refer to the following information for the next five questions.

A 5-kg mass moving along the x-axis passes through the origin with an initial velocity of 3 m/sec. Its potential energy as a function of its position is given in the graph shown below. How much total energy does the mass have as it passes through the origin?

 Between 2.5 meters and 5 meters is the mass gaining speed or losing speed?

 How fast is it moving at 7.5 meters?

 How much potential energy would have to be present for the mass to stop moving?

 At what position would this occur if the graph were to continue past 15 meters?

Force and Position Relationships

To look into this more carefully, let's re-examine some important graphs for a vibrating spring. Notice that the position and acceleration/force graphs are 180º out-of-phase: when the spring's displacement from its equilibrium is UP, the restoring force and acceleration are DOWN and vice-versa.

Our next graphs draw our attention to the spring's displacement, energy modes and restoring force. Notice that

• when the spring is either in a state of maximum extension or compression its potential energy is also a maximum
• when the spring's displacement is DOWN the restoring force is UP
• when the potential energy function has a negative slope, the restoring force is positive and vice-versa
• when the restoring force is zero, the potential energy is zero
• at any point in the cycle, the total energy is constant, U + K = Umax = Kmax

Force Functions

Our next step will be to show that a function representing the instantaneous values of the restoring force can be expressed as the negative of the derivative of our potential energy function Remember our two relationships involving work  The work done by a conservative force decreases an object's potential energy while it is increasing its kinetic energy   Defining the initial potential energy Uo = 0, gives us Using the calculus, we see that our desired expression of the instantaneous restoring force being equal to the negative derivative of the potential energy function. Let's practice this relationship with an example.

Refer to the following information for the next three questions.

Based on the following potential energy graph, calculate the force acting on the system during each segment.
 0 £ x < 1

 1 < x < 3

 3 < x £ 6

 segment U(x) 0 ≤ x < 11 < x < 33 < x ≤ 6 4 - 3.5x ½(x - 2)2 ½x - 1 States of Equilibrium

On a potential energy graph, when the function's derivative is equal to zero, then the net force acting on the system is equal to zero. When an object is located at one of these positions or in one of these regions it is said to be in a state of equilibrium: stable, unstable, dynamic, and static (or neutral).

On the following diagram,
• x3 and x5 are points of stable equilibrium - bowls. If the system is slightly displaced to either side the forces on either side will return the object back to these positions.
• x4 is a position of unstable equilibrium - a crest or peak. If the object is displaced ever so slightly from this position, the internal forces on either side will act to encourage further displacement instead of returning it back to x4.
• x6 is a position of either dynamic or static (neutral) equilibrium - a plateau. Since there is no net force acting on the object it must either possess only potential energy and be at rest or, it also possesses kinetic energy and must be moving at a constant velocity. If a mass were to be released from rest where the grey line meets our potential energy curve, why would the region marked off by the grey line be called an energy well? Related Documents