The act of displacing an object on an ideal spring by a distance y from its equilibrium position requires work, which is recovered and changed to KE when the object is released. The restoring force is therefore conservative.
What is the work required to move an object from equilibrium to a point .69 meters from equilibrium, against a restoring force governed by the constant k = 64 Newtons/meter?
The force required to move the object against the restoring force is parallel to the displacement, so the work required from one position to another is equal to the average force multiplied by the distance moved. The force is a linear function of position so the average force is equal to the average of the forces at the two positions.
To move from equilibrium to .69 meters from equilibrium, the average force is the average of the ( 64 Newton/meter) ( .69 meter) = 44.16 Newton force at this position and the force at equilibrium, which is 0. The average force is therefore 22.08 Newtons. The distance is the .69 meters from equilibrium to the position, so the work is
This will be the potential energy at the .69 meter position.
Moving from the .69 meter position to the 1.26 meter position requires moving through a distance of ( 1.26 - .69) meters; the average force is the average of the 44.16 Newton force at .69 meters and the 1.26 meter ( 64 N/m) = 80.64 Newton force experienced at the 1.26 meter position. The work is thus
This results in a total potential energy of
{Note that the same PE would have been found by averaging the force at equilibrium and at the extreme 1.26 meter position, then multiplying by the 1.26 meters the object has been moved from equilibrium. You should verify this.}
After being released, the object moves back to the .365 meter position, where its potential energy is found by the previous method to be
At the equilibrium point, there is no potential energy; the original 93.85001 Joules of potential energy has therefore been converted completely to kinetic energy and at this point we have
Note that the total of the kinetic and potential energies at every point is 93.85001 Joules.
The work required to move against a linear restoring force F = - k y, from equilibrium to displacement y from equilibrium is
where Fave is the average force that must be exerted to move the object at constant velocity at the instant its displacement is y. This work is done on the object by the restoring force as it returns from position y to the equilibrium point, as is manifested as KE. We therefore say that the PE at position y is .5 k y^2.
The PE at positions y1 and y2 is respectively PE1 = .5 k y1^2 and PE2 = .5 k y2^2; the work required to get from y1 to y2 can be derived by using the average force and distance between these points. However this is unnecessary, since by conservation of energy the work is just the PE difference
This work is equal to the KE gained while accelerating from y2 to y1 under the influence of the restoring force.
The maximum PE will occur at the extreme point, where y = A or -A; the PE will be
Since the PE at any point is .5 k y^2, the PE loss from an extreme point to position y is
This PE loss is the KE at position y.
When y = 0, corresponding to the object being at equilibrium, we have
The figure below shows an object at positions y1 and y2 between the equilibrium and extreme positions. The PE is shown at every indicated position, as is the work from y1 to y2, from y2 to extreme position A, and from y1 to extreme position A. The KE at y1 and at y2 is also shown; note that it is equal to the work `dW from the point to the extreme point.