Springs and Potential Energy

[BlueEight]

Homework Statement

An ideal spring is suspended from a ceiling with a wombat attached to the end. The wombat is slowly lowered until the upward force exerted by the spring on the wombat balances the weight of the wombat. Show that the loss of gravitational potential energy equals twice the gain in the elastic potential energy of the spring-wombat-Earth system. WHY are these two quantities NOT equal?

Homework Equations

Change in gravitational potential = -mg(X_f - X_i)
Change in elastic potential = .5k(X_f² - X_i²)

The Attempt at a Solution

I realize that it is due to the fact that a person slowly lowers the wombat, thus acting as a non-conservative force, but do not know how to derive the solution.

 

[BlueEight]

I realize that the gravitational potential energy must be greater, but why is is exactly twice, no matter the mass of the object?

 

[kerimek]

To derive the solution, we can use the principle of conservation of energy. This states that the total energy of a closed system remains constant, and can only be transferred or transformed between different forms.

In this situation, we have a closed system consisting of the spring, wombat, and Earth. Initially, the only form of energy present is gravitational potential energy, given by mgh, where m is the mass of the wombat, g is the acceleration due to gravity, and h is the height at which the wombat is suspended.

As the wombat is slowly lowered, the force of gravity is doing work on the system, converting gravitational potential energy into kinetic energy. However, the spring is also being stretched, storing elastic potential energy.

At the point where the spring is fully stretched and the wombat is balanced, the system has reached equilibrium and the forces are balanced. At this point, all the gravitational potential energy has been converted into elastic potential energy.

Using the equations given in the homework statement, we can express the change in gravitational potential energy as -mg(hf-hi), where hi is the initial height and hf is the final height. Similarly, the change in elastic potential energy can be expressed as .5k(Xf^2-Xi^2), where Xi is the initial length of the spring and Xf is the final length.

Since the system is in equilibrium, the final height and final length are equal, thus we can rewrite the equations as -mg(hf-hi) = .5k(Xf^2-Xi^2). Rearranging, we get -2mg(hf-hi) = k(Xf^2-Xi^2).

This shows that the loss in gravitational potential energy, which is equal to the work done by the force of gravity, is equal to the gain in elastic potential energy, which is equal to the work done by the spring.

However, in reality, the two quantities are not exactly equal due to the presence of non-conservative forces, such as air resistance and friction, which may cause some energy to be lost from the system. Additionally, the process of slowly lowering the wombat may not be perfectly efficient, leading to some energy loss.

In summary, the two quantities are not exactly equal due to the presence of non-conservative forces and the limitations of the process, but in an ideal, frictionless and perfectly efficient system, they would be equal.