13.9 Potential energy  (Page 3/4)

5: Potential energy is defined only in the context of conservative force. For the sake of clarity, we should also realize that the conservative force doing work is internal to the system, whose potential energy is determined. We can rewrite the defining relation of potential energy that work refers to work by conservative as :

$$\begin{array}{l}\Delta U=-W_C\end{array}$$

Change in potential energy for constant and variable forces

The relationship as given above forms the basic relation that is used to develop specific relations for different force systems. For constant conservative force, work is defined as :

$$\begin{array}{l}W=F_{C}r\cos \theta \end{array}$$

Substituting in the equation, the change in potential energy involving constant force is :

$$\begin{array}{l}\Rightarrow \Delta U=-F_{C}r\cos \theta \end{array}$$

For variable force having displacement in the same direction, the work is obtained as :

$$\begin{array}{l}\Rightarrow W=\int F_C\left(r\right)dr\end{array}$$

The change in potential energy is equal to the integral evaluated over two appropriate limits corresponding to displacement :

$$\begin{array}{l}\Rightarrow \Delta U=-\int F_C\left(r\right)dr\end{array}$$

In mechanical context, we deal with gravitational and elastic potential energies. Elastic potential energy involves any configuration, which can be deformed elastically like spring. In this module, we will drive specific expressions of potential energy for these two systems.

Gravitational potential energy

We shall evaluate expression for potential energy for the vertical motion of a ball, which is thrown up with a certain speed. The gravity "-mg" works against the motion. Importantly, gravity is a constant force and as such we can use the corresponding form for the change in potential energy. For a displacement from a height $y_1$ to $y_2$ , the change in potential energy for the "Earth - object" system is :

$$\begin{array}{l}\Rightarrow \Delta U=-(-mg)(y_2-y_1)\cos 0^0\\ \Rightarrow \Delta U=mg(y_2-y_1)\end{array}$$

The change in potential energy is, thus, directly proportional to the vertical displacement. It is clear that potential energy of "Earth - ball" system increases with the separation between the elements of the system . Evaluation of change in gravitational potential energy requires evaluation of change in displacement.

Motion of an object on Earth involves "Earth - object" as the system. Thus, we may drop reference to the system and may assign potential energy to the object itself as the one element of the system i.e. Earth is common to all systems that we deal with on Earth. This is how we say that an object (not system) placed on a table has potential energy of 10 J. We must, however, be aware of the actual context, when referring potential energy to an object.

We should also understand here that we have obtained this relation of potential energy for the case of vertical motion. What would be the potential energy if the motion is not vertical - like for motion on a smooth incline? The basic form of expression for the potential energy is :

$$\begin{array}{l}\Rightarrow \Delta U=-Fr\cos \theta \end{array}$$

Here, we realize that the right hand expression yields to zero for θ = 90°. Now, any linear displacement, which is not vertical, can be treated as having a vertical and horizontal component. The enclosed angle between gravity, which is always directed vertically downward, and the horizontal component of displacement is 90°. As such, the expression of change in potential energy evaluates to zero for horizontal component of displacement. In other words, gravitational potential energy is independent of horizontal component of displacement and only depends on vertical component of displacement.