13.9 Potential energy  (Page 4/4)

This is an important result as it renders the calculation of gravitational potential lot easier and independent of actual path of motion. The change in gravitational potential energy is directly proportional to the change in vertical altitude. As a matter of fact, this result follows from the fact that the work done by the gravitational force is directly proportional to the change in vertical displacement.

Reference potential energy

In order to assign unique value of potential energy for a specific configuration, we require to identify reference gravitational potential energy, which has predefined value - preferably zero value. Let us consider the expression of change in potential energy again,

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

The important aspect of this relation is that change in displacement is not dependent on the coordinate system. Whatever be the reference, the difference in displacement $(y_2-y_1)$ remains same. Thus, we are at liberty as far as choosing a gravitational potential reference (and preferably assign zero value to it). Let us consider here the motion of ball which rises from $y_1=2m$ to $y_2=5m$ with reference to ground. Here,

$$\begin{array}{l}\Rightarrow \Delta U=mg(y_2-y_1)\\ \Rightarrow \Delta U=mg(5-2)=3mg\end{array}$$

Let us, now consider if we measure displacement from a reference, which is 10 m above the ground. Then,

$$\begin{array}{l}{y}_1=-(10-2)=-8m\\ y_2=-(10-5)=-5m\end{array}$$

and

$$\begin{array}{l}\Rightarrow \Delta U=mg(y_2-y_1)\\ \Rightarrow \Delta U=mg\{-5-(-8\left)\right\}=3mg\end{array}$$

If we choose ground as zero gravitational potential, then

$$\begin{array}{l}{y}_1=0,U_1=0,y_2=h\mathrm{(say)},\mathrm{and}{U}_2=U\end{array}$$

Gravitational potential energy of an object (i.e. "Earth - ball" system) at a vertical height "h" is :

$$\begin{array}{l}\Rightarrow U_2-U_1=mg(y_2-y_1)\\ \Rightarrow U-0=mg(h-0)\\ \Rightarrow U=mgh\end{array}$$

We must be aware that choice of reference does not affect the change in potential energy, but changes the value of potential energy corresponding to a particular configuration. For example, potential energy of the object at ground in ground reference is :

$$\begin{array}{l}{U}_g=mgh=mgx0=0\end{array}$$

The potential energy of the object at ground in reference 10 m above the ground is :

$$\begin{array}{l}{\mathrm{U\text{'}}}_g=mgh=mgx-10=-10mg\end{array}$$

Thus, we must be consistent with the choice of reference potential, when analyzing a situation.

Spring potential energy

Spring force transfers energy like gravitational force through the system of potential energy. In this sense, spring force is also a conservative force. When we stretch or compress a spring, the coils of the spring are accordingly extended or compressed, reflecting a change in the arrangement of the elements within the spring and the block attached to it. The spring and the block attached to the spring together form the system here as we consider that there is no friction involved between block and the surface. The relative change in the spatial arrangement of this system indicates a change in the potential energy of the system.

When we give a jerk to the block right ways, the spring is stretched. The coils of the spring tend to restore the un-stretched condition. In the process of this restoration effort on the part of the spring elements, spring force is applied on the block. This force tries to pull back the spring to the origin or the position corresponding to relaxed condition. The spring force is a variable force. Therefore, we use the corresponding expression of potential energy :

$$\begin{array}{l}\Rightarrow \Delta U=-\int F\left(x\right)dx\end{array}$$

$$\begin{array}{l}\Rightarrow \Delta U=-\int (-kx)dx\end{array}$$

Integrating between two positions from the origin of the spring,

$$\begin{array}{l}\Rightarrow \Delta U={\int }_{x_i}^{x_f}kxdx\\ \Rightarrow \Delta U=k[\frac{x^2}{2}{]}_{x_i}^{x_f}\\ \Rightarrow \Rightarrow \Delta U=\frac{1}{2}k({x_f}^2-{x_i}^2)\end{array}$$

Considering zero potential reference at the origin, $x_i=0,U_i=0,x_f=x\mathrm{(say)},U_f=U\mathrm{(say)},$ we have :

$$\begin{array}{l}\Rightarrow U=\frac{1}{2}k{x}^2\end{array}$$

The important aspect of this relation is that potential energy of the block (i.e. "spring - block" system) is positive for both extended and compressed position as it involves squared terms of "x". Also, note that this relation is similar like that of work except that negative sign is dropped.