Potential Energy
Potential energy refers to the energy a body possesses due to its position or configuration in a force field.
This energy is connected to how objects in a system are arranged and the forces they exert on each other. If the arrangement or configuration changes, so does the potential energy.
There are two main types of potential energy:
- Gravitational potential energy.
- Elastic potential energy.
For instance, when you wind the spring of a watch, energy is stored in the spring due to its new configuration. As the spring unwinds, it powers the movement of the watch’s hands, demonstrating the stored energy’s ability to do work.
Similarly, compressing or stretching a spring stores energy, which is released when the spring returns to its original state.
This same concept applies to the compressed spring in a pistol. When fired, the energy stored in the spring propels the bullet forward at high speed.
Likewise, when a stretched bowstring is released, the stored energy pushes the arrow forward with great velocity.
Gravitational Potential Energy
Gravitational potential energy is the energy a body holds due to its position relative to the Earth’s surface.
For example, during the 1996 Olympics, when a weightlifter raised heavy weights above his head, he increased the separation between the weights and the ground, which altered the system’s gravitational potential energy. The work done by the lifter to change the position of the weights also changed their potential energy within the Earth-weight system.
In fact, the work required to lift a body from the Earth’s surface to a higher position is stored as gravitational potential energy. Once the body is allowed to fall, this energy can be converted into other forms of work. A practical application is found in dams, where water stored at great heights is used to generate electricity by driving turbines at hydroelectric power plants.
To figure out the gravitational potential energy, let’s assume the following:
$ \displaystyle m=$ mass of a body
$ \displaystyle g=$ acceleration due to gravity on the surface of Earth
$ \displaystyle h=$ height through which the body is lifted
If the height $h$ is relatively small and $g$ stays consistent, then the force required to overcome gravity is:
$ \displaystyle F\text{ }=\text{ }mg\text{ }\ldots (1)$
Since the movement happens in the same direction as the applied force, the work done can be expressed as:
\[
\text{Work done} = \text{Force} \times \text{Distance}
\]
\[
W = (F) \times h = mgh
\]
Thus, the gravitational potential energy (P.E.) of the body becomes:
Gravitational P.E. of the body $ \displaystyle =W=mgh\text{ }\ldots (2)$
At ground level, or $h = 0$,
\[
\therefore \text{Gravitational P.E.} = 0
\]
Gravitational potential energy clearly depends on both the mass of the object and its height above Earth’s surface. The higher or heavier the object, the greater the energy.
It’s also important to note that gravitational potential energy connected to an object-Earth system relies solely on its vertical position in relation to a reference point—not the horizontal position.
Key Points to Keep in Mind Regarding Potential Energy
1. Potential energy only applies to conservative forces. It doesn’t exist for non-conservative forces. For instance, when we push an object up a frictionless incline of height $h$, the work done on it is equal to $mgh$. This is the potential energy stored in the object. If there’s friction, some of the work is lost to heat, and only $mgh$ remains as potential energy.
2. Potential energy is relative to the frame of reference. It can be positive or negative. Generally, in a conservative field, potential energy is considered the negative of the work done to move an object from a reference point to a given position.
3. Objects in motion might or might not possess potential energy.
4. The formula $mgh$ represents the gravitational potential energy of a system involving the Earth and an object. The height $h$ is measured from a specific reference level that you choose in the problem. If the Earth’s surface is used as the reference, then gravitational potential energy increases linearly with height. Conventionally, potential energy is considered **positive** when an object returns to its original position on its own. It’s termed **negative** when external work is required to bring it back.
5. Energy is also stored in a system when a body is formed by bringing together small parts from faraway distances. This is called self-potential energy.
Conversion of potential energy into kinetic energy
When a body is freely falling due to the force of gravity, the total sum of its kinetic and potential energies remains constant. As the object descends, its potential energy decreases, while its kinetic energy increases by the same amount. Visually, the changes in kinetic and potential energy are represented by the dotted lines in the figure, showing that their sum, $\text{K.E.} + \text{P.E.}$, always equals $Mgh$ as illustrated by the horizontal line.
The oscillatory movement of a simple pendulum is another good example of how potential energy converts into kinetic energy and vice versa, demonstrating the conservation of mechanical energy.
Example 1: Find the velocity of the pendulum’s bob at its lowest point if it rises to a height of 10 cm. Given $g = 980 \, \text{cm/s}^2$.
Solution:
Consider the bob of a simple pendulum with mass $M$ that is initially at its lowest position $O$. When the bob swings up to a maximum height at point $A$, it reaches a height $h$ above point $O$. At point $A$, the energy is entirely potential, $Mgh$. As the bob swings back down, the potential energy converts to kinetic energy at point $O$. The energy at point $O$ is entirely kinetic, $\frac{1}{2} M v^2$.
By applying the law of conservation of energy:
\[
\dfrac{1}{2} M v^2 = Mgh
\]
Or, solving for $v$:
\[
v = \sqrt{2gh}
\]
Here, $h = 10 \, \text{cm}$, $g = 980 \, \text{cm/s}^2$.
\[
v = \sqrt{2 \times 980 \times 10}
\]
\[
v = 140 \, \text{cm/s}
\]
Example 2: A pendulum bob is released from position $A$. The pendulum is $1.5 \, \text{m}$ in length. What is the speed of the bob when it reaches its lowest point $B$, assuming 5% of its initial energy is lost to air resistance?
Solution:
Let $M$ represent the bob’s mass. The potential energy at point $A$ is:
\[
Mgh = M \times 9.8 \times 1.5 \, \text{J}
\]
At the lowest point $B$, the bob’s kinetic energy equals its potential energy at point $A$ minus the energy lost to air resistance. Since 5% of the energy is lost, the remaining energy is 95% of the initial energy:
K.E. of the bob at $\displaystyle B=M\times 9.8\times 1.5\times \dfrac{{95}}{{100}}\text{J}$
If $v$ is the speed of the bob at point $B$, the kinetic energy can be expressed as:
\[
\frac{1}{2} M v^2 = M \times 9.8 \times 1.5 \times \dfrac{95}{100}
\]
Solving for $v$:
\[
v = \left( 2 \times 9.8 \times 1.5 \times \dfrac{95}{100} \right)^{1/2}
\]
\[
v = 5.285 \, \text{m/s}
\]
