Potential Energy

What do you do everyday? No matter how easy or difficult the things we do everyday, these could be done if we have energy. Energy is the ability to do work. It can be categorized in different types and one of these is Mechanical Energy. Mechanical energy has two types: Potential energy and Kinetic Energy. In this page, we will focus only on Potential Energy.

Potential Energy is the energy possessed by an object due to its position. It comes in two types: Gravitational Potential Energy and Elastic Potential Energy.

Gravitational Potential Energy (GPE)

Gravitational potential energy is the energy stored in an object because of its position (vertical separation from the Earth). It depends on the height of the object relative to some reference point. A reference point is the area or position where the height of the object is zero meter. If we consider the ground as the reference point, this means, anything above the ground is positive and anything below it is negative. Mathematically, GPE can be stated as

$$GPE=mgh$$

where,

$$GPE=\text{Gravitational Potential Energy}\;\text{(in Joules, J)}\\ m = \text{mass (in kg)}\\ g = \text{acceleration due to gravity}\;(9.8 m/s^2)\\ h=\text{height relative to the reference point (in m)}$$

Thus, gravitational potential energy has 3 values : negative, zero, and positive.

Example 1.

Solve for the change in GPE of a 450-kg roller coaster car starting at 38 m above the ground and ends the ride 5 m above the ground.

Given:

$$m=450\;kg\\ h_i=38 \;m\;\text{(initial height)}\\ h_f=5\;m\;\text{(final height)}\\ g=9.8\;m/s^2$$

The reference point is the ground. Since the roller coaster car is above the ground, GPE must be positive. Hence,

$$\text{GPE}=mgh\\ \; \;\;\;\quad=mg(h_f-h_i)\\ \; \;\;\;\quad=(450\;kg)(9.8\;m/s^2)(5\;m-38\;m)\\ \; \;\;\;\quad=(450\;kg)(9.8\;m/s^2)(-33\;m)\\ \; \;\;\;\quad=-145,530\;J$$

Note: Negative sign only indicates that the velocity of the roller coaster car slows down or eventuallr comes to a stop.

Example 2.

A 60-kg loaded box is pulled at a distance of 15 m up an inclined plane 20° from the ground. Calculate the change in GPE.

Given:

$$m=60\;kg\\ l=15\;m\\ \theta=20^\circ$$

First, solve for the height (h) using trigonometric ratio $$\sin \theta=h/l.$$ Thus,

$$h=l\sin\theta\\ \;\;=(15 \;m)(\sin 20^\circ)\\ \;\;=(15 \;m)(0.34)\\ \;\;=5.1 \;m$$

We can now solve for GPE using h,

$$\text{GPE}=mgh\\ \;\;\;\;\;\;\;=(60\;kg)(9.8\;m/s^2)(5.1\;m)\\ \;\;\;\;\;\;\;=2,998.8\;J$$

Elastic Potential Energy(EPE)

Elastic potential energy is the energy stored in elastic materials like springs. It depends on the amount of stretch on the material - the more stretch, the more stored energy. The EPE of a spring can be expressed in a simple equation. It can also store EPE through stretching and compression.

Spring follows the Hooke's Law which states that the amount of force is directly proportional to the amount of stretch or compression. Therefore, the further the spring is compressed, the more force is needed. This is also true in stretching a spring. Mathematically, the law can be stated as $$F_{spring}=kx.$$

Spring constant (k) is a property of springs which determines the proportion of the stretching and compression. The equation relating the amount of elastic potential energy of a spring to the amount of stretch or compression can be expressed as $$\text{EPE}=1/2 \,kx^2$$ where k is the spring constant in N/m and is the amount of stretch or compression in m.

A reference point is also important in EPE. The position where the spring is neither compressed nor streched is the reference point. This only means that when a spring is moved to the right as in stretching, x has a positive value. When a spring is compressed, has a negative value. But, it is shown in the equation that x is squared resulting to a positive value, thus, either a spring is stretched or compressed, the EPE stored is always postive. The spring is said to be in its equilibrium stage at the reference point giving EPE as zero.

Thus, elastic potential energy has only two values: positive and zero.

Example 1.

A spring is stretched with a force $$F=(50\;N/m)x$$. If the spring is stretched to 3 m, what is its elastic potential energy?

Since $$F=(50\;N/m)x$$ and $$F=kx$$, then $$k=50\;N/m$$. The elastic potential energy is

$$EPE=\frac12 kx^2\\ \;\quad\quad=\frac12(50\;N/m)(3\;m)\\ \;\quad\quad=75\;J$$

The EPE is positive since the spring is stretched.

Example 2.

A spring with a force constant of $$1.5\times 10^3$$is compressed to a distance of 1.3-cm. If the spring is released back to its equilibrium position, what is the change in elastic potential energy?

Given:

$$k=1.5 \times 10^3\;N/m\\ x_i=1.3 \;cm=0.013\;m\\ x_f=0$$

Since the spring is released back to its equilibrium position, the final distance is zero. We can compute for the change in EPE as

$$\text{EPE}=\frac12 kx^2\\ \;\;\;\, \quad=\frac12(1.5 \times 10^3\;N/m)(0-0.013\;m)^2\\ \;\;\;\, \quad=\frac12(1.5 \times 10^3\;N/m)(0.000169)\\ \;\;\;\, \quad=0.13\;J$$

The change in EPE is 0.13 J. To denote that the spring lost EPE from its initial position to its equilibrium position, we could add negative sign to the final answer, -0.13 J.