Potential Energy - Definition, Formula, Examples, FAQs

Understanding potential energy helps us see how stored energy can be converted into motion and used in everyday activities. This concept falls under the broader category of work, energy, and power, which is a crucial chapter in Class 11 physics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and National Eligibility Entrance Test (NEET). Over the last ten years of the JEE Main exam (from 2013 to 2023), almost eight questions have been asked on this concept, and for NEET two questions were asked about this concept. This article delves into what is potential energy, change in the potential energy formula, potential energy examples, potential energy stored when a particle is placed against gravity, the formula of potential energy stored in a spring, the potential energy formula for three dimensions, potential energy curve, types of equilibrium and solved examples.

What is Potential Energy

Potential energy definition: Potential energy is the energy stored in an object because of its position or state. Potential energy is defined only for conservative forces. In the space occupied by conservative forces, every point is associated with a certain energy which is called the energy of position or potential energy. Think of a roller coaster at the top of a hill: it has a lot of potential energy due to its height. When it starts to descend, this energy transforms into kinetic energy, causing the roller coaster to speed up. Another example is a stretched rubber band; it holds potential energy that can be released to launch a paper airplane.

Change In Potential Energy Formula

Change in potential energy between any two points is defined as the work done by the associated conservative force

in displacing the particle between these two points without any change in kinetic energy.

$
U_i-U_f=\int_{r_i}^{r_f} \vec{f} \cdot \overrightarrow{d s} ....(1)
$

Where,

• $U_f-$ final potential energy
• $U_i$ - initial potential energy
• $f$ - force
• $d s$ - small displacement
• $r_i-$ initial position
• $r_f -$ final position

We can define a unique value of potential energy only by assigning some arbitrary value to a fixed point called the reference point.

Whenever and wherever possible, we take the reference point at infinite and assume potential energy to be zero there.

i.e; if take $r_i=\infty$ and $r_f=r$ then from equation (1)
$
U_r=-\int_{\infty}^r \vec{f} \cdot \overrightarrow{d r}=-W
$

In the case of conservative force (field), potential energy is equal to the negative work done in shifting the body from the reference position to the given position.

Potential Energy Examples or Types

Potential energy generally is of three types:

1. Elastic potential energy

When a body is in its natural shape, its potential energy corresponding to the molecular forces is minimal. But When the body is deformed, internal forces appear and work has to be done against these forces.

Thus, the potential energy of the body is increased. This is called elastic potential energy.

2. Electric potential energy

It is the amount of work done by external forces in bringing a body from $\infty$ to a given point against electric force.
Or It is defined as negative work done by the electric force in bringing a body from $\infty$ to that point.

• It is a Scalar quantity

• SI Unit: Joule

• Dimension : $\left[M L^2 T^{-2}\right]$

3. Gravitational potential energy

Gravitational potential energy is the amount of work done in bringing a body from $\infty$ to that point against gravitational force.

• It is a Scalar quantity

• SI Unit: Joule

• Dimension : $\left[M L^2 T^{-2}\right]$

Potential Energy Stored When a Particle is Placed Against Gravity

$$
U=-\int f d x=-\int(\mathrm{mg}) d x \cos 180^{\circ}
$$

where,

• m is the mass of the body
• g is the acceleration due to gravity
• dx is the small displacement

Potential Energy Stored In The Spring

Restoring force, $$f=-k x$$ (or spring force)

Where $\mathbf{k}$ is called the spring constant.

Work done by restoring the force
$$
W=-\frac{1}{2} k x^2
$$
Potential Energy
$$
U=\frac{1}{2} k x^2
$$

Where

• $K=$ spring constant
• $x=$ elongation or compression of spring from a natural position.

The Relation Between Conservative Force And Change In Potential Energy

For only conservative fields $\mathrm{F}$ equals the Negative of the rate of change of potential energy with respect to position.
$$
F=\frac{-d U}{d r}
$$

Potential Energy Formula for Three-Dimension

For only conservative fields $F$ equals the negative gradient $(-\vec{\nabla})$ of the potential energy.
$$
F=-\vec{\nabla} U
$$

Where $\vec{\nabla}$ is del operator

And,
$
\vec{\nabla}=\frac{d}{d x} \vec{i}+\frac{d}{d y} \vec{j}+\frac{d}{d z} \vec{k}
$

So,
$
F=-\left[\frac{d U}{d x} \vec{i}+\frac{d U}{d y} \vec{j}+\frac{d U}{d z} \vec{k}\right]
$

Where,

$\frac{d U}{d x}=$ Partial derivative of $\mathrm{U}$ w.r.t. $\mathrm{x}$ (keeping $\mathrm{y}$ and $\mathrm{z}$ constant)
$
\begin{aligned}
& \frac{d U}{d y}=\text { Partial derivative of } \mathrm{U} \text { w.r.t. } \mathrm{y} \text { (keeping } \mathrm{x} \text { and } \mathrm{z} \text { constant) } \\
& \frac{d U}{d z}=\text { Partial derivative of } \mathrm{U} \text { w.r.t. } \mathrm{Z} \text { (keeping } \mathrm{x} \text { and } \mathrm{y} \text { constant) }
\end{aligned}
$

Potential Energy Curve

A graph plotted between the potential energy of a particle and its displacement from the center of force is called a potential energy curve.

The figure shows a graph of the potential energy function U(x) for one-dimensional motion. As we know the negative gradient of the potential energy gives force.

$\frac{d U}{d x}=FS$

Nature of Force

There are mainly three types of force which are listed below:

1. Attractive force

• If $\frac{d U}{d x}$ is positive (means on increasing x, U is increasing ), then F is negative in direction i.e. force is attractive.

• In the graph, this is represented in region BC.

2. Repulsive force

• ​​​​​​If $\frac{d U}{d x}$ is negative (means on increasing x, U is decreasing), then F is positive in direction i.e. force is repulsive.

• In the graph, this is represented in the region AB.

3. Zero force

• If $\frac{d U}{d x}$ is zero (means on increasing x, U is not changing ) then F is zero

• Points B, C, and D represent the point of zero force.

• These points can be termed as a position of equilibrium.

Types of Equilibrium

If the net force acting on a particle is zero, it is said to be in equilibrium. Means For equilibrium $\frac{d U}{d x}$=0

Equilibrium of particles can be of three types.

1. Stable equilibrium

• When a particle is displaced slightly from a position, then a force acting on it brings it back to the initial position, it is said to be in the stable equilibrium position.

• $\frac{d^2 U}{d x^2}>0$ is positive. i.e; the rate of change of $\frac{d U}{d x}$ is positive

• Potential energy is minimal.

• For example: A marble is placed at the bottom of a hemispherical bowl.

2. Unstable equilibrium

• When a particle is displaced slightly from a position, then a force acting on it tries to displace the particle further away from the equilibrium position, it is said to be in unstable equilibrium.

• $\frac{d^2 U}{d x^2}$ is negative

i.e; rate of change of $\frac{d U}{d x}$ is negative

• Potential energy is maximum.

• For example: A marble balanced on top of a hemispherical bowl.

3. Neutral equilibrium

• When a particle is slightly displaced from a position then it does not experience any force acting on it and continues to be in equilibrium in the displaced position, it is said to be in neutral equilibrium. i.e; the rate of change of $\frac{d U}{d x}$ is zero.

• Potential energy is constant.

• For example: A marble is placed on a horizontal table.

Also read:

• NCERT Exemplar Solutions for All Subjects
• NCERT Notes For All Subjects
• NCERT Solutions for All Subjects

Solved Example Based On Potential Energy

Now let's understand the above concept by some solved examples:

Example 1: A spring of force constant 800 N/m has an extension of 5 cm. The work done in extending it from 5 cm to 15 cm is

1) 16 J

2) 8 J

3) 32 J

4) 24 J

Solution:

Potential Energy -

$\begin{aligned} & U_f-U_i=\int_{r_i}^{r_f} \vec{f} \cdot \overrightarrow{d s} \\ & \text {-wherein } \\ & U_f-\text { final potential energy } \\ & U_i-\text { initial potential energy } \\ & f-\text { force } \\ & d s-\text { small displacement } \\ & r_i-\text { initial position } \\ & r_f-\text { final position }\end{aligned}$

Work done = change in energy stored in the spring.

$\begin{aligned} &=\mathrm{u}_{\mathrm{f}}-\mathrm{u}_{\mathrm{i}} \\ & \omega=\frac{1}{2} k x_2^2-\frac{1}{2} k x_i^2 \\ &=\frac{1}{2} \times 800 \times\left(15 \times 10^{-2}\right)^2-\frac{1}{2} \times 800 \times\left(5 \times 10^{-2}\right)^2 \\ &=400 \times 10^{-4}(225-25)=8 \mathrm{~J}\end{aligned}$

Hence, the answer is the option (2).

Example 2: A block of mass ‘ m ‘ is attached to a spring in the natural length of spring constant ‘ k ‘. The other end A of the spring is moved with a constant velocity v away from the block. Find the maximum extension in the spring.

1) $\frac{1}{4} \sqrt{\frac{m v^2}{k}}$
2) $\sqrt{\frac{m v^2}{k}}$
3) $\frac{1}{2} \sqrt{\frac{m v^2}{k}}$
4) $2 \sqrt{\frac{m v^2}{k}}$

Solution:

Potential Energy stored in the spring -

$
U=\frac{1}{2} k x^2
$
- wherein
$K=$ spring constant
$\mathrm{x}=$ elongation or compression of spring from the natural position
During maximum extension, the block will come to rest
Now, by energy conservation,
$
\frac{1}{2} m v^2=\frac{1}{2} k x_{\max }^2 \quad \therefore X_{\max }=\sqrt{\frac{m v^2}{k}}
$

Example 3: The potential energy function for the force between two atoms in a diatomic molecule is approximately given by $U(x)=\frac{a}{x^{12}}-\frac{b}{x^6}$, where $a$ and $b$ are constants and $x$ is the distance between the atoms. If the dissociation energy of the molecule is $D=\left[U(x=\infty)-U_{\text {ot equilibrium }}\right]$, then $D$ is:

Solution:

$
\begin{aligned}
F & =\frac{-d U}{d r} \\
U & =\frac{a}{x^{12}}-\frac{b}{x^6}
\end{aligned}
$

At equilibrium $F=-d U / d x=0$
$
\begin{aligned}
& \Rightarrow a \frac{-12}{x^{13}}-b \frac{-6}{x^7}=0 \\
& \text { or, } x^6=\frac{12 a}{6 b} \text { or } x=\left(\frac{2 a}{b}\right)^{1 / 6} \\
& U_{\text {equilibrium }}=\frac{a}{\left(\frac{2 a}{b}\right)^2}-\frac{b}{\left(\frac{2 a}{b}\right)}=\frac{-b^2}{4 a} \\
&
\end{aligned}
$
$
D=U(x=\infty)-U_{\text {equilibrium }}=0-\left(-\frac{b^2}{4 a}\right)=+\frac{b^2}{4 a}
$

Example 4: The potential energy of a particle in a conservative field is given by $U=20 \frac{x y}{z}$ then, the Force acting on the particle is:
1) $\left(\frac{20 y}{z}\right) \hat{i}+\left(\frac{20 x}{z}\right) \hat{j}+\left(\frac{20 x y}{z^2}\right) \hat{k}$
2) $-\left(\frac{20 y}{z}\right) \hat{i}-\left(\frac{20 x}{z}\right) \hat{j}+\left(\frac{20 x y}{z^2}\right) \hat{k}$
3) $-\left(\frac{20 y}{z}\right) \hat{i}-\left(\frac{20 x}{z}\right) \hat{j}-\left(\frac{20 x y}{z^2}\right) \hat{k}$
4) $\left(\frac{20 y}{z}\right) \hat{i}+\left(\frac{20 x}{z}\right) \hat{j}-\left(\frac{20 x y}{z^2}\right) \hat{k}$

Solution

For the Conservative Force field, the force acting on the system -
$
f=\frac{-d U(x)}{d x}
$
- wherein

The negative of differentiation of $U(x)$ concerning $x$.
$
\begin{aligned}
\mathrm{U} & =20 \mathrm{xy} / \mathrm{z} \\
\mathrm{F} & =-\mathrm{du} / \mathrm{dr} \\
& =-\frac{\delta u}{\delta x} \hat{i}-\frac{\delta u}{\delta y} \hat{j}-\frac{\delta u}{\delta z} \hat{k} \\
& =-\frac{20 y}{z} \hat{i}-\frac{20 x}{z} \hat{j}+\frac{20 x y}{z^2} \hat{k}
\end{aligned}
$

Hence, the answer is the option (2).

Example 5: A block of mass M is attached to the lower end of a vertical spring. The spring is hung from a ceiling and has a force constant value of k. The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be:

1) 2 Mg/k

2) 4 Mg/k

3) Mg/2k

4) Mg/k

Solution:

The ball is at rest, it has maximum potential energy. When the ball is released from rest with the spring at its normal (unstretched) length it loses some potential energy and the energy of the spring increases. Hence, the loss in the potential energy of the ball is equal to the gain in the potential energy of the spring.

$\begin{aligned} & m g x=(1 \div 2) k x^2 \\ & x=\frac{2 m g}{k}\end{aligned}$

Hence, the answer is the option (1).