Summary:
This is part of Lectures of the Topic Work, Energy and Power. You can check all lectures by clicking on the category link
1. The Prerequisite: Conservative Forces
Before defining potential energy ($U$), students must understand conservative forces. A force is conservative if the work it does on an object moving between two points is independent of the path taken.
- Examples: Gravitational force, electrostatic force, ideal spring force.
- Counter-examples (Non-conservative): Friction, air resistance, viscous drag.
Core Rule: Potential energy is only defined for conservative forces. You cannot have “frictional potential energy.”
2. Defining Potential Energy ($U$)
Potential energy is the energy stored in a system due to its configuration or position within a conservative force field.
We cannot define absolute potential energy; we can only define a change in potential energy ($\Delta U$). The fundamental physics definition is:
The change in potential energy of a system is equal to the negative of the work done by the conservative force.
$$\Delta U = U_f – U_i = -W_c$$
Or in integral form:
$$\Delta U = -\int_{\vec{r}_i}^{\vec{r}_f} \vec{F}_c \cdot d\vec{r}$$
Why the negative sign? If gravity does positive work on a falling apple (speeding it up), the system must be losing stored potential energy.
3. Gravitational Potential Energy
Gravitational PE is the energy associated with the separation of two masses.
Case A: Near the Earth’s Surface (Uniform Field)
If an object of mass $m$ is raised to a height $h$ that is very small compared to the Earth’s radius ($h \ll R_e$), the gravitational force $mg$ is effectively constant.
Assuming the reference point ($U=0$) is at the ground:
$$U = mgh$$
Case B: Universal Gravitation (General Case)
For large distances (like satellites or planets), the force of gravity follows the inverse-square law: $F = \frac{GMm}{r^2}$.
By convention, we set $U = 0$ at infinity ($r = \infty$). The potential energy of a two-mass system separated by a distance $r$ is:
$$U = -\frac{GMm}{r}$$
(Note: It is always negative because gravity is an attractive force; you must do external work to separate the masses to infinity).
4. The Relation Between Force and Potential Energy
If we know the potential energy field $U(x, y, z)$, we can find the conservative force generating it. Since $dU = -F \cdot dx$ (in 1D):
$$F_x = -\frac{dU}{dx}$$
The force is the negative gradient (slope) of the potential energy curve. In three dimensions, this requires partial derivatives:
$$\vec{F} = -\left( \frac{\partial U}{\partial x}\hat{i} + \frac{\partial U}{\partial y}\hat{j} + \frac{\partial U}{\partial z}\hat{k} \right) = -\nabla U$$
5. Equilibrium and Potential Energy Curves
When the net force on a particle is zero, it is in equilibrium. Since $F = -dU/dx$, equilibrium occurs wherever the slope of the $U-x$ graph is zero ($\frac{dU}{dx} = 0$).
There are three types of equilibrium, entirely dictated by the curvature (second derivative) of the PE graph:
| Type | Condition | U−x Graph Shape | Physical Meaning |
| Stable | $\frac{d^2U}{dx^2} > 0$ | Local Minimum (Valley) | If slightly displaced, a restoring force pushes it back. (e.g., marble in a bowl) |
| Unstable | $\frac{d^2U}{dx^2} < 0$ | Local Maximum (Hill) | If slightly displaced, force pushes it further away. (e.g., marble on an inverted bowl) |
| Neutral | $\frac{d^2U}{dx^2} = 0$ | Flat Line | If displaced, it stays in the new position. (e.g., marble on a flat table) |
JEE-Mains/NEET Practice Questions
Question 1: Definition of Potential Energy
A particle is taken from point A to point B under the influence of a conservative force field. The kinetic energy of the particle increases by $20 \text{ J}$. If no non-conservative forces act on the system, what is the change in its potential energy?
(a) $+20 \text{ J}$
(b) $-20 \text{ J}$
(c) $0 \text{ J}$
(d) Cannot be determined
Solution: (b)
By the conservation of mechanical energy (which applies when only conservative forces do work), $\Delta K + \Delta U = 0$.
Therefore, $\Delta U = -\Delta K = -20 \text{ J}$. The system lost potential energy to gain kinetic energy.
Question 2: Gradient in 1D
The potential energy of a particle in a conservative force field is given by $U(x) = 2x^3 – 3x^2 \text{ J}$, where $x$ is in meters. What is the force acting on the particle at $x = 2 \text{ m}$?
(a) $-12 \text{ N}$
(b) $12 \text{ N}$
(c) $-24 \text{ N}$
(d) $24 \text{ N}$
Solution: (a)
Force is the negative derivative of potential energy: $F = -\frac{dU}{dx}$.
$F = -\frac{d}{dx}(2x^3 – 3x^2) = -(6x^2 – 6x)$
At $x = 2 \text{ m}$: $F = -(6(2)^2 – 6(2)) = -(24 – 12) = -12 \text{ N}$.
Question 3: Identifying Equilibrium
The potential energy of a conservative system is given by $U = ax^2 – bx$, where $a$ and $b$ are positive constants. The position of stable equilibrium is:
(a) $x = \frac{b}{a}$
(b) $x = \frac{b}{2a}$
(c) $x = \frac{2b}{a}$
(d) $x = 0$
Solution: (b)
For equilibrium, net force must be zero: $F = -\frac{dU}{dx} = 0$.
$\frac{dU}{dx} = 2ax – b = 0 \implies x = \frac{b}{2a}$.
To verify it is stable, check the second derivative: $\frac{d^2U}{dx^2} = 2a$. Since $a$ is positive, the second derivative is positive ($> 0$), meaning this position is a local minimum, which corresponds to stable equilibrium.
Question 4: 2D Force Field
A potential energy field is given by $U(x,y) = 3x^2y – y^3$. Find the force vector $\vec{F}$ acting on a particle at the coordinate $(1, 2)$.
(a) $-12\hat{i} + 9\hat{j}$
(b) $-12\hat{i} – 9\hat{j}$
(c) $12\hat{i} + 9\hat{j}$
(d) $6\hat{i} – 9\hat{j}$
Solution: (a)
Use partial derivatives: $\vec{F} = -\frac{\partial U}{\partial x}\hat{i} – \frac{\partial U}{\partial y}\hat{j}$.
$\frac{\partial U}{\partial x} = 6xy$
$\frac{\partial U}{\partial y} = 3x^2 – 3y^2$
At $(1, 2)$:
$F_x = -(6)(1)(2) = -12$
$F_y = -[3(1)^2 – 3(2)^2] = -[3 – 12] = -(-9) = +9$
Therefore, $\vec{F} = -12\hat{i} + 9\hat{j}$.
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