Prerequisites
Before studying this topic, review:
- Work Done by Forces - Work by gravity and springs
- Kinetic Energy - Energy concepts
What is Potential Energy?
Potential energy is energy stored in a system due to the position or configuration of its parts. Unlike kinetic energy (energy of motion), potential energy represents the potential to do work.
Key Characteristics
- Depends on position or configuration
- Associated with conservative forces only
- Can be positive, negative, or zero
- Reference point is arbitrary (only changes matter)
Gravitational Potential Energy
Near Earth’s Surface
For a body of mass $m$ at height $h$ above a reference level:
$$\boxed{U_g = mgh}$$where:
- $m$ = mass (kg)
- $g$ = acceleration due to gravity (m/s²)
- $h$ = height above reference (m)
The choice of reference (where $h = 0$) is arbitrary. Usually choose:
- Ground level
- Starting position
- Lowest point in the problem
Interactive Demo: Visualize Potential Energy
Explore how potential energy changes with height and spring compression.
General Gravitational PE
For two masses $M$ and $m$ separated by distance $r$ (as discussed in Gravitation):
$$U = -\frac{GMm}{r}$$The negative sign indicates a bound system (energy needed to separate).
Spring Potential Energy
For a spring stretched or compressed by $x$ from its natural length:
$$\boxed{U_s = \frac{1}{2}kx^2}$$where:
- $k$ = spring constant (N/m)
- $x$ = extension or compression (m)
Properties
- Always positive (or zero at natural length)
- Symmetric for extension and compression
- Maximum when spring is most deformed
- Zero at natural length ($x = 0$)
Relation Between Force and Potential Energy
For conservative forces, there’s a fundamental relationship:
$$\boxed{F = -\frac{dU}{dx}}$$In three dimensions:
$$\vec{F} = -\nabla U = -\left(\frac{\partial U}{\partial x}\hat{i} + \frac{\partial U}{\partial y}\hat{j} + \frac{\partial U}{\partial z}\hat{k}\right)$$Interpretation
- Negative sign: Force points in direction of decreasing PE
- Force is stronger where PE changes rapidly (steep slope)
- At equilibrium points ($F = 0$), PE has maximum, minimum, or inflection
Examples
- Gravity ($U = mgh$): $F = -mg$ (force downward)
- Spring ($U = \frac{1}{2}kx^2$): $F = -kx$ (restoring force)
Work Done and Potential Energy Change
For conservative forces:
$$\boxed{W_{conservative} = -\Delta U = U_i - U_f}$$Proof
$$W = \int F\,dx = \int -\frac{dU}{dx}dx = -(U_f - U_i) = -\Delta U$$Equilibrium Analysis Using PE
This analysis connects to Newton’s Laws where $F = 0$ at equilibrium.
Types of Equilibrium
| Type | $F = 0$ | $\frac{d^2U}{dx^2}$ | PE Curve | Stability |
|---|---|---|---|---|
| Stable | Yes | $> 0$ | Minimum | Returns when displaced |
| Unstable | Yes | $< 0$ | Maximum | Moves away when displaced |
| Neutral | Yes | $= 0$ | Flat | Stays at new position |
Example: PE Curve Analysis
For $U(x) = x^3 - 3x$:
- $F = -\frac{dU}{dx} = -(3x^2 - 3) = 3 - 3x^2$
- Equilibrium at $x = \pm 1$ (where $F = 0$)
- $\frac{d^2U}{dx^2} = 6x$
- At $x = 1$: $\frac{d^2U}{dx^2} = 6 > 0$ → Stable
- At $x = -1$: $\frac{d^2U}{dx^2} = -6 < 0$ → Unstable
Common Potential Energy Functions
| System | Potential Energy | Force |
|---|---|---|
| Gravitational (near Earth) | $mgh$ | $-mg$ |
| Spring | $\frac{1}{2}kx^2$ | $-kx$ |
| Gravitational (general) | $-\frac{GMm}{r}$ | $-\frac{GMm}{r^2}$ |
| Electric | $\frac{kq_1q_2}{r}$ | $\frac{kq_1q_2}{r^2}$ |
Worked Examples
Example 1: Spring Compression
Solution:
$$U = \frac{1}{2}kx^2 = \frac{1}{2}(400)(0.15)^2 = \frac{1}{2}(400)(0.0225) = \boxed{4.5 \text{ J}}$$Example 2: Force from PE
Solution:
$$F = -\frac{dU}{dx} = -(6x + 5)$$At $x = 2$:
$$F = -(6 \times 2 + 5) = -17 \text{ N}$$The force is $\boxed{17 \text{ N}}$ in the negative x-direction.
Example 3: Equilibrium Points
Solution:
Find equilibrium ($F = 0$):
$$F = -\frac{dU}{dx} = -(4x^3 - 8x) = -4x(x^2 - 2) = 0$$ $$x = 0, \pm\sqrt{2}$$Check stability ($\frac{d^2U}{dx^2}$):
$$\frac{d^2U}{dx^2} = 12x^2 - 8$$
- At $x = 0$: $12(0) - 8 = -8 < 0$ → Unstable
- At $x = \pm\sqrt{2}$: $12(2) - 8 = 16 > 0$ → Stable
Practice Problems
A 2 kg ball is dropped from 5 m. What is its PE at the top and bottom? (Take bottom as reference)
A spring (k = 200 N/m) stretched from 10 cm to 30 cm. Calculate the change in PE.
Given $U(x) = x^2 - 6x + 10$ J, find the position of stable equilibrium and the force at x = 5 m.