Potential Energy

A stretched spring or a raised brick has the capacity to do work — energy stored by virtue of position or configuration. This is potential energy (PE), defined only for conservative forces.

Concept

For a conservative force $\vec{F}$, the potential energy $U(\vec{r})$ is defined by

$U(\vec{r}) - U(\vec{r}_0) = -\int_{\vec{r}_0}^{\vec{r}} \vec{F}\cdot d\vec{r}$

where $\vec{r}_0$ is an arbitrary reference. Equivalently,

$\vec{F} = -\nabla U \quad\text{(in 1D: } F = -\frac{dU}{dx}\text{)}$

The sign tells us that forces point toward lower potential energy — systems tend to "roll downhill" on the $U$ landscape.

Gravitational PE (near Earth's surface):

$U_g = mgh$

with $h$ measured from a chosen zero.

Spring PE (Hooke's law):

$U_s = \tfrac{1}{2}kx^2$

where $x$ is the displacement from natural length.

Gravitational PE (general, at distance $r$ from a mass $M$):

$U(r) = -\frac{GMm}{r}$

with zero at infinity.

Derivation

Spring PE. The spring force is $F = -kx$. Work done by the spring as the spring stretches from 0 to $x$:

$W_{\text{spring}} = \int_0^x (-kx')\,dx' = -\tfrac{1}{2}kx^2$

By definition $\Delta U = -W$, so taking $U(0) = 0$:

$U_s(x) = \tfrac{1}{2}kx^2$

Gravitational PE near surface. Force is $-mg\hat{j}$, and displacement is $h\hat{j}$ upward:

$W_g = (-mg)(h) = -mgh$

$U_g = -W_g = mgh$

Worked Example

A spring of stiffness 200 N/m is compressed by 0.1 m and released, launching a 0.5 kg ball horizontally on a frictionless surface. Find the speed of the ball when it leaves the spring.

Solution:

Initial PE stored in spring: $U_i = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(200)(0.01) = 1 \, \text{J}$.

When ball leaves the spring (spring back to natural length), all PE is converted to KE:

$\tfrac{1}{2}mv^2 = 1 \Rightarrow v = \sqrt{2/0.5} = \sqrt{4} = 2 \, \text{m/s}$

Common Confusions

• Zero level is arbitrary. Only differences in PE have physical meaning. Choosing the ground or table top as zero doesn't change physics.
• Sign of $U_g = -GMm/r$. With $U(\infty) = 0$, gravitational PE is negative because work must be done against gravity to move a body to infinity.
• PE is a property of the system, not of one body alone. "PE of the Earth-ball system."
• Forces point downhill in $U$. $F = -dU/dx$ — minima of $U$ are stable equilibria.

Key Takeaways

• $U(\vec{r}) = -\int_{\vec{r}_0}^{\vec{r}} \vec{F}\cdot d\vec{r}$, defined for conservative forces.
• $\vec{F} = -\nabla U$; forces point from higher to lower PE.
• Gravitational PE (near surface): $mgh$.
• Spring PE: $\tfrac{1}{2}kx^2$.
• Gravitational PE (general): $-GMm/r$.