Conservative and Non-Conservative Forces

Some forces have a magical property: the work they do depends only on the starting and ending points, not on the route taken. These are conservative forces, and they allow us to define potential energy. Others (like friction) lose this property — they are non-conservative.

Concept

A force $\vec{F}$ is conservative if any one of these equivalent conditions holds:

1. The work done by $\vec{F}$ depends only on initial and final positions, not on the path taken.
2. The work done by $\vec{F}$ on any closed path is zero: $\oint \vec{F}\cdot d\vec{r} = 0$.
3. $\vec{F}$ can be derived from a scalar potential energy: $\vec{F} = -\nabla U$ (in 1D: $F = -dU/dx$).
4. $\vec{F}$ is "curl-free": $\nabla \times \vec{F} = 0$.

Examples of conservative forces:

• Gravitational force.
• Spring (elastic) force.
• Electrostatic force.

Non-conservative forces: Work depends on path; work on a closed loop is non-zero.

• Kinetic friction.
• Air resistance.
• Viscous drag.

A non-conservative force cannot be associated with a potential energy. Its work is typically converted into thermal energy (frictional heating).

Derivation

Why path-independence ↔ potential energy exists.

If $W_{AB}$ is the same for every path from $A$ to $B$, define

$U(B) - U(A) = -W_{A\to B}$

This is well-defined (path-independent). Then taking $B$ infinitesimally close to $A$ at position $\vec{r}$, $dU = -\vec{F}\cdot d\vec{r}$, giving

$\vec{F} = -\nabla U$

Conversely, if $\vec{F} = -\nabla U$, then

$W_{A\to B} = \int_A^B \vec{F}\cdot d\vec{r} = -\int_A^B \nabla U \cdot d\vec{r} = -(U_B - U_A)$

— path-independent. Closing the loop gives zero net work.

Worked Example

A block slides down a frictionless incline of height 5 m, then along a frictionless horizontal floor for 10 m. Compare the work done by gravity if (a) it took this two-segment path, (b) it had fallen straight down 5 m.

Solution:

In both cases, gravity does

$W_g = mgh = m \cdot g \cdot 5$

because gravity is conservative: only the vertical drop matters. The horizontal segment contributes zero work (gravity perpendicular to displacement).

Now repeat with friction $\mu = 0.2$ on the floor only (length 10 m), mass 2 kg, $g=10$:

Two-segment path: friction does $W_f = -\mu m g \times 10 = -0.2 \times 2 \times 10 \times 10 = -40 \, \text{J}$.

Direct drop: no friction acts. $W_f = 0$.

Friction's work depends on path — it's non-conservative.

Common Confusions

• "Friction reverses on the return trip, so net work is zero." No — friction always opposes motion. On the return trip, friction reverses direction along with the displacement, so the work is again negative. Total over a closed loop is negative.
• Normal force is technically conservative-ish. Normal force is perpendicular to motion (does no work), so it's trivially conservative — but no potential energy is associated with it.
• Tension can be conservative or not depending on whether it does net work. In an ideal pulley, internal tension does zero work; for a stretching string, it can store elastic PE.
• Air drag is non-conservative, even at low speed.

Key Takeaways

• Conservative force: work depends only on endpoints; $\oint \vec{F}\cdot d\vec{r} = 0$; $\vec{F} = -\nabla U$ exists.
• Examples of conservative: gravity, spring, electrostatic.
• Non-conservative: friction, air drag — path-dependent, no PE.
• Potential energy can be defined only for conservative forces.
• Total mechanical energy is conserved when only conservative forces do work.