Not all forces are conservative. Friction, air drag, and other dissipative forces depend on velocity and cannot be derived from a potential energy function. When non-conservative forces act, mechanical energy is not conserved — it is converted to heat, sound, and deformation. This section develops methods to account for dissipation and shows how to apply energy conservation even when friction is present.

Characteristics of non-conservative forces

A force is non-conservative if:

  1. Path-dependent: The work done depends on the path taken, not just the start and end points
  2. Dissipative: The force removes mechanical energy from the system, converting it to heat
  3. Velocity-dependent: The force depends on \( \mathbf{v} \), not just on position \( \mathbf{r} \)
  4. Cannot be expressed as a gradient: There is no potential energy function \( U \) such that \( \mathbf{F} = -\nabla U \)

Examples of non-conservative forces:

Why are they non-conservative? Consider a ball rolling on a rough floor from point A to point B:

Different paths dissipate different amounts of energy. This violates the hallmark of conservative forces (path independence).

The work–energy theorem with friction

When friction and other non-conservative forces act, the generalized work–energy theorem is:

\[ W_{\text{non-conservative}} = \Delta E_{\text{mech}} = \Delta(KE + U) \]

Equivalently:

\[ E_f = E_i + W_{\text{non-conservative}} \]

Since friction opposes motion, \( W_{\text{friction}} < 0 \), and:

\[ E_f < E_i \]

The mechanical energy decreases. The "lost" energy is converted to internal energy (heat):

\[ Q = -W_{\text{friction}} = E_i - E_f \]

Example: Block sliding on a rough surface

A \( 10 \) kg block is pushed across a rough floor with initial velocity \( v_i = 5 \) m/s. The coefficient of kinetic friction is \( \mu_k = 0.3 \). How far does it slide before stopping?

Friction force: \( f_k = \mu_k mg = 0.3 \times 10 \times 9.8 = 29.4 \) N

Work done by friction: \( W_f = -f_k \times d = -29.4 d \) (negative because friction opposes motion)

Energy equation:

\[ E_f = E_i + W_f \]

At the stopping point, \( v_f = 0 \), so \( E_f = 0 \) (taking ground as reference for potential energy):

\[ 0 = \frac{1}{2} \times 10 \times 5^2 + (-29.4 d) \]

\[ 0 = 125 - 29.4 d \]

\[ d = \frac{125}{29.4} \approx 4.3 \text{ m} \]

The block slides about 4.3 m before friction brings it to rest.

Heat generated: \( Q = f_k \times d = 29.4 \times 4.3 = 126 \) J (approximately equal to the initial kinetic energy).

Air drag and terminal velocity

When an object falls through the air, it experiences both gravitational force and air drag. The drag force typically has the form:

\[ F_{\text{drag}} = -bv \quad \text{(linear drag, low speeds)} \]

or

\[ F_{\text{drag}} = -cv^2 \quad \text{(quadratic drag, high speeds)} \]

where b and c are positive constants (drag coefficients).

Terminal velocity

Eventually, the drag force balances gravity, and the net force becomes zero. The object then falls at constant terminal velocity \( v_{\text{term}} \).

For linear drag:

\[ mg = bv_{\text{term}} \implies v_{\text{term}} = \frac{mg}{b} \]

For quadratic drag:

\[ mg = cv_{\text{term}}^2 \implies v_{\text{term}} = \sqrt{\frac{mg}{c}} \]

Example: Falling raindrop

A raindrop of mass \( m = 0.05 \) g = \( 5 \times 10^{-5} \) kg falls through air. Assuming quadratic drag with \( c = 2 \times 10^{-4} \) kg/m, what is its terminal velocity?

\[ v_{\text{term}} = \sqrt{\frac{mg}{c}} = \sqrt{\frac{5 \times 10^{-5} \times 9.8}{2 \times 10^{-4}}} \]

\[ = \sqrt{\frac{4.9 \times 10^{-4}}{2 \times 10^{-4}}} = \sqrt{2.45} \approx 1.57 \text{ m/s} \]

A typical raindrop reaches terminal velocity of about 1.6 m/s, which is slow enough not to hurt you (or the plant below).

Energy considerations with friction

When a system has both conservative and non-conservative forces, we can write:

\[ \Delta E_{\text{mech}} = W_{\text{non-conservative}} \]

Or rearranging:

\[ (KE_f + U_f) - (KE_i + U_i) = -Q \]

where \( Q \) is the energy dissipated (heat generated).

This is a generalization of mechanical energy conservation: mechanical energy decreases by exactly the amount dissipated to heat.

Example: Pendulum with air resistance

A \( 0.5 \) kg pendulum bob swings from an initial height \( h = 0.5 \) m (at rest) to a final height \( h_f = 0.3 \) m (at rest again, after oscillating and losing energy to air drag). How much energy was dissipated?

Initial mechanical energy: \[ E_i = mgh = 0.5 \times 9.8 \times 0.5 = 2.45 \text{ J} \]

Final mechanical energy: \[ E_f = mgh_f = 0.5 \times 9.8 \times 0.3 = 1.47 \text{ J} \]

Energy dissipated: \[ Q = E_i - E_f = 2.45 - 1.47 = 0.98 \text{ J} \]

About 40% of the initial energy was dissipated to air resistance and internal friction in the pivot.

Connection to thermodynamics

The dissipation of mechanical energy into heat is the bridge between mechanics and thermodynamics. The first law of thermodynamics states:

\[ \Delta U_{\text{internal}} = Q - W \]

where \( U_{\text{internal}} \) is the internal (thermal) energy. When friction dissipates mechanical energy, that energy becomes internal (thermal) energy of the system and surroundings, increasing the total entropy of the universe (second law of thermodynamics).

What you should be able to do

After this section, you should be able to:

Preview of §8.6: Energy Diagrams and Qualitative Analysis

Energy diagrams provide a visual way to analyze motion without solving equations of motion. §8.6 develops this technique in detail, showing how to identify allowed and forbidden regions, find turning points, and qualitatively predict motion from the shape of the potential energy curve.

📚 See also: Halliday Vol 1, Ch 8, §8.6 — Non-Conservative Forces and the Work–Energy Theorem. 📖 Open reference: OpenStax University Physics Vol 1 — Chapter 8.6: Non-Conservative Forces. 🎓 Video: MIT 8.01 Lecture 15 (Walter Lewin) — "Friction and Non-Conservative Forces".

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