Work Done Against Friction in a Resistive Medium: MDCAT Physics notes

Conservative and non-conservative forces

A force is conservative if the work it does in moving a body between two points is independent of the path, and the work done round any closed path is zero. The energy spent against it is stored as potential energy and can be fully recovered.

A force is non-conservative if the work depends on the path taken. Energy spent against it is not recoverable as mechanical energy; it turns mainly into heat (and some sound).

Conservative	Non-conservative
Gravitational force	Friction
Electric force	Air resistance (drag)
Elastic (spring) force	Viscous force in a fluid
Work round a closed path = 0	Work round a closed path $\neq 0$

Energy with friction present

The law of conservation of energy still holds, but part of the energy is used up in doing work against friction. If a body of mass $m$ falls from height $h$ through a resistive medium (air) that exerts an average force $f$, then

$$\text{loss of P.E.} = \text{gain of K.E.} + \text{work against friction}$$

$$mgh = \tfrac12 mv^2 + fh$$

The work done against friction is simply force × distance: $W_f = fh$ (or $fS$ for a distance $S$ along a surface). It is a product, not a sum or ratio.

For a body sliding a distance $S$ down a rough incline of height $h$:

$$mgh = \tfrac12 mv^2 + fS, \qquad f = \mu N = \mu mg\cos\theta$$

Using kinetic energy loss

When a body starts and ends at the same level, its potential energy is unchanged. Any drop in kinetic energy must equal the work done against the resistance:

$$W_f = \tfrac12 m(v_i^2 - v_f^2)$$

Worked example. A 0.2 kg ball is thrown at $20\ \mathrm{m\,s^{-1}}$ and returns to the same height at $15\ \mathrm{m\,s^{-1}}$. Work against air resistance $= \tfrac12(0.2)(400 - 225) = 0.1 \times 175 = 17.5\ \mathrm{J}$.

Worked example. A 2 kg stone falls 10 m through air and reaches the ground at $12\ \mathrm{m\,s^{-1}}$ ($g = 10\ \mathrm{m\,s^{-2}}$). P.E. lost $= 200$ J, K.E. gained $= \tfrac12(2)(144) = 144$ J, so work against air $= 56$ J and average resistance $f = 56/10 = 5.6$ N.

Key formulas

• $W_f = fS$ (work against friction).
• $mgh = \tfrac12 mv^2 + fh$ (falling through a resistive medium).
• Same starting and final level: $W_f = \tfrac12 m(v_i^2 - v_f^2)$.
• $f = \mu N$.

Common MDCAT traps

• Friction is the non-conservative force; electric, spring and gravitational forces are conservative.
• Work against friction is $fh$ (product), not $f + h$ or $f/h$.
• Convert grams to kilograms before using $\tfrac12 mv^2$; 10 g is 0.01 kg.
• Subtract the squares of speeds, not the speeds themselves: $v_i^2 - v_f^2 \neq (v_i - v_f)^2$.
• Energy lost to friction is not destroyed; it becomes heat.

Quick revision

• Conservative force: path-independent work, zero over a closed loop.
• Friction and air drag are non-conservative.
• Work against friction = $f \times$ distance.
• $mgh = \tfrac12 mv^2 + fh$.
• With friction, mechanical energy decreases but total energy is conserved.