Classical-Mechanics · Unit 17 · Video 5 · Interactive Practice

Dissipation and the System Boundary: Where Friction's Energy Goes

IKey Formulas

Formula	Name	What you need
$\|W_{\text{friction}}\| = f_k\,d$	Kinetic energy the sliding object loses (point-particle model)	The kinetic friction force and the sliding distance
$\Delta E_{\text{thermal}} = \Delta E_{\text{thermal}}^{\text{block}} + \Delta E_{\text{thermal}}^{\text{surface}} = -\Delta K$	Dissipation inside the object $+$ surface system	System boundary enclosing both bodies, $W^{\text{ext}} = 0$ , $\Delta U = 0$
$0 = \Delta E_{\text{chemical}} + \Delta E_{\text{thermal}} + \Delta E_{\text{mechanical}}$	Closed-system energy balance	A boundary with nothing crossing it
$-\Delta E_{\text{chemical}} = \Delta E_{\text{thermal}} + \Delta K$	Source energy on level ground	$\Delta U = 0$ : no springs, no change in height

Key Insight: Dissipation destroys nothing — it converts coherent molecular motion into random molecular motion in both bodies. Redrawing the boundary changes no physics, only the bookkeeping: outside the boundary friction is an external force acting at an ill-defined contact; inside it, friction is internal and the system's total energy is unchanged.

IIVisualization 1 — Coherent Motion Becomes Random Motion

The block's $18\ \text{J}$ does not vanish: follow it into the random motion of the molecules of both bodies.

Slide 0.00 s Thermal split (block / surface) 50/50

💡 The conversion runs one way only: aligned molecular motion becomes random motion, but random molecular motion never spontaneously re-aligns into a $3.0\ \text{m/s}$ shove — which is exactly what makes a dissipative process irreversible.

IIIVisualization 2 — Where You Draw the Boundary

One skid, two boundaries: friction is external here and internal there, and only the bookkeeping changes.

1 · Friction is external

$\vec{f}_k$ crosses the boundary, so the surroundings act on the system.

2 · The point-particle bookkeeping still works

|W_{\text{friction}}| = f_k\,d = (9.0)(2.0) = 18\ \text{J}

which is the kinetic energy the object loses.

2 · The ledger closes

W^{\text{ext}} = 0 \;\Rightarrow\; \Delta E_{\text{sys}} = 0

\Delta E_{\text{thermal}} = -\Delta K = +18\ \text{J}

3 · One question stays open

Mechanics fixes the total but not the branching ratio at the interface; splitting that $18\ \text{J}$ between block molecules and floor molecules takes a thermal equation of state.

IVVisualization 3 — The Walking Paradox

The planted foot never slides, so friction does no work — yet the walker's kinetic energy climbs.

Stride 0.00 s

💡 Any slippage of the planted foot points backward, opposite the motion, so slippage could only drain energy — it can never be the source of the walker's kinetic energy.

VQuiz Questions

Problem 1 · Following the Skid's Energy

Given: a $4.0\ \text{kg}$ block slides across a rough level floor at $3.0\ \text{m/s}$ and stops after $d = 2.0\ \text{m}$ against a kinetic friction force $f_k = 9.0\ \text{N}$ , with the system taken to be the block and the floor — find the thermal energy generated, $\Delta E_{\text{thermal}}$ .

✅ Correct! Both routes agree:

f_k d = 18\ \text{J}

and

K_i = \tfrac{1}{2}mv_i^2 = 18\ \text{J}

❌ That is $\tfrac{1}{2}f_k d$ . The factor

\tfrac{1}{2}

belongs to a force that grows with displacement, like a spring;

f_k

is constant over the whole skid.

❌ That is $mv_i^2$ , not $\tfrac{1}{2}mv_i^2$ . Check it against the other route:

f_k d = (9.0)(2.0) = 18\ \text{J}

❌ Nothing was destroyed. The kinetic energy became random molecular kinetic energy in the block and the floor — both bodies end up warmer.

❌ Not quite. With friction internal to the system,

\Delta E_{\text{sys}} = 0

, so

\Delta E_{\text{thermal}} = -\Delta K

Show solution

With the boundary around block and floor, friction is an internal force and no external force does work, so the system's total energy is unchanged:

0 = \Delta E_{\text{sys}} = \Delta K + \Delta E_{\text{thermal}}

The block starts with

K_i = \tfrac{1}{2}mv_i^2 = \tfrac{1}{2}(4.0)(3.0)^2 = 18\ \text{J}

and ends at rest, so $\Delta K = -18\ \text{J}$ and

\Delta E_{\text{thermal}} = -\Delta K = 18\ \text{J}

The friction route gives the same number, which is why the block stops in exactly $2.0\ \text{m}$ : $f_k d = (9.0)(2.0) = 18\ \text{J}$ .

Problem 2 · How the Thermal Energy Divides

Given: the same skid, in which $18\ \text{J}$ of thermal energy appears — find how much of it ends up in the block itself.

✅ Correct! Both bodies warm, and the total is fixed at

18\ \text{J}

, but nothing in the mechanics fixes the branching ratio at the interface.

❌ Equal shares are a guess, not a result. A

50/50

split satisfies energy conservation — but so does every other split, which is exactly the point.

❌ The floor warms too. Run a hand along the skid mark: the surface molecules gained random kinetic energy as well.

❌ Friction acts on both bodies. The equal-and-opposite pair dissipates energy into the block and the floor at once.

❌ Not quite. Conservation of energy pins the total, not the share each body receives.

Show solution

Energy conservation for the block $+$ floor system gives one equation for two unknowns:

\Delta E_{\text{thermal}}^{\text{block}} + \Delta E_{\text{thermal}}^{\text{surface}} = 18\ \text{J}

Any pair of non-negative shares summing to $18\ \text{J}$ satisfies it. The dissipation happens right at the interface between the two parts, and mechanics alone does not say how the new thermal energy divides. Settling it requires an additional model — a thermal equation of state describing how the dissipated energy distributes itself among the constituent parts of the system, which depends on properties of the two materials.

Problem 3 · The Walker's Ledger

Given: a $60\ \text{kg}$ person starts from rest and reaches $1.5\ \text{m/s}$ on level ground while the body converts $270\ \text{J}$ of chemical energy; treat person $+$ air $+$ ground as a closed system with $\Delta U = 0$ — find $\Delta K$ and $\Delta E_{\text{thermal}}$ .

What is the change in kinetic energy?

What is the change in thermal energy?

✅ Correct! Of the

270\ \text{J}

released,

67.5\ \text{J}

became motion and the remaining

202.5\ \text{J}

became heat in the body and the ground.

❌ You dropped the $\tfrac{1}{2}$ .

mv^2 = 135\ \text{J}

; the kinetic energy is

\tfrac{1}{2}mv^2

❌ Check the kinetic energy. The person starts from rest, so

\Delta K = \tfrac{1}{2}mv_f^2 - 0

with

m = 60\ \text{kg}

and

v_f = 1.5\ \text{m/s}

❌ That would leave nothing for the motion. The released energy splits:

-\Delta E_{\text{chemical}} = \Delta E_{\text{thermal}} + \Delta K

❌ Sign slip.

\Delta K

is subtracted from the released energy, not added to it:

\Delta E_{\text{thermal}} = 270 - 67.5

❌ Check the thermal term. Rearrange the balance:

\Delta E_{\text{thermal}} = -\Delta E_{\text{chemical}} - \Delta K

Show solution

For the closed system, no energy enters or leaves:

0 = \Delta E_{\text{chemical}} + \Delta E_{\text{thermal}} + \Delta E_{\text{mechanical}}

Level ground and no springs give $\Delta E_{\text{mechanical}} = \Delta K$ , so moving the chemical term across:

-\Delta E_{\text{chemical}} = \Delta E_{\text{thermal}} + \Delta K

Burning fuel makes $\Delta E_{\text{chemical}} = -270\ \text{J}$ , so the energy released is $-\Delta E_{\text{chemical}} = +270\ \text{J}$ . The kinetic term is

\Delta K = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2 = \tfrac{1}{2}(60)(1.5)^2 - 0 = 67.5\ \text{J}

and therefore

\Delta E_{\text{thermal}} = 270 - 67.5 = 202.5\ \text{J}

The forward friction force from the ground never appears in this ledger: it does no work, because the planted foot's contact point undergoes no displacement. The source was internal all along.

Problem 4 · Pushed at Constant Speed

Given: you push the same $4.0\ \text{kg}$ box horizontally at constant speed for $d = 2.0\ \text{m}$ across the same floor ( $f_k = 9.0\ \text{N}$ ), with the system taken to be the box and the floor and no heat flow, $Q = 0$ — find the external work $W^{\text{ext}}$ and the thermal energy generated.

✅ Correct! Here the pusher pays the

18\ \text{J}

instead of the box's own kinetic energy — the boundary now has an external force doing work across it.

❌ The energy has to land somewhere.

\Delta K = 0

and

\Delta U = 0

, so the whole

18\ \text{J}

that crossed the boundary shows up as thermal energy.

❌ Constant speed means $\Delta K = 0$ , not zero work. Your force is

9.0\ \text{N}

forward through

2.0\ \text{m}

, so it does

+18\ \text{J}

on the system.

❌ You counted the friction force twice. At constant speed the applied force equals

f_k = 9.0\ \text{N}

, it is not added to it.

❌ Not quite. Use the first law for the system:

W^{\text{ext}} + Q = \Delta E_{\text{sys}} = \Delta K + \Delta U + \Delta E_{\text{thermal}}

Show solution

At constant speed the box has zero acceleration, so the applied force balances kinetic friction, $F = f_k = 9.0\ \text{N}$ , and the surroundings do

W^{\text{ext}} = F d = (9.0)(2.0) = 18\ \text{J}

on the system. The first law for the box $+$ floor system, with $Q = 0$ , gives

W^{\text{ext}} + Q = \Delta E_{\text{sys}} = \Delta K + \Delta U + \Delta E_{\text{thermal}}

18\ \text{J} = 0 + 0 + \Delta E_{\text{thermal}} \;\Rightarrow\; \Delta E_{\text{thermal}} = 18\ \text{J}

Compare with Problem 1: the same $18\ \text{J}$ of thermal energy, but a different payer. There the boundary was closed and the box's own kinetic energy funded the heating; here energy crosses the boundary as external work and the kinetic energy never changes. As before, how the $18\ \text{J}$ divides between box and floor still needs a thermal model.

Solved: 0 / 4