Classical-Mechanics · Unit 17 · Video 4 · Interactive Practice

Non-Conservative Forces: When Mechanical Energy Is Not Conserved

IKey Formulas

Formula	Name	What you need
$W = W_c + W_{\text{nc}}, \qquad \Delta U = -W_c$	Splitting the internal work	The internal forces, sorted by path dependence
$W_{\text{nc}} = \Delta K + \Delta U = \Delta E_m$	Non-conservative work	$K$ and $U$ at the initial and final states
$\Delta E_{\text{system}} = \Delta E_m - W_{\text{nc}} = 0$	Total energy of a closed system	Nothing crossing the boundary
$W^{\text{ext}} + Q = \Delta E_{\text{sys}}$	First law of thermodynamics	The work and the heat crossing the boundary, with signs

Key Insight: mechanical energy is one column of the ledger, not the whole book. $W_{\text{nc}}$ measures exactly how much of it made the one-way trip into non-recoverable energy — and the total never moves.

IIVisualization 1 — Why Friction Has No Potential Energy

Same start, same finish, two different routes: does friction remove the same energy along both?

Detour height (m) 0.00

💡 No potential energy function can be written for friction: the stored value would have to depend on which route the book took to arrive.

IIIVisualization 2 — The Ledger: $W_{\text{nc}} = \Delta E_m$

Kinetic energy drains away and no potential energy rises to catch it — so where does it go?

Time (s) 0.00

💡 Run the film backwards and the table would cool itself to hurl the book across the surface — every total still balances, yet it never happens. That one-way character is what makes the process irreversible.

IVVisualization 3 — Two Channels Across the Boundary

Unseal the system: external work and heat carry energy in or out, and their signs decide which way.

External work (J) +40 Heat flow Q (J) −10

VQuiz Questions

Problem 1 · Reading the Result

Given: a $2.0\ \text{kg}$ book slides straight across a level table carrying $K_i = 60.0\ \text{J}$ ; friction does $-49.0\ \text{J}$ of work on it and no potential energy changes — find $\Delta E_m$ and the kinetic energy at the finish.

✅ Correct! Mechanical energy is not conserved here — it changed by exactly the non-conservative work, and the

49.0\ \text{J}

is now non-recoverable.

❌ Total energy is conserved; mechanical energy is not.

W_{\text{nc}} = \Delta E_m

, and a nonzero

W_{\text{nc}}

makes

\Delta E_m

nonzero.

❌ Check the sign. Friction opposes the sliding at every instant, so

W_{\text{nc}} < 0

— and

\Delta E_m

inherits that sign.

❌ That is the energy removed, not the energy left. Subtract it from

K_i

K_f = 60.0 - 49.0

❌ Not quite. Start from

W_{\text{nc}} = \Delta K + \Delta U

with

\Delta U = 0

on a level table.

Show solution

The table is level, so no potential energy changes and $\Delta U = 0$ :

W_{\text{nc}} = \Delta K + \Delta U = \Delta K = \Delta E_m = -49.0\ \text{J}

K_f = K_i + \Delta K = 60.0 - 49.0 = 11.0\ \text{J}

Nothing has been lost. The non-recoverable column gained $-W_{\text{nc}} = +49.0\ \text{J}$ , so for the closed system

\Delta E_{\text{system}} = \Delta E_m - W_{\text{nc}} = -49.0 - (-49.0) = 0

The book's motion was traded for warmth in the book and the table — a transfer, not a loss.

Problem 2 · The Sign Convention

Given: a system does $30\ \text{J}$ of work on its surroundings while $80\ \text{J}$ of thermal energy flows into it — find $\Delta E_{\text{sys}}$ and $\Delta E_{\text{surroundings}}$ .

✅ Correct! The heat flowing in outweighs the work flowing out, and the surroundings' books close on the opposite entry.

❌ $W^{\text{ext}}$ is the work done by the surroundings on the system. Here the system does the work, so

W^{\text{ext}} = -30\ \text{J}

, not

+30\ \text{J}

❌ Both channels ran backwards.

Q = +80\ \text{J}

flows in and only

30\ \text{J}

of work leaves, so the system ends up richer.

❌ Both sides cannot gain.

\Delta E_{\text{system}} = -\Delta E_{\text{surroundings}}

: whatever one gains, the other loses.

❌ Not quite. Assign a sign to each channel first, then add:

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

Show solution

The system does the work, so the external-work channel runs outward and carries a minus sign; the heat channel runs inward and carries a plus sign:

W^{\text{ext}} = -30\ \text{J}, \qquad Q = +80\ \text{J}

\Delta E_{\text{sys}} = W^{\text{ext}} + Q = -30 + 80 = +50\ \text{J}

The surroundings supplied the difference:

\Delta E_{\text{surroundings}} = -\Delta E_{\text{sys}} = -50\ \text{J}

The two channels are physically different — $W^{\text{ext}}$ moves coherent, organized motion, $Q$ moves random thermal motion — but the first law adds them as plain signed numbers.

Problem 3 · A Rough Ramp

Given: a $2.0\ \text{kg}$ block starts from rest and slides $L = 3.0\ \text{m}$ down a ramp, descending $h = 1.5\ \text{m}$ in height, while a constant friction force $f = 3.0\ \text{N}$ acts along the ramp ( $g = 9.8\ \text{m/s}^2$ ) — find $\Delta E_m$ and the speed at the bottom.

What is the change in mechanical energy?

What is the speed at the bottom?

✅ Correct! Only the friction work shows up in

\Delta E_m

; the potential energy released is still fully accounted for inside the mechanical column.

❌ The potential energy drop does not belong in $\Delta E_m$ as a loss.

\Delta E_m = W_{\text{nc}}

alone —

\Delta U

is already inside

E_m

, trading against

\Delta K

❌ That is $\Delta U$ , not $\Delta E_m$ . The potential energy released goes into kinetic energy; only friction leaves the mechanical column.

❌ The factor of $\tfrac12$ was dropped.

K_f = \tfrac12 m v_f^2

, so

v_f = \sqrt{2K_f/m}

, not

\sqrt{K_f/m}

❌ That is the frictionless answer.

5.42\ \text{m/s}

comes from

K_f = 29.4\ \text{J}

; friction removed

9.0\ \text{J}

of that first.

❌ The ramp length is not the height.

\Delta U = -mgh

uses the

1.5\ \text{m}

drop; the

3.0\ \text{m}

belongs in

W_{\text{nc}} = -fL

❌ Not quite. Mechanical energy changes only by the non-conservative work:

\Delta E_m = W_{\text{nc}} = -fL

❌ Not quite. Get

K_f

from

\Delta K = \Delta E_m - \Delta U

, then invert

K_f = \tfrac12 m v_f^2

Show solution

Step 1 — the two work terms. Gravity is conservative and its work is tracked by $\Delta U$ ; friction is the only non-conservative force:

\Delta U = -mgh = -(2.0)(9.8)(1.5) = -29.4\ \text{J}, \qquad W_{\text{nc}} = -fL = -(3.0)(3.0) = -9.0\ \text{J}

Step 2 — the change in mechanical energy.

\Delta E_m = W_{\text{nc}} = -9.0\ \text{J}

Step 3 — the kinetic energy. From $W_{\text{nc}} = \Delta K + \Delta U$ ,

\Delta K = W_{\text{nc}} - \Delta U = -9.0 - (-29.4) = 20.4\ \text{J}

The block started from rest, so $K_f = 20.4\ \text{J}$ and

v_f = \sqrt{\frac{2K_f}{m}} = \sqrt{\frac{2(20.4)}{2.0}} = \sqrt{20.4} = 4.52\ \text{m/s}

Check: with no friction the block would arrive at $\sqrt{29.4} = 5.42\ \text{m/s}$ . The missing $9.0\ \text{J}$ is not gone — it now sits in the non-recoverable column, warming the block and the ramp.

Problem 4 · Around a Closed Loop

Given: a block slides $L = 4.0\ \text{m}$ down a rough incline from $A$ to $B$ , dropping $2.0\ \text{m}$ in height, and is then pushed back up the same incline to $A$ ; a constant friction force $f = 6.0\ \text{N}$ opposes the sliding on both legs — find the work each force does around the complete loop $A \to B \to A$ .

What work does gravity do around the loop?

What work does friction do around the loop?

✅ Correct! The conservative force returns everything it took; friction keeps every joule it took, so the hand had to supply

48\ \text{J}

of external work to restore the same state.

❌ The two legs have opposite signs. Gravity does

+mgh

going down and

-mgh

coming up, and those cancel exactly.

❌ The mass is not needed. Whatever

mgh

is, the descent and the climb contribute

+mgh

and

-mgh

around this loop.

❌ Zero around a closed path is the signature of a conservative force. Friction fails that test — that failure is what makes it non-conservative.

❌ That is one leg only. The block slides

4.0\ \text{m}

down and

4.0\ \text{m}

back up, so friction acts over

8.0\ \text{m}

of path.

❌ Friction never does positive work on the sliding block. It reverses with the motion, so it opposes the displacement on both legs and both contributions are negative.

❌ Not quite. Gravity's work depends only on the endpoints, and the loop ends where it began.

❌ Not quite. Friction's work depends on the path length:

W_{\text{nc}} = -f \times (\text{total distance slid})

Show solution

Gravity. Its work depends only on the height change, and the loop returns the block to its starting height:

W^g = (+mgh) + (-mgh) = 0

This is the closed-loop signature of a conservative force, and it is what allows $U^g$ to exist at all.

Friction. It reverses direction with the motion, so it opposes the displacement on both legs and its contributions add instead of cancelling:

W_{\text{nc}} = -f(L + L) = -(6.0)(8.0) = -48\ \text{J}

What it costs. The block ends in exactly the state it started in, so $\Delta E_m = 0$ over the full loop — yet $48\ \text{J}$ of non-recoverable energy was created. The system is not closed: the hand pushing the block back up did external work

W^{\text{ext}} = \Delta E_{\text{sys}} = +48\ \text{J} \quad (\text{with } Q = 0)

and that is precisely the price of the round trip. A conservative force lends energy and takes the loan back; a non-conservative force must be paid from outside every single lap.

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Non-Conservative Forces: When Mechanical Energy Is Not Conserved

IKey Formulas

IIVisualization 1 — Why Friction Has No Potential Energy

IIIVisualization 2 — The Ledger: Wnc=ΔEmW_{\text{nc}} = \Delta E_mWnc​=ΔEm​

IVVisualization 3 — Two Channels Across the Boundary

VQuiz Questions

IIIVisualization 2 — The Ledger: $W_{\text{nc}} = \Delta E_m$