Conservative & Non-Conservative Forces

Orbital mechanics is a luxury purchase made possible by gravity's conservative books: mission planners compute a spacecraft's speed anywhere on its trajectory from position alone — no path history — which is why gravity-assist flybys can be designed years ahead to the second. The same license powers molecular dynamics: simulated proteins fold under force fields defined as conservative potentials, so the computer can budget energy exactly across billions of steps. And engineering's war on the non-conservative family has a scoreboard: regenerative braking in electric vehicles intercepts kinetic energy before friction brakes can burn it, recovering 60–70% of it into the battery — an explicit refusal to pay the odometer-biller. The remaining, unavoidable tolls (tyre rolling resistance, aerodynamic drag) are why your car's range figure exists at all.

Watch gravity run its books on a pendulum. Pull the bob aside, release, and let it swing for a minute: up, down, up, down, thousands of centimetres of path — yet every time the bob returns to the same height, its speed is the same as the last visit. Gravity charged the rises and refunded the falls exactly; over any closed loop of the swing, net work zero. Now watch friction run its books on a box you slide around a warehouse floor and back to the start: every metre of the trip was charged at full rate, no leg refunded anything, and the box comes home with less kinetic energy than it left with. Same loop, opposite verdicts.

Why does the loop verdict matter so much? Because of what the previous module quietly assumed. Potential energy assigns one number per position — the shelf stores $mgh$, full stop. That only works if the work to reach the shelf is the same by every route; otherwise the "balance at the shelf" would depend on the crate's biography, and the whole idea of a stored, position-labelled energy dies. Path independence is not a bonus property of gravity and springs — it is their license to own accounts. Friction's application was rejected on exactly these grounds. A useful mental sort as you read any problem: for each force acting, ask "what does this force bill by?" Height → gravity, conservative. Deformation → spring, conservative. Path length → friction and drag, non-conservative. Nothing at all (always perpendicular, or no displacement) → workless constraint: normal forces on fixed surfaces, tensions in inextensible strings doing circular duty. The workless family needs no classification drama — they never touch the energy books.

The definition, in exam-ready words:

where: $W_{A\to B}$ — the force's own work (not the net work) between the two configurations. Read it as a billing policy: a conservative force is allowed to look only at the endpoints when writing its invoice. Gravity's invoice, $-mg(h_B - h_A)$, literally contains only endpoint data; so does the spring's, $\tfrac12 k x_A^2 - \tfrac12 k x_B^2$. The loop form, which is the same statement wearing test-equipment:

where: the loop is any round trip whatsoever. Equivalence is two lines (derivation A), and the loop version is the practical instrument — you do not need to check infinitely many paths; one dishonest loop convicts a force forever, and friction is convicted by every loop there is. The license the definition grants:

where: $U$ — the potential energy of the previous module. Read the logic direction carefully: conservative force $\Rightarrow$ potential energy exists. Non-conservative $\Rightarrow$ no $U$, ever; friction's work appears in energy accounts only as a loss line, $-f \times (\text{path length})$, computed fresh for each problem's actual route.

1 · Board level — gravity audited on two routes (CBSE/ICSE pattern, 3 marks)

A $4.0$ kg crate is moved from the floor to a platform $2.5$ m high, (a) straight up by a lift, (b) along a $6.5$ m frictionless ramp. Compute gravity's work on each route and state the conclusion. ($g = 10$ m/s².)

Step 1. (a) Straight up: $W_{\rm grav} = -mgh = -4.0 \times 10 \times 2.5 = \boxed{-100\ \mathrm{J}}$.

Step 2. (b) Along the ramp: the along-ramp component of gravity is $mg\sin\theta$ with $\sin\theta = 2.5/6.5$, opposing the motion for the whole $6.5$ m: $W = -(mg \times \tfrac{2.5}{6.5}) \times 6.5 = -mg \times 2.5 = \boxed{-100\ \mathrm{J}}$. The ramp length cancels — it always does.

Step 3. Conclusion, stated in the marks-scheme sentence: gravity's work is path-independent; gravity is a conservative force, which is why a potential energy $U = mgh$ can be defined.

2 · Board level — the round trip that doesn't cancel (ICSE/AP-1 pattern)

A $2.0$ kg block is pushed $3.0$ m up a $30°$ rough incline ($\mu_k = 0.3$) and then slides back down to the start. Find the total work done over the round trip by (a) gravity, (b) friction. ($g = 10$ m/s², $\cos 30° = 0.87$.)

Step 1. (a) Gravity: up leg $-mgh = -2 \times 10 \times (3\sin 30°) = -30$ J; down leg $+30$ J. Round trip: $\boxed{0}$ — conservative, refunded in full.

Step 2. (b) Friction: $f = \mu_k mg\cos 30° = 0.3 \times 20 \times 0.87 = 5.2$ N. It opposes the motion on BOTH legs: $W = -5.2 \times 3.0 = -15.7$ J up, and $-15.7$ J again down. Round trip: $\boxed{-31\ \mathrm{J}}$ (2 s.f.).

Step 3. The two answers side by side are the module: the conservative force's loop reads zero; the non-conservative force's loop reads minus-two-legs. Those 31 J are now warmth in the incline's surface — collected by no account, refundable never.

3 · Competitive level — one force, two paths, verdict and damages (JEE-Main pattern, ~2.5 minutes)

A force acting on a particle does $+25$ J along path 1 from A to B, and $+40$ J along path 2 between the same points. (a) Can this force be conservative? (b) If the particle is taken from A to B along path 1 and returned along path 2, how much work does the force do on the loop? (c) A student proposes a potential energy $U$ for this force with $U_A - U_B = 25$ J. What goes wrong?

Step 1. (a) Conservative forces bill identically on all paths; $25 \ne 40$, so $\boxed{\text{no}}$ — one disagreeing pair of paths convicts.

Step 2. (b) Loop = out along 1 ($+25$ J), back along 2 reversed. For a position-determined force, reversing path 2 negates its work: $-40$ J. Loop total: $25 - 40 = \boxed{-15\ \mathrm{J}}$.

Step 3. (c) The proposed account matches path 1 but is falsified by path 2, which insists $U_A - U_B = 40$ J. One position pair, two demanded balances — no single-valued $U$ exists. That is precisely why the conservative condition is the entry ticket to potential energy. Timing note: ~2.5 min. Part (b)'s designed error is $25 + 40 = 65$ J (forgetting the reversal flips the sign) or $25 - 40$ reported as $+15$ (subtraction order). Write the loop as "out + (−back)" mechanically and both die.

4 · Competitive level — the warehouse loop, full ledger (JEE-Main/NEET pattern, ~3.5 minutes)

A $10$ kg carton is dragged at constant speed around a rectangular route ($8.0$ m × $4.0$ m) on a level floor ($\mu_k = 0.40$) and returned exactly to its start. ($g = 10$ m/s².) Find, for the complete loop: (a) friction's work, (b) gravity's and the normal force's works, (c) the work done by the person, and (d) state where the person's energy has gone.

Step 1. (a) Perimeter $L = 2(8+4) = 24$ m; $f = \mu_k mg = 0.40 \times 100 = 40$ N, opposing every step: $W_{\rm fric} = -40 \times 24 = \boxed{-960\ \mathrm{J}}$.

Step 2. (b) Gravity and normal are perpendicular to every step of the level route: $\boxed{0}$ each. (Also: gravity is conservative and the loop is closed — either reason earns the mark.)

Step 3. (c) Constant speed → zero net work over the loop (work–energy theorem). So $W_{\rm person} + W_{\rm fric} = 0$: $W_{\rm person} = \boxed{+960\ \mathrm{J}}$.

Step 4. (d) The carton ends where it started, at the same speed: no kinetic energy gained, no potential energy banked (height unchanged). All 960 J passed through the carton into heat at the sliding surfaces — the signature of a non-conservative loop. A conservative world would have handed the person their joules back at the finish line. Timing note: ~3.5 min. The conceptual mark is (d): "work done against friction" is not stored anywhere — it is gone as heat. Options claiming the carton "has 960 J of energy due to the work done on it" test exactly this.

Classifying by vibes. "It slows things down → non-conservative" fails both ways: gravity slows every rising ball (conservative), and a belt's friction speeds up the parcel (still non-conservative). The only honest tests are path-comparison and the loop.

One leg's sign as evidence. Negative work on some stretch proves nothing — see the pitfall box. Ask for the round trip.

Refunding friction. On up-and-back trips, friction's two legs ADD ($-2fd$); they never cancel. Writing $0$ for friction's round trip because "it reverses direction" inverts the truth: reversing with the motion is exactly why both legs charge.

Force of the person/engine treated as conservative. Applied forces bill however the agent pleases — no account exists. In energy chains they appear as sources (positive) or brakes (negative), computed from the problem, never from a stored $U$.

Normal force and tension dragged into the fight. On fixed surfaces and taut whirling strings these do zero work on every path — there is nothing to classify and no energy line to write. (A lift floor's normal force on a passenger does work — but that is the moving-surface exception from the Work module, not a conservativeness question.)

"Work done against friction is stored as heat energy of the block." Half-marks at best: the energy leaves the mechanical books entirely, warming both surfaces; it is not recoverable, not a potential energy, and not retrievable by reversing the motion. Say dissipated.

Spring force suspected because it changes direction. The spring pushes when compressed and pulls when stretched — but its invoice still reads endpoint deformations only: $\tfrac12 kx_i^2 - \tfrac12 kx_f^2$. Conservative. Direction changes with position are fine; changes with direction of travel (friction's habit) are the crime.

Loop-test misfires. The loop must return the configuration (position — and for springs, deformation) to the start. Out-and-back along the same line IS a legitimate loop, and friction already fails on it.

Texts. NCERT Physics Class 11, Work, Energy and Power chapter (conservative and non-conservative forces) · HC Verma, Concepts of Physics Vol. 1, Ch. 8 · Halliday–Resnick–Walker Ch. 8 (the classification this module downshifts).

Provenance. All results are standard. A verifier should recompute: the 3-4-5 gravity loop (0) and friction loop ($-24$ J); example 1's ramp cancellation ($-100$ J both routes); example 2's pair ($0$ vs $-31$ J); example 3's loop ($-15$ J); example 4's ledger ($-960$, $0$, $+960$ J).

Prerequisites. Work Done by a Force · Potential Energy: Gravitational & Elastic. Leads to. Conservation of Mechanical Energy & Efficiency · Potential-Energy Diagrams & Equilibrium · ARCHETYPE: Energy Accounting on Rough Inclines · Conservative Forces & Potential Energy (UG deep dive).