Gravity, spring = conservative. Friction = non-conservative. Path independence.

Conservative vs Non-Conservative Forces

Question

A block is moved from point A to point B along three different paths — a straight line, a curved path, and a zigzag route. In each case, calculate the work done by (a) gravity and (b) friction. What do you observe?

This is a conceptual-plus-numerical favourite in NCERT Class 11 and appears regularly in JEE Main as a single-correct MCQ testing whether you truly understand path independence.

Solution — Step by Step

Gravity only cares about vertical displacement — the height difference between A and B. If the block drops by $h = 2\,\text{m}$ and mass $m = 1\,\text{kg}$ :

W_{\text{gravity}} = mgh = 1 \times 10 \times 2 = 20\,\text{J}

This is the same for all three paths. Gravity doesn’t know (or care) whether you took the scenic route.

Friction acts opposite to motion, so its work equals $-f \times d$ where $d$ is the actual path length. The three paths have different lengths — say $d_1 = 5\,\text{m}$ , $d_2 = 8\,\text{m}$ , $d_3 = 12\,\text{m}$ — so the work done by friction is different every time.

With $f = 3\,\text{N}$ : work by friction is $-15\,\text{J}$ , $-24\,\text{J}$ , and $-36\,\text{J}$ respectively. The path matters.

A force is conservative if the work it does depends only on the start and end points — not the path taken. Gravity satisfies this. Friction does not.

The formal test: if you go from A to B and back to A, a conservative force does zero net work over any closed loop. Friction always does negative work both ways, so net work is never zero.

Force	Work done	Path-dependent?	Type
Gravity	$mgh$	No	Conservative
Spring force	$\frac{1}{2}kx^2$	No	Conservative
Friction	$-f \cdot d$	Yes	Non-conservative
Air drag	$-bv^2 \cdot d$	Yes	Non-conservative

Answer: Gravity is conservative (same work on all paths). Friction is non-conservative (work depends on path length).

Why This Works

Conservative forces are tied to a potential energy. When gravity does positive work, gravitational PE decreases by the same amount — energy is just converting form, not disappearing. The total mechanical energy stays constant. That’s why we can write $\Delta KE + \Delta PE = 0$ for gravity.

Non-conservative forces like friction convert mechanical energy into heat. That energy is gone from the mechanical system — it can’t be recovered. This is why we can’t define a “friction potential energy.” The work-energy theorem still holds ( $W_{\text{net}} = \Delta KE$ ), but we have to account for friction’s work separately.

The mathematical fingerprint of a conservative force: if you express it as a vector $\vec{F}$ , then $\nabla \times \vec{F} = 0$ (curl is zero). For JEE Advanced, knowing this criterion is useful, though the conceptual definition is enough for JEE Main and NEET.

Alternative Method — Closed Loop Test

Instead of comparing paths A→B, test by going A→B→A (a round trip).

For gravity: going down does $+mgh$ , coming back up does $-mgh$ . Net work over the closed loop = zero. Conservative confirmed.

For friction: going one way does $-f \cdot d$ , coming back does $-f \cdot d$ again (friction always opposes motion). Net work = $-2fd \neq 0$ . Non-conservative confirmed.

The closed-loop test is faster for MCQs. If you can show net work = 0 for any closed path, the force is conservative. Spring force passes this test too — stretch it and release, and it does zero net work over a complete oscillation cycle.

Common Mistake

Students often write “friction does work only when there is motion” and then incorrectly conclude friction is conservative because “it always opposes motion.” The direction argument is irrelevant to the conservative/non-conservative classification. What matters is whether work depends on the path, not on the direction. Friction is non-conservative precisely because longer paths mean more work done against friction — same start, same end, different work.

A related slip: writing $W_{\text{gravity}} = mgs\cos\theta$ using the actual path length $s$ instead of the vertical height $h$ . This gives the right answer only for a straight inclined path. The correct formula is always $W_{\text{gravity}} = mg \times \Delta h$ , regardless of path shape.

Question

Solution — Step by Step

Why This Works

Alternative Method — Closed Loop Test

Common Mistake

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