Question bank — Conservative forces — path-independent work, potential energy defined

False — gravity does negative work when you rise but positive work when you fall; the sign of work per trip does not decide conservativeness, path-dependence does.

False — you need loop work to vanish for every possible closed loop; one lucky loop proves nothing.

True — $U=mgy$ is negative below your chosen $y=0$ reference; only differences $\Delta U$ are physical, so the sign of $U$ itself is a bookkeeping choice.

False — it equals the decrease: $W_{\text{cons}}=-\Delta U=U_A-U_B$, so as a ball falls, $U$ drops and gravity does positive work.

False — it does zero work always (like the magnetic force or normal force), so no energy is lost, but "zero work" is a different property from "path-independent"; such forces simply have no associated potential energy and are neither the point of the definition.

False — force is $-dU/dx$ and the derivative of a constant is zero, so the physics is untouched; this freedom is exactly why we may pick any $U=0$ reference.

False — over an out-and-back loop friction does $-2fd\ne 0$ because it opposes motion on both legs; returning to the start does not undo the loss.

False — it is determined only up to an additive constant fixed by your reference choice; the same physical point can carry different $U$ values under different references.

False — apply the 2D test $\partial F_x/\partial y$ vs $\partial F_y/\partial x$: here $\partial (-y)/\partial y=-1$ but $\partial (x)/\partial x=+1$, they disagree, so the field swirls and its loop work is nonzero.

True — a conservative force is $\vec{F}=-\nabla U$, and mixed second partials of $U$ are equal (${\partial }^{2}U/\partial x\partial y={\partial }^{2}U/\partial y\partial x$), forcing this curl-free condition.

Wrong — gravitational work depends only on the height change $h$, not the path; both routes store the same $mgh$. Geometrically, only the vertical projection of each step counts, so all paths with the same endpoints share one answer — the visual signature of a conservative field.

Sign error — the correct relation is $F=-dU/dx$ so force points downhill on the $U$ graph toward lower $U$; test on a spring: $U=\frac{1}{2}k{x}^2$ must give the restoring $F=-kx$, which only the minus sign delivers.

Wrong test — conservativeness is decided by path-dependence, not the force's name; ideal tension on an inextensible string does zero work and simply carries no potential energy.

Nonsense — friction has no potential-energy function because its loop work is nonzero; the $50\text{ J}$ became heat, not stored recoverable energy.

Sign slip — the return leg is traversed in the opposite sense, so it is $-W_2$; the loop work is $W_1-W_2$, which is zero for gravity because $W_1=W_2$.

Conflates two ideas — doing zero work is not the same as being conservative; the normal force isn't assigned a potential energy at all, it simply never appears in the energy budget.

Wrong loop chosen — on a straight out-and-back it may cancel, but on a circle the force always points along the motion, giving positive work every step and a nonzero loop, which exposes it as non-conservative.

So that force points down the potential slope toward smaller $U$, matching our intuition that a ball rolls downhill; a plus sign would predict objects fleeing up the hill.

This is the fundamental theorem for gradients: if $\vec{F}=-\nabla U$, then ${\int }_A^B\vec{F}\cdot d\vec{r}=U_A-U_B$ no matter the path, exactly as ${\int }_a^b{g}^{\prime }(x)dx=g(b)-g(a)$ ignores what happens between the endpoints.

Friction always opposes the instantaneous motion, so its work is $-fds$ where $ds$ is the arc length (distance actually travelled), accumulating over every meter; gravity's work only tracks net vertical drop, so wiggles cancel.

Experiments measure work and force, and both depend only on how $U$ changes ($-dU/dx$ and $-\Delta U$); the absolute level is set by an arbitrary reference, so it never shows up physically.

Gravity points vertically, so $\vec{F}\cdot d\vec{r}=-mgdy$; horizontal displacement has $dy=0$, giving zero dot product regardless of how far sideways you move.

A non-conservative force can accidentally give zero on a symmetric loop while giving nonzero on another; only all loops vanishing guarantees a consistent potential exists.

By Green's theorem the loop work equals the area integral of $(\partial F_y/\partial x-\partial F_x/\partial y)$; if this "swirl" is zero everywhere, every loop integral vanishes and a potential exists.

Shifting the reference adds the same constant to $U$ everywhere, which cancels in every $\Delta U$ and vanishes under the derivative in $F=-dU/dx$.

The force is zero — this is an equilibrium point (a flat spot on the $U$ curve); whether it is stable or unstable depends on whether $U$ curves up or down around it.

Exactly zero by definition, since $U=mgy$ with $y=0$ there; this is a choice, not a physical fact about that location.

No — the conservative part contributes zero, but friction contributes a negative amount, so the total loop work is negative, revealing the non-conservative piece.

No — work is not force times net displacement but the path integral $\int \vec{f}\cdot d\vec{r}$; friction accumulates loss along the whole path even though net displacement vanishes.

All of it, $\frac{1}{2}k{x}^2$, because the spring force is conservative and its loop work is zero — the stored energy converts fully back to kinetic.

Yes, trivially conservative (all loop work is zero), and its potential energy is any constant — a perfectly flat $U$, consistent with $F=-dU/dx=0$.

It does zero work on every path (always perpendicular to velocity), so it trivially has zero loop work, but it is velocity-dependent and not derivable from a position-only $U$, so it sits outside the usual conservative-force framework.

It gives $2\pi$ (twice the enclosed area by Green's theorem), a nonzero value — concrete proof that this swirling field is non-conservative even though it looks smooth.