= Potential energy
{title2=$\mathbf F=-\nabla V$}
{wiki}
<Potential energy> assigns <energy> to a system's arrangement, so a change of arrangement accounts for the <work> done by a <conservative force>. For a time-independent <potential energy> $V(\mathbf r)$,
$$
\mathbf F=-\nabla V,
\qquad W_{\mathrm{conservative}}=V_i-V_f=-\Delta V.
$$
The <gradient> points toward increasing <potential energy>; the <force> points toward decreasing <potential energy>. Only differences matter: adding a constant to $V$ changes neither the <force> nor its <work>. This endpoint rule follows by substituting $-\nabla V$ into the <line integral> for <work>. https://openstax.org/books/university-physics-volume-1/pages/8-2-conservative-and-non-conservative-forces[OpenStax University Physics 1, §8.2: Conservative forces and potential energy]{ref}
Near <Earth>'s surface, stipulate constant <gravitational acceleration> $g=9.81\,\mathrm{m/s^2}$ and measure height $h$ upward from a chosen zero. Then <gravitational potential energy> is $V=mgh$. Raising a $2\,\mathrm{kg}$ <mass> by $0.5\,\mathrm m$ increases its <gravitational potential energy> by $9.81\,\mathrm J$, while <gravity> does $-9.81\,\mathrm J$ of <work>. These are computed model values, not measurements of a particular lift. An ideal <spring> obeying <Hooke's law> instead has $V=kx^2/2$, with <displacement> $x$ from its relaxed state and <spring constant> $k$: $k=100\,\mathrm{N/m}$ and $x=0.2\,\mathrm m$ give $V=2\,\mathrm J$ relative to $x=0$. https://openstax.org/books/university-physics-volume-1/pages/8-1-potential-energy-of-a-system[OpenStax University Physics 1, §8.1: Gravity and spring examples]{ref}
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