= Solution
Put $V=k_1x_1$ and $F_*=Nf_0x_2/\Delta$. The <constant-force spatially gated motor> produces
$$
F_m(u)=F_*\left(1-e^{-V/u}\right),\qquad
\boxed{F_{\max}=\lim_{u\to0^+}F_m(u)=\frac{Nf_0x_2}{\Delta}.}
$$
The limit is approached after the periodic steady state is established; at exactly $u=0$ arbitrary initially bound heads in inactive phase intervals need not equilibrate. The curve decreases from $F_{\max}$ and has formal large-$u$ tail $F_{\max}V/u$. Extrapolation of that tail beyond the small-$u$ detachment assumptions is not controlled.
\b[Sign inconsistency in the printed force balance.] For a positive <viscous drag coefficient> $\gamma$, the signed <drag force> is $-\gamma u$. The printed $F_{\rm ext}+F_m+\gamma u=0$ cannot give the requested positive no-load contraction speed: both last terms are positive when $u>0$. With $F_{\rm ext}=-F_{\max}/2$, the left side starts at $F_{\max}/2>0$ and may have zero, one or two positive roots depending on parameters; the requested usual load construction is not implied by that equation.
The physical <linear-drag load curve for a spatially gated motor> uses
$$
\boxed{F_{\rm ext}+F_m(u)-\gamma u=0.}
$$
On the same axes plot the positive drag magnitude $D(u)=\gamma u$ and the decreasing motor <force> $F_m(u)$. The no-load speed $u_0$ is their intersection. For opposing load $F_{\max}/2$, intersect $F_m(u)$ with the parallel raised line $D(u)+F_{\max}/2$; equivalently intersect $D(u)$ with $F_m(u)-F_{\max}/2$. Drop verticals to the speed axis. The loaded speed obeys $0<u_{1/2}<u_0$. Each intersection is unique because $F_m$ decreases and $D$ increases. The signed drag is also indicated below the axis in the sketch.
\Image[/past-exam-of-the-flows-fluctuations-and-complexity-and-symmetry-and-relativity-paper-of-part-b-of-the-physics-course-of-the-university-of-oxford/2011-motor-load.png]
{title=Motor-force and drag construction after correcting the printed drag sign}
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