= Solution
During slow peeling in <atomic force microscopy>, $n$ adjusts to minimize the <Helmholtz free energy> at each imposed height. Define the positive binding gain $\mu=k_BT[\epsilon b/\xi-(b/\xi)^2]$ for the adsorbed segment. The <stationary point> satisfies
$$
0=\left(\frac{\partial F}{\partial n}\right)_z
=\mu-\frac{3k_BTz^2}{2b^2n^2},\qquad
n_*(z)=\frac zb\sqrt{\frac{3k_BT}{2\mu}}.
$$
Since $\partial_n^2F=3k_BTz^2/(b^2n^3)>0$, this is a minimum while $0<n_*<N$. The upward pulling <force> is the height <derivative> of the minimized <free energy>. The implicit $n_*(z)$ term vanishes because $\partial_nF=0$, giving
$$
f=\frac{dF(n_*(z),z)}{dz}
=\frac{3k_BTz}{b^2n_*}
=\frac{\sqrt{6k_BT\mu}}b.
$$
Using $\xi=2b/\epsilon$,
$$
\boxed{f_{\mathrm{peel}}=\sqrt{\frac32}\,\frac{\epsilon k_BT}{b}.}
$$
It is independent of $z$: extra pulling height releases more <monomers> while leaving the fractional extension of the stretched part unchanged. Equivalently $F_{\min}=-N\mu+f_{\mathrm{peel}}z$. This is the <equilibrium peeling force of an adsorbed Gaussian chain>.
The plateau ends when $n_*=N$, at
$$
z_{\mathrm{det}}=Nb\sqrt{\frac{2\mu}{3k_BT}}=\frac{Nb\epsilon}{\sqrt6}.
$$
Thereafter all <monomers> are detached and the <Gaussian chain> force is $f=3k_BTz/(Nb^2)$, agreeing continuously with the plateau at $z_{\mathrm{det}}$. Very near the start and final release, finite-chain and junction effects lie outside the blob approximation.
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