= Solution
No slip and impermeability give dimensional $u=-U$, $w=0$ on the upward-moving wall, so $\boxed{\widehat u=-1,\ \widehat w=0}$ there. For the interface normal, $h_x=\delta\widehat h_{\widehat x}$, and hence
$$
\mathbf n=(\delta\widehat h_{\widehat x},-1)+O(\delta^2)
$$
in its $x,z$ components. The relevant stresses scale as
$$
\sigma_{xx}=-\Pi\widehat p+O(\mu U/L),\qquad
\sigma_{xz}=\frac{\mu U}{\delta L}\widehat u_{\widehat z}+O(\delta\mu U/L),
\qquad\sigma_{zz}=-\Pi\widehat p+O(\mu U/L).
$$
Multiplying by the inward normal gives the leading <traction>
$$
\boxed{\mathbf t\sim-\frac{\mu U}{\delta L}
(\widehat p\widehat h_{\widehat x}+\widehat u_{\widehat z})\mathbf i
+\frac{\mu U}{\delta^2L}\widehat p\mathbf k.}
$$
Its leading normal component vanishes only if $\widehat p=0$ at the interface; the tangential component then requires $\widehat u_{\widehat z}=0$.
The <free surface> is material. Applying the <material derivative> to $z-h(x,t)=0$ gives $w=h_t+uh_x$, which in the scaled variables becomes
$$
\boxed{\widehat w=\widehat h_{\widehat t}+\widehat u\widehat h_{\widehat x},\qquad
\widehat p=0,\qquad\widehat u_{\widehat z}=0
\quad\text{at }\widehat z=\widehat h.}
$$
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