= Solution
Approximate the gap by steady <Couette flow>, neglecting edge effects and additional losses from membrane deformation. Its <velocity> gradient is $U/h$, so the <shear stress> is $\eta U/h$. The <force> on the <gliding bacterium> and the rate of <viscous dissipation> are
$$
F_d=\eta A\frac Uh,\qquad P=F_dU=\eta A\frac{U^2}{h}.
$$
The same power follows by integrating $\eta(U/h)^2$ over gap volume $Ah$. In SI units, $A=10^{-10}\,\mathrm{m^2}$, $h=10^{-7}\,\mathrm m$, and $U=10^{-6}\,\mathrm{m\,s^{-1}}$. Thus
$$
\boxed{\dot\gamma=10\,\mathrm{s^{-1}},\quad F_d=10^{-12}\,\mathrm N,\quad P=10^{-18}\,\mathrm W.}
$$
An <energy> needs a duration or distance: \b[$E=10^{-18}\Delta t\,\mathrm J$ for $\Delta t$ in seconds], equivalently $E=F_ds$ for travel distance $s$. Moving one micrometre takes one second and dissipates $10^{-18}\,\mathrm J$. This is the mechanical cost of the sheared film. A numerical biological efficiency would additionally require the motor's input power and the losses associated with its peristaltic actuation.
Back to article page
