Hill's force-velocity relation for a fully activated muscle (or for a single…

PHYSICS · Multi-part · Fermi University Physics · Biophysics & Medical Physics · Molecular machines, transport and bioenergetics · Molecular machines, transport and bioenergetics

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Problem

Hill's force-velocity relation for a fully activated muscle (or for a single actomyosin cross-bridge ensemble) can be writtenv(F)=b (F0−F)F+a,v(F)=\frac{b\,(F_0-F)}{F+a},where F0F_0 is the isometric (zero-velocity) force, FF is the load, and aa (a force) and bb (a velocity) are positive constants. This relation is an empirical fit to measurements, quoted here. ((a)) The mechanical power delivered against the load is P(F)=Fv(F)P(F)=Fv(F). Find the load F∗F^* at which this power is largest, in terms of aa and F0F_0. ((b)) Find how F∗/F0F^*/F_0 behaves as a/F0→0a/F_0\to0 and as a/F0→∞a/F_0\to\infty. In which limit does Hill's relation become a straight line from its top speed at zero load to zero speed at F0F_0, and why does the power then peak at exactly half the isometric force? ((c)) A different muscle has an isometric force F0=150 NF_0=150\,\mathrm{N} and a Hill constant a=60 Na=60\,\mathrm{N} (so a/F0=0.40a/F_0=0.40). Find, in newtons and as a fraction of F0F_0, the load at which this muscle delivers its maximum mechanical power.

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