posted by Adam on .
(a) If the length of the Achilles tendon increases 0.55 cm when the force exerted on it by the muscle increases from 2200 N to 6000 N, what is the "spring constant" of the tendon?
(b) How much work is done by the muscle in stretching the tendon 0.55 cm as the force increases from 2200 N to 6000 N?
Could any one please explain this to me I don't know how to do it.Thank you.
Spring constant is defined as k = ∆F/∆x, change in force divided by change in spring length. ∆F = 6000-2200=3800N, and ∆x = 0.55 cm (0.0055 m), so k = 3800/0.0055 N/m =6.9•10^5 N/m
The energy stored in a deflected spring is
kΔx²/2 = 6.9•10^5•3•10^-5/2 =10.35 J ; this must equal the work done in changing the spring length. So for the muscle, W = 10.35 J