Physics
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?
N/m
(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?
J
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 = 60002200=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