If the length of the Achilles tendon increases 0.52 cm

If the length of the Achilles tendon increases 0.52 cm when the force exerted on it by the muscle increases from 2200 N to 5500 N, what is the \”spring constant\” of the tendon?
Answer in N/m .

(b) How much work is done by the muscle in stretching the tendon 0.52 cm as the force increases from 2200 N to 5500 N?
Answer in J

Solution

a)del x ( change in length) = 0.52 cm

del F ( change in force) = 5500-2200 = 3300 N

k ( spring constant) = del F / del x

k= 3300 / 0.52*10-2 = 634.61*103 N/m

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