Question

For small stretches, the Achilles tendon can be modeled as an ideal spring. Experiments using a particular tendon showed that it stretched 2.66 mm when a 125-kg mass was hung from it. (a) Find the spring constant of this tendon. (b) How much would it have to stretch to store 50.0 J of energy

Answer 1

Answer:

(a) Spring constant K = 460526.316 N/m

(b) Spring is stretched by 14.73 mm

Explanation:

We have given mass m = 125 kg

Acceleration due to gravity $g=9.8m/sec^2$

Spring is stretched to x = 2.66 m = 0.00266 m

(a) Force on spring is equal to F = Kx

And force due to gravity is F = mg

These two forces will be equal

So mg = Kx

So $125\times 9.8=K\times 0.00266$

K = 460526.316 N/m

(b) Energy stored is given as E = 50 J

Energy stored in spring is equal to $E=\frac{1}{2}Kx^2$

So $50=\frac{1}{2}\times 460526.316\times x^2$

x = 14.73 mm

Answer 2

Answer:

Explanation:

extension in spring, Δx = 2.66 mm

mass, m = 125 kg

(a) Let K is the spring constant.

F = mg = K Δx

125 x 9.8 = K (2.66 x 10^-3)

K = 460.5 x 10^3 N / m

(b) Energy, U = 50 J

Let the spring is stretched by a distance of Δs.

U = 0.5 K Δs²

50 = 0.5 x 460.5 x 1000 x Δs²

Δs = 0.0147 m

Δs = 1.47 cm