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A cylindrically shaped piece of collagen (a substance found in the body in connective tissue) is being stretched by a force that increases from 0 to 3.0×102N.0 \text { to } 3.0 \times 10 ^ { - 2 } \mathrm { N }. The length and radius of the collagen are, respectively, 2.5 and 0.091 cm, and Young's modulus is 3.1×106N/m23.1 \times 10 ^ { 6 } \mathrm { N } / \mathrm { m } ^ { 2 }

If the stretching obeys Hooke's law, what is the spring constant k for collagen?

Solution

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Answered 7 months ago
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Given values:

r=0.091 cmr=0.091 \ \text{cm}
L0=2.5102 mL_0=2.5 \cdot 10^{-2} \ \text{m}
Y=3.1106 Nm2Y=3.1 \cdot 10^{6} \ \dfrac{\text{N}}{\text{m}^2}
Fa=3.0102 NF_{a}=3.0 \cdot 10^{-2} \ \text{N}

a) We can start from the Hooke’s law. So, the force that must be applied to stretch or compress an ideal spring is given by following equation.

Fa=kx(Hooke’s law.)\begin{aligned} F_{a}&=k \cdot x \tag{Hooke's law.}\\ \end{aligned}

In this equation, with k is the marked the spring constant and with x is the marked displacement of the spring from its unstrained length.

From the Hooke’s law, spring constant is:

k=Fx(Equation 1.)\begin{aligned} k=\frac{F}{x} \tag{Equation 1.}\\ \end{aligned}

As the force in this case is defined as:

F=YAL0,F=\dfrac{YA}{L_0},

equation 1 becomes:

k=YAL0(Equation 2.)\begin{aligned} k=\dfrac{YA}{L_0} \tag{Equation 2.}\\ \end{aligned}

Continue derivation,

k=Yπr2L0(Equation 3.)\begin{aligned} k=\dfrac{Y \pi r^2}{L_0} \tag{Equation 3.}\\\\ \end{aligned}

Substitute values into equation 3 and solve,

k=Yπr2L0       Equation 3.Substitute values.k=(3.1106 Nm2)π(0.091102 m)22.5102 mk=322.59 Nm\begin{aligned} k &=\dfrac{Y \pi r^2}{L_0} ~~~~~~~ \text{Equation 3.}\\ & \text{Substitute values.} \\ k &=\dfrac{\bigg(3.1 \cdot 10^{6} \ \dfrac{\text{N}}{\text{m}^2} \bigg) \pi (0.091 \cdot 10^{-2} \ \text{m})^2}{2.5 \cdot 10^{-2} \ \text{m}} \\ k &=\boxed{322.59 \ \dfrac{\text{N}}{\text{m}} } \\ \end{aligned}

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