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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 $0 \text { to } 3.0 \times 10 ^ { - 2 } \mathrm { N }.$ The length and radius of the collagen are, respectively, 2.5 and 0.091 cm, and Young's modulus is $3.1 \times 10 ^ { 6 } \mathrm { N } / \mathrm { m } ^ { 2 }$If the stretching obeys Hooke's law, what is the spring constant k for collagen?

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Given values:

$r=0.091 \ \text{cm}$
$L_0=2.5 \cdot 10^{-2} \ \text{m}$
$Y=3.1 \cdot 10^{6} \ \dfrac{\text{N}}{\text{m}^2}$
$F_{a}=3.0 \cdot 10^{-2} \ \text{N}$

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.

\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:

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

As the force in this case is defined as:

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

equation 1 becomes:

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

Continue derivation,

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

Substitute values into equation 3 and solve,

\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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