# A cylindrically shaped piece of collagen (a substance found in the body in connective tissue) is...

## Question:

A cylindrically shaped piece of collagen (a substance found in the body in connective tissue) is being stretched by a force that increases from {eq}0 {/eq} to {eq}0.0436 \, N {/eq}. The length and radius of the collagen are, respectively, {eq}2.90\, cm {/eq} and {eq}0.0815\, cm {/eq}, and Young's modulus is {eq}3.1 \times 10^6\, N/m^2 {/eq}.

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

(b) How much work is done by the variable force that stretches the collagen?

## Collagen

The force applied to the collagen produces tension in the collagen. This results in an increase in the length of the collagen. Since the collagen is said to follow Hooke's Law, so the increase in length is proportional to the force.

## Answer and Explanation: 1

**Part a**

The spring constant **k** for the collagen is calculated as:

{eq}k = \dfrac{EA}{L_0} {/eq}

where E is the Young's Modulus.

{eq}k = \dfrac{3.1 \times 10^6 \times (\pi \times (0.0815 \times 10^{-2})^2)}{2.9 \times 10^{-2}} {/eq}

{eq}k =223 N/m {/eq}

**Part b**

The work done by the variable force in stretching the collagen is:

{eq}W = F_{avg} \times \Delta L {/eq}

{eq}W =\dfrac{F}{2} \times \dfrac{F}{k} {/eq}

{eq}W =\dfrac{0.0436}{2} \times \dfrac{0.0436}{223} = 4.26 \times 10^{-6} \ J {/eq}

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Chapter 4 / Lesson 19After watching this video, you will be able to explain what Hooke's Law is and use the equation for Hooke's Law to solve problems. A short quiz will follow.