91Ó°ÊÓ

Resilin is a rubber-like protein that helps insects to fly more efficiently. The resilin, attached from the wing to the body, is relaxed when the wing is down and is extended when the wing is up. As the wing is brought up, some elastic energy is stored in the resilin. The wing is then brought back down with little muscular energy, since the potential energy in the resilin is converted back into kinetic energy. Resilin has a Young's modulus of $1.7 \times 10^{6} \mathrm{N} / \mathrm{m}^{2}$ (a) If an insect wing has resilin with a relaxed length of \(1.0 \mathrm{cm}\) and with a cross-sectional area of \(1.0 \mathrm{mm}^{2},\) how much force must the wings exert to extend the resilin to \(4.0 \mathrm{cm} ?\) (b) How much energy is stored in the resilin?

Step by step solution

01

Find extension (e) of the resilin after stretching.

We're given the initial relaxed length L0 and the final extended length L. We can find the extension e by subtracting the initial length from the final length. e = L - L0 Given that L0 = 1.0 cm and L = 4.0 cm: e = 4.0 cm - 1.0 cm = 3.0 cm To work in SI units, we need to convert e from cm to meters: e = 3.0 cm * (1 m/100 cm) = 0.03 m

02

Use Young's modulus to calculate stress (σ).

By definition, Young's modulus (Y) is the ratio of stress (σ) to strain (ε). Therefore, σ = Y * ε. Here, we still need to find the strain (ε). By definition, strain (ε) is the ratio of the extension (e) to the initial length (L0). Therefore, ε = e/L0. Given that e = 0.03 m and L0 = 1.0 cm (converted to meters becomes 0.01 m), we find: ε = 0.03 m / 0.01 m = 3 Now, we can calculate stress (σ) using Young's modulus (Y): σ = Y * ε Given that Y = \(1.7 \times 10^{6} \mathrm{N/m^2}\) and ε = 3, we find: σ = \(1.7 \times 10^{6} \mathrm{N/m^2}\) * 3 = \(5.1 \times 10^{6} \mathrm{N/m^2}\)

03

Calculate the force (F) exerted by the wings.

Now that we have the stress (σ), we can calculate the force (F) exerted by the wings to extend the resilin to 4.0 cm. Force can be defined as the product of the stress (σ) and the cross-sectional area (A). F = σ * A Given that σ = \(5.1 \times 10^{6} \mathrm{N/m^2}\) and A = \(1.0 \mathrm{mm^2}\) (converted to square meters becomes \(1.0 \times 10^{-6} \mathrm{m^2}\)), we find: F = \(5.1 \times 10^{6} \mathrm{N/m^2} * 1.0 \times 10^{-6} \mathrm{m^2} = 5.1 \mathrm{N}\)

04

Calculate the energy stored in the resilin.

To calculate the energy stored in the resilin, we can use the formula for the elastic potential energy (U) in an elastomeric material: U = (1/2) * F * e Given that F = 5.1 N and e = 0.03 m, we find: U = (1/2) * 5.1 N * 0.03 m = 0.0765 J The energy stored in the resilin is 0.0765 Joules. To summarize: a) The force that must be exerted by the wings to extend the resilin to 4.0 cm is 5.1 N. b) The amount of energy stored in the resilin is 0.0765 J.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!