Question

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 1.70MPa. f an insect wing has resilin with a relaxed length of 1.00 cm and a cross-sectional area of 1.00 mm², how much force must the wings axert to extend the resilin to 3.68 cm ?

Answer 1

The force required to stretch resilin from a relaxed length of 1.00 cm to a stretched length of 3.68 cm, given a Young's modulus of 1.70 MPa and a cross-sectional area of 1.00 mm², is approximately 4.556 N.

Step-by-step explanation:

The question pertains to the force needed to extend a piece of resilin from its relaxed state to a stretched state. This is a practical application of Hooke's Law, which describes the behavior of springs and elastic materials. The Young's modulus (E) given is essential to relate the stress and strain in a linear elastic material like resilin.

For a piece of resilin, the stress (σ) is the force applied (F) divided by the cross-sectional area (A), and the strain (ε) is the fractional change in length (ΔL over L0). The Young's modulus (E) relates the stress and strain by σ = Eε. To find the force needed, we rearrange the formula to F = EεA.

Given:

• Young's modulus (E) = 1.70 MPa = 1.70 × 106 Pa
• Relaxed length (L0) = 1.00 cm
• Stretched length = 3.68 cm
• Change in length (ΔL) = Stretched length - Relaxed length = 2.68 cm
• Cross-sectional area (A) = 1.00 mm² = 1.00 × 10-6 m²

Calculating the strain, ε = ΔL/L0 = 2.68 cm / 1.00 cm = 2.68.

Applying the values:
F = (1.70 × 106 Pa)(2.68)(1.00 × 10-6 m²)

Therefore, the force required to extend the resilin to 3.68 cm is:

F = 4.556 N (approximately)