Answer:
K = 3.07 x 10⁷ N/m = 30.7 MN/m
Δx = 2.93 x 10⁻⁶ m = 0.00293 mm
Explanation:
In order to find out the effective spring constant of the straw, we can use the following equation:
K = AE/L
where,
K = Effective Spring Constant = ?
A = Cross-Sectional Area = π(Do - Di)²/4 = π(0.014² - 0.011²)/4 = 9.5 x 10⁻⁵ m²
L = Length of Straw = 21 cm = 0.21 m
E = Young's Modulus = 68 GPa = 68 x 10⁹ Pa
Therefore,
K = (9.5 x 10⁻⁵ m²)(68 x 10⁹ Pa)/(0.21 m)
K = 3.07 x 10⁷ N/m
Now, we find stress to check the elastic limit:
Stress = Force/Area = 90 N/9.5 x10⁻⁵ m²
Stress = 0.95 MPa < Tensile Strength (7 MPa)
Thus, the straw is still in elastic limit and hook's law can be applied:
Δx = F/K = (90 N)/(3.07 x 10⁷ N/m)
Δx = 2.93 x 10⁻⁶ m = 0.00293 mm
