A wire behaves like a spring with spring constant AY/l0. For a 75cm copper wire of diameter 1.29mm, it's 348528.29 N/m. A weight of 435.66 N stretches it by 1.25mm.
A Wire as a Spring
a) Spring Constant:
Consider a wire of length l0 and cross-sectional area A stretched by a length Δl due to a weight W. Hooke's law states that the force (F) exerted by the wire is proportional to the change in length:
F = k * Δl
where k is the spring constant.
The stress (σ) on the wire is defined as the force per unit area:
σ = F / A
The strain (ε) is the ratio of the change in length to the original length:
ε = Δl / l0
Stress and strain are related by Young's modulus (Y) of the material:
σ = Y * ε
Substituting the expressions for stress and strain into Hooke's law:
F / A = Y * Δl / l0
Solving for the force F:
F = AY * Δl / l0
Therefore, the wire behaves like a spring with a force constant of:
k = AY / l0
b) Force Constant for Copper Wire:
Given:
l0 = 75.0 cm = 0.750 m
d = 1.29 mm = 1.29 x 10^-3 m
A = π * d^2 / 4 = 1.30 x 10^-6 m^2
Y for copper = 2.0 x 10^11 Pa
Using the formula for spring constant:
k = Y * A / l0 = (2.0 x 10^11 Pa) * (1.30 x 10^-6 m^2) / (0.750 m) = 348528.29 N/m
c) Weight Needed for 1.25mm Stretch:
Given:
Δl = 1.25 mm = 1.25 x 10^-3 m
Using the formula for force due to stretch:
F = k * Δl = 348528.29 N/m * (1.25 x 10^-3 m) = 435.66 N
This force is equal to the weight W needed to stretch the wire by 1.25mm.