a) Hooke's law:
F = kΔx
F = spring force, k = spring constant, Δx = change of spring length
Given values:
F = 30N, Δx = 0.1m
Plug in and solve for k:
30 = k(0.1)
k = 300N/m
b) Apply the work-energy theorem; whatever work you put into deforming the spring becomes stored as spring potential energy:
W = 0.5kΔx²
W = work, k = spring constant, Δx = change of spring length
Given values:
k = 300N/m (from part a), Δx = 0.5m
Plug in and solve for W:
W = 0.5(300)(0.5)²
W = 37.5J
c) Apply the work-energy theorem here:
W = 0.5kΔx²
Given values:
k = 300N/m (from part a), Δx = 0.4m
Plug in and solve for W:
W = 0.5(300)(0.4)²
W = 24J
d) To find how much additional work needs to be done to stretch the spring an additional 0.1m if it's already been stretched 0.1m, find the potential energies for when the spring is stretched 0.2m and 0.1m and subtract them:
W = 0.5kΔx₂² - 0.5kΔx₁² = 0.5k(Δx₂² - Δx₁²)
Given values:
k = 300N/m, Δx₂ = 0.2m, Δx₁ = 0.1m
Plug in and solve for W:
W = 0.5(300)(0.2² - 0.1²)
W = 4.5J
