Answer:
The value of the spring constant of this spring is 1000 N/m
Explanation:
Given;
equilibrium length of the spring, L = 10.0 cm
new length of the spring, L₀ = 14 cm
applied force on the spring, F = 40 N
extension of the spring due to applied force, e = L₀ - L = 14 cm - 10 cm = 4 cm
From Hook's law
Force applied to a spring is directly proportional to the extension produced, provided the elastic limit is not exceeded.
F ∝ e
F = ke
where;
k is the spring constant
k = F / e
k = 40 / 0.04
k = 1000 N/m
Therefore, the value of the spring constant of this spring is 1000 N/m
The answer to this question is Helium
Hook's law states that
where F is the applied force, k is the spring constant, and 
is the change in length.
Plugging your values, we have
