Answer:
k = 2400 N/m
Explanation:
Given that,
Force acting on the spring, F = 12 N
The spring stretches 5 mm = 0.005 m
We need to find the force constant of the spring.
The force acting on the spring is given by Hooke's law as follows :
F = kx
Where k is force constant
[tex]k=\dfrac{F}{x}\\\\k=\dfrac{12\ N}{0.005\ m}\\\\=2400\ N/m[/tex]
So, the force constant of the spring is 2400 N/m.
