A $12\ \mathrm{cm}$‑long spring has a force constant $(k)$ of $400\ \mathrm{\displaystyle \frac{N}{m}}$.
How much force is required to stretch the spring to a length of $14\ \mathrm{cm}$?
$F=8\ \mathrm{N}$
The displacement of the spring has a magnitude of $14-12=2\ \mathrm{cm}=0.02\ \mathrm{m}$ so, according to Hooke’s Law, the spring exerts a force of magnitude
$$F=kx$$
$$F=\left(400\ \mathrm{\frac{N}{m}}\right)(0.02\ \mathrm{m})$$
$$F=8\ \mathrm{N}$$
Therefore, we’d have to exert this much force to keep the spring in this stretched state.