# If the length of a 15 cm spring increases to 42 cm when a 4 kg weight is hanging from it, what is the spring's constant?

I found $145.2 \frac{N}{m}$
I tried using Hooke's Law to describe the elastic force ${F}_{e l}$, and Newton's law to describe the equilibrium of forces:
According to our choice of axis the displacemeht can be considered negative (below the chosen origin) $y = - h = - 0.27 m$; so we get:
$k = \frac{m g}{- y} = \frac{4 \cdot 9.8}{-} \left(- 0.27\right) = 145.2 \frac{N}{m}$