# If the length of a 20 cm spring increases to 56 cm when a 15 kg weight is hanging from it, what is the spring's constant?

$F = m g = 15 k g \times 9 , 8 \frac{m}{{s}^{2}} = 147 N$
For a spring $F = K \left({X}_{2} - {X}_{1}\right)$ so $K = \left(\frac{F}{{X}_{2} - {X}_{1}}\right) = \frac{147 N}{\left(56 - 20\right) c m} = \frac{147 N}{0 , 36 m} = 408 \frac{N}{m}$