If the length of a 15 cm spring increases to 28 cm when a 2 kg weight is hanging from it, what is the spring's constant?

Apr 15, 2016

$k = 150 , 92 \text{ N/m}$

Explanation:

$m = 2 k g$
$g = 9 , 81 \frac{N}{k g}$

$G = m \cdot g = 2 \cdot 9 , 81 = 19 , 62 N \text{ Force acting to spring}$
$k : \text{spring constant}$
$\Delta x = 28 - 15 = 13 c m = 0 , 13 m$
$F = G = k \cdot \Delta x$

$19 , 62 = k \cdot 0 , 13$

$k = \frac{19 , 62}{0 , 13}$

$k = 150 , 92 \text{ N/m}$