# If the length of a 23 cm spring increases to 46 cm when a 16 kg weight is hanging from it, what is the spring's constant?

Apr 26, 2017

$682 N {m}^{- 1}$
Using Hooke's Law, $F = k x$,
$\Delta F = k \Delta x$
$\left(\Delta m\right) g = k \Delta x$
$\left(16\right) \left(9.81\right) = k \left(\left(46 - 23\right) \cdot {10}^{- 2}\right)$
$k = 682 N {m}^{- 1}$