# If the length of a 27 cm spring increases to 43 cm when a 7 kg weight is hanging from it, what is the spring's constant?

$428 , 75 N / m$
From Hooke's Law, $F = k x$ from which $k = \frac{F}{x}$, where x is the displacement of the spring from its equilibrium position.
$\therefore k = \frac{7 \times 9 , 8}{\left(43 - 27\right) \times {10}^{- 2} m} = 428 , 75 N / m$