# If the length of a 46 cm spring increases to 77 cm when a 16 kg weight is hanging from it, what is the spring's constant?

Hooke's Law can be rearranged as $k = \frac{F}{x}$. In this case $F$ is the weight force, $m g = 16 \cdot 9.8 = 156.8$ $N$. The displacement, $x$ is $0.77 - 0.46 = 0.31$ $m$. So $k = \frac{156.8}{0.31} = 505.8$ $N k {g}^{-} 1$ (or $k {g}^{s} - 2$).
Not much more explanation needed, except to note that it was necessary to convert from $c m$ to $m$, the correct SI unit for length and distance.
The units of the spring constant can be expressed as $N k {g}^{-} 1$ or as $k g {s}^{-} 2$.