Question

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?

Answer 1

Hooke's Law can be rearranged as `k=F/x`. In this case `F` is the weight force, `mg=16*9.8=156.8` `N`. The displacement, `x` is `0.77-0.46=0.31` `m`. So `k=156.8/0.31=505.8` `Nkg^-1` (or `kg^s-2`).

Explanation:

Not much more explanation needed, except to note that it was necessary to convert from `cm` to `m`, the correct SI unit for length and distance.

The units of the spring constant can be expressed as `Nkg^-1` or as `kgs^-2`.