Question

If the length of a `13 cm` spring increases to `45 cm` when a `4 kg` weight is hanging from it, what is the spring's constant?

Answer 1

`122.5` newtons per meter.

Explanation:

We use Hooke's law, which states that,

`F=kx`

where:

• `F` is the force applied in newtons

• `k` is the spring's constant

• `x` is the extension in meters

Here we have `x=45 \ "cm"-13 \ "cm"=32 \ "cm"=0.32 \ "m"`, `F=4 \ "kg"*9.8 \ "m/s"^2=39.2 \ "N"`.

So, the spring's constant is:

`k=F/x`

`=(39.2 \ "N")/(0.32 \ "m")`

`=122.5 \ "N/m"`