Question

If the length of a `63 cm` spring increases to `98 cm` when a `15 kg` weight is hanging from it, what is the spring's constant?

Answer 1

`420` newtons per meter

Explanation:

We use Hooke's law, which states that,

`F=kx`

where:

• `F` is the force exerted on the spring in newtons

• `k` is the spring constant, usually in newtons per meter

• `x` is the extension of the spring in meters

Here:

• `F=15 \ "kg"*9.8 \ "m/s"^2=147 \ "N"`

• `x=(98 \ "cm"-63 \ "cm")*(1 \ "m")/(100 \ "cm")=0.35 \ "m"`

`:.x=F/k=(147 \ "N")/(0.35 \ "m")`

`=420 \ "N/m"`