Question

If the length of a `45 cm` spring increases to `86 cm` when a `1 kg` weight is hanging from it, what is the spring's constant?

Answer 1

Approximately `24` newtons per meter

Explanation:

Well, it looks to me that the spring will increase in length towards the direction of the weight hanging from it, so Hooke's law here will be:

`F=kx`

• `F` is the force exerted in newtons

• `k` is the spring constant

• `x` is the extension of the spring

The force here is the object's weight, which is `1 \ "kg"*9.8 \ "m/s"^2=9.8 \ "N"`.

Therefore, the spring constant is:

`k=F/x`

`=(9.8 \ "N")/(86 \ "cm"-45 \ "cm")`

`=(9.8 \ "N")/(41 \ "cm")`

`~~0.24 \ "N/cm"`

`=24 \ "N/m"`