Question

If the length of a `45 cm` spring increases to `63 cm` when a `7 kg` weight is hanging from it, what is the spring's constant?

Answer 1

Approximately `381.1` 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 constant in `"kg/s"^2` or `"N/m"`

• `x` is the extension in meters

Here, the force is the weight of the object, which is equal to:

`7 \ "kg"*9.8 \ "m/s"^2=68.6 \ "N"`.

The spring constant is then:

`k=F/x`

`=(68.6 \ "N")/(0.63 \ "m"-0.45 \ "m")`

`=(68.6 \ "N")/(0.18 \ "m")`

`~~381.1 \ "N/m"`