Question

If the length of a `23 cm` spring increases to `96 cm` when a `5 kg` weight is hanging from it, what is the spring's constant?

Answer 1

I got `67N/m`

Explanation:

We can use Hooke's Law:
`F_(el)=-kx`
Newton's Second Law...
and the diagram:

At equilibrium:
`F_(el)-W=0`
`-kx-mg=0`
using our data and from our choice of axes (changing in meters):

`-k(-0.73)-(5*9.8)=0`

`k=(5*9.8)/0.73=67N/m`