Question

If the length of a `13 cm` spring increases to `40 cm` when a `7 kg` weight is hanging from it, what is the spring's constant?

Answer 1

I got: `k=254N/m`

Explanation:

Consider that at equilibrium the elastic force `F_(el)` (given by Hooke's Law: `F_(el)=-kx`) and the weight `W` must balance to give, into Newton's Second Law, zero acceleration (the sistem is at rest).
We get:
`F_(el)+W=ma=0`
`-kx-mg=0`
(eleasic force upwards and weight downwards)
The displacement `x` is considered negative (downwards) so we have:
`-k[-(0.4-0.13)]-(7*9.8)=0`
(I changed into meters)
rearranging we get:
`k=254N/m`