Question

If the length of a `27 cm` spring increases to `43 cm` when a `7 kg` weight is hanging from it, what is the spring's constant?

Answer 1

`428,75N//m`

Explanation:

From Hooke's Law, `F=kx` from which `k=F/x`, where x is the displacement of the spring from its equilibrium position.

`therefore k=(7xx9,8)/((43-27)xx10^(-2)m)=428,75N//m`