Question

If the length of a `23 cm` spring increases to `46 cm` when a `16 kg` weight is hanging from it, what is the spring's constant?

Answer 1

`682 N m^(-1)`

Explanation:

Using Hooke's Law, `F=kx`,

`Delta F = k Delta x`
`(Delta m)g = k Delta x`
`(16)(9.81)= k ((46-23)*10^(-2))`
`k= 682Nm^(-1)`