Question

If the length of a `15 cm` spring increases to `42 cm` when a `4 kg` weight is hanging from it, what is the spring's constant?

Answer 1

I found `145.2N/m`

Explanation:

I tried using Hooke's Law to describe the elastic force `F_(el)`, and Newton's law to describe the equilibrium of forces:

According to our choice of axis the displacemeht can be considered negative (below the chosen origin) `y=-h=-0.27m`; so we get:
`k=(mg)/(-y)=(4*9.8)/-(-0.27)=145.2N/m`