Question

If the length of a `21 cm` spring increases to `45 cm` when a `2 kg` weight is hanging from it, what is the spring's constant?

Answer 1

I got `k=81.7N/m`

Explanation:

We can us Hooke's Law to relate the elastic force of a spring of elastic constant `k` after being elongated of `x`:
`F_(el)=-kx`
Also, Newton's Second Law will help us to consider the forces involved through:
`SigmavecF=m veca`

let us see:

So at the equilibrium the elastic force and the weight (`W=mg`) will give zero acceleration:
`-kx-mg=0`
But the elongation is directed downwards so we write, using numbers:
`-k(-0.24)-(2*9.8)=0`
So:
`k=(2*9.8)/0.24=81.7N/m`
I changed the lengths into meters because I am used to this but if you need it in N/cm use `24cm` instead of `0.24m` in the above.