Question

If a spring has a constant of `12 (kg)/s^2`, how much work will it take to extend the spring by `76 cm`?

Answer 1

`4,56J`

Explanation:

I assume the units of the spring constant you meant was `12N//m` since from Hooke's Law, `F=kx=>k=F/x` and so the dimensions of k is N/m.

In this case, we use the formula for energy stored in a compressed or elongated spring. `W=1/2kx^2`, (this may be derived from calculus - ask me if you require the derivation as well), to obtain

`W=1/2xx12N//mxx0,76m`

`=4,56J`