Question

If a spring has a constant of `12 (kg)/s^2`, how much work will it take to extend the spring by `45 cm`?

Answer 1

`W = (12 times 0.45^2)/2`[J]

Explanation:

The elementary work is
`dw=K x dx` and the total work done during the elongation is given by
`W=int_0^Delta K x dx = 1/2K Delta^2 = (12 times 0.45^2)/2`[J]