Question

If a spring has a constant of `8 (kg)/s^2`, how much work will it take to extend the spring by `40cm`?

Answer 1

Consider your system,

Recall Hooke's Law,

`F_"s" = -kx`, and by extension

`F_"ext" = -F_"s" = kx` (we're looking to find this)

where `k` is the constant you describe and `x` is the displacement of spring.

Moreover, recall that work is represented by,

`W = F*d`

Now,

`=> W = (1/2kx)*x = 1/2kx^2`

(the reason this is halved is because we're averaging the force)

Hence,

`W_"ext" = 1/2 * (8"kg")/("s"^2) * (0.40"m")^2 approx 0.64"J"`

To be sure, the elastic potential energy is equal to the work done to stretch it that far (which shares the same equation as that derived).