Question

If a spring has a constant of `7 (kg)/s^2`, how much work will it take to extend the spring by `22 cm`?

Answer 1

Given,

`k = (7"N")/"s"`

`Deltax = 0.22"m"`

Recall how external force is related to these variables,

`F_"ext" = kx`

, and let's use the average force in determining the work done over displacement `Deltax`, such that,

`W_"ext" = barFDeltax = (1/2kx)*x = 1/2kx^2`

This is familiar to elastic potential energy, which we are essentially working against in the positive direction. Hence, the work required to do this is,

`W_"ext" = 1/2kx^2 approx 0.17"J"`