Question

Can anyone help me understand why so many of the questions about springs use the units `kgs^-2` rather than `Nm^-1` for the spring constant?

Given the definition of the newton, dimensional analysis shows the two to be equivalent, but `Nm^-1` makes much more sense: how many `m` does the spring expand or compress by when a force of `x N` is applied?

Answer 1

Don't know, but the question is interesting.

Explanation:

As physicist I do not know the answer.
I can just guess: back in time there was the tentative to define a new International System of measurements removing the names of people. So no more Newtons, Gauss, Tesla, etc.
Maybe the `kg/s^{-2}` is used by some textbook that belongs to this tentative.
Of course this crazy idea was discarded because we love too much our great scientists to avoid their name form the units.
I agree with you that for a spring `N/m` is much more clear.

Answer 2

Because they are equivalent units, and testing on units is an entirely valid choice for a professor/teacher.

Yes, you could simply say "how far does the spring shift from its equilibrium position when a force of `x` `"N"` is applied?", but that's too easy; all that asks is for you to somehow use `F_"spring" = -kx`.

However:

`"N"/"m" = (("kg"cdotcancel("m"))/"s"^2)/(cancel("m")) = "kg"/"s"^2`

So either way, it doesn't really matter.

Although, the spring constant has been used in other contexts that warrant its `"kg"/"s"^2` units...

`\mathbf(omega = sqrt(k/m))`

where:

• `omega` is the angular frequency in `"s"^(-1)`.
• `k` is the spring constant in `"kg"/"s"^2`.
• `m` is the mass in `"kg"`.

You might see this when working with oscillatory systems like pendulums, though springs count as well.

No force is discussed for such an equation, and thus, it makes more sense to use `"kg"/"s"^2` than `"N"/"m"`, even though they are equivalent units.