# If the length of a 21 cm spring increases to 57 cm when a 5 kg weight is hanging from it, what is the spring's constant?

Dec 26, 2015

We should expect $k$ to be somewhere in the hundreds for typical springs.

This is simply looking at the equation

$\setminus m a t h b f \left(F = - k \Delta y\right)$

...for when an object is hanging off a spring.

So, the only force acting on the spring is the force $\setminus m a t h b f \left({F}_{g}\right)$ due to gravity $\setminus m a t h b f \left(g\right)$. The generic force $F$, then, is ${F}_{g}$. Thus, with $g < 0$ (where down is negative):

${F}_{g} = - k \Delta y$

$m g = - k \left({y}_{f} - {y}_{i}\right)$

$\frac{m g}{{y}_{i} - {y}_{f}} = k$

color(blue)(k) = (("5 kg")(-"9.80665" cancel"m""/s"^2))/("0.21" cancel"m" - "0.57" cancel"m")

$= \textcolor{b l u e}{{\text{136.20 kg/s}}^{2}}$ or $\textcolor{b l u e}{\text{N/m}}$