If the length of a 25 cm spring increases to 67 cm when a 15 kg weight is hanging from it, what is the spring's constant?

Apr 11, 2016

$k = \frac{15000}{42} = \frac{7500}{21} = \frac{2500}{7} \frac{\text{Newtons}}{m e t e r s}$ or
$k = 357.14$ $\text{Newtons"/"meter}$

Explanation:

${F}_{e l} = k \cdot x$
${F}_{e l}$ is the force upon the spring, the "elastic" force;
$k$ is the spring constant;
$x$ is the deformity of the spring;
${F}_{w}$ is the force from the weight;
$g$ is gravity;
$m$ is mass.

Since the length of the spring is $25 c m$, the deformity is $67 - 25$, or $42 c m$. Using the International Sistem (or S.I.), the deformity must be measured in meters, since the constant($k$) is measured in Newtons/meter. So $x = 0.42$ $m e t e r s$.

The force applied to the spring is the weight force from the weight hanging from the spring. So ${F}_{w} = {F}_{e l}$. Now, we apply the math:
$m \cdot g = k \cdot x$
$15 \cdot 10 = k \cdot 0.42$
$k = \frac{150}{\frac{42}{100}}$
$k = \frac{150}{1} \cdot \frac{100}{42}$
$k = \frac{15000}{42} = \frac{7500}{21} = \frac{2500}{7} \frac{\text{Newtons}}{m e t e r s}$ or
$k = 357.14$ $\text{Newtons"/"meter}$