If the length of a 49 cm spring increases to 91 cm when a 5 kg weight is hanging from it, what is the spring's constant?

Jan 21, 2016

Spring constant $k = 166.67 N {m}^{-} 1$
After stretching, when the spring force and the force of mass equalize each other, there exists an equilibrium where the applied forces are equal in magnitude and opposite in direction. Taking this into consideration we see that for a spring being stretched by a mass attracted by gravity, $k \setminus \vec{\left(\setminus \delta x\right)} + m \setminus \vec{g} = 0$ which implies they are equal in magnitude by opposite in direction.
Disregarding the direction, taking account only the magnitude, we see that $k \left(\delta x\right) = m g$
Given, $m = 5 k g$, $g = 9.8 m {s}^{- 2}$, $\delta x = {x}_{2} - {x}_{1} = 91 - 49 = 42 c m = 0.42 m$
So, $k = m \frac{g}{\delta x} = \frac{5}{0.42} \cdot 9.8 = 11.9 \cdot 5 = 59.5 \cong 116.67 N {m}^{-} 1$