# If the length of a 24 cm spring increases to 47 cm when a 5 kg weight is hanging from it, what is the spring's constant?

Jan 18, 2016

Better to rewrite the question in metres: "$0.24 m$ spring increases $0.47 m$..." Spring constant, $k = \frac{F}{\Delta d} = \frac{m \cdot g}{\Delta d} = \frac{5 \cdot 9.8}{0.47 - 0.24} = 21.3 N {m}^{-} 1$

#### Explanation:

The spring constant, $k$, of a spring is expressed in newton per metre $\left(N {m}^{-} 1\right)$: for each additional $N$ of force the spring expands by $k \left(m\right)$.

It's important to use correct SI units for the quantities: centimetre is not a base SI unit, metre is, so convert the $c m$ to $m$.

I used $\Delta d$ to represent the change in length of the spring - final length of $0.47 m$ minus initial length $0.24 m$.

The force acting on the spring is the weight force of the mass, which is its mass, $5 k g$ times $g = 9.8 N k {g}^{-} 1$ (more often written as $m {s}^{-} 2$, but this is an exactly equivalent unit that makes more sense in this context).

Calculating all this leads to a spring constant equal to $21.3 N {m}^{-} 1$.