# If the length of a 13 cm spring increases to 45 cm when a 4 kg weight is hanging from it, what is the spring's constant?

Jul 1, 2018

$122.5$ newtons per meter.

#### Explanation:

We use Hooke's law, which states that,

$F = k x$

where:

• $F$ is the force applied in newtons

• $k$ is the spring's constant

• $x$ is the extension in meters

Here we have $x = 45 \setminus \text{cm"-13 \ "cm"=32 \ "cm"=0.32 \ "m}$, $F = 4 \setminus \text{kg"*9.8 \ "m/s"^2=39.2 \ "N}$.

So, the spring's constant is:

$k = \frac{F}{x}$

$= \left(39.2 \setminus \text{N")/(0.32 \ "m}\right)$

$= 122.5 \setminus \text{N/m}$