# If the length of a 21 cm spring increases to 37 cm when a 7 kg weight is hanging from it, what is the spring's constant?

Apr 10, 2018

I get $428.75 \setminus \text{N/m}$.

#### Explanation:

We use Hooke's law, which states that,

$F = k x$

• $k$ is the spring constant in newtons per meter

• $x$ is the extension of the spring in meters

So here, we got:

$x = 37 \setminus \text{cm"-21 \ "cm}$

$= 16 \setminus \text{cm}$

$= 0.16 \setminus \text{m}$

The force is the weight of the weight thingy, which is $7 \setminus \text{kg"*9.8 \ "m/s"^2=68.6 \ "N}$.

And so,

$68.6 \setminus \text{N"=k*0.16 \ "m}$

$k = \left(68.6 \setminus \text{N")/(0.16 \ "m}\right)$

$= 428.75 \setminus \text{N/m}$