# If the length of a 49 cm spring increases to 73 cm when a 5 kg weight is hanging from it, what is the spring's constant?

May 29, 2018

About $204.17$ kilograms per second squared.

#### Explanation:

We use Hooke's law, which states that,

$F = k x$

where:

• $F$ is the force in newtons

• $k$ is the spring constant in newtons per meter or kilograms per second squared

• $x$ is the extension in meters

So, we get:

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

$= \frac{W}{x} \setminus \left(\because \text{the force here is the weight's weight}\right)$

$= \left(5 \setminus \text{kg"*9.8 \ "m/s"^2)/(73 \ "cm"-49 \ "cm}\right)$

$= \left(49 \setminus \text{kg m/s"^2)/(24 \ "cm}\right)$

$= \left(49 \setminus \text{kg m/s"^2)/(0.24 \ "m") \ (because 1 \ "m"=100 \ "cm}\right)$

$\approx 204.17 \setminus {\text{kg/s}}^{2}$