# If the length of a 45 cm spring increases to 86 cm when a 1 kg weight is hanging from it, what is the spring's constant?

May 5, 2018

Approximately $24$ newtons per meter

#### Explanation:

Well, it looks to me that the spring will increase in length towards the direction of the weight hanging from it, so Hooke's law here will be:

$F = k x$

• $F$ is the force exerted in newtons

• $k$ is the spring constant

• $x$ is the extension of the spring

The force here is the object's weight, which is $1 \setminus \text{kg"*9.8 \ "m/s"^2=9.8 \ "N}$.

Therefore, the spring constant is:

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

$= \left(9.8 \setminus \text{N")/(86 \ "cm"-45 \ "cm}\right)$

$= \left(9.8 \setminus \text{N")/(41 \ "cm}\right)$

$\approx 0.24 \setminus \text{N/cm}$

$= 24 \setminus \text{N/m}$