# If the length of a 45 cm spring increases to 63 cm when a 7 kg weight is hanging from it, what is the spring's constant?

Jun 11, 2018

#### Answer:

Approximately $381.1$ 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 constant in ${\text{kg/s}}^{2}$ or $\text{N/m}$

• $x$ is the extension in meters

Here, the force is the weight of the object, which is equal to:

$7 \setminus \text{kg"*9.8 \ "m/s"^2=68.6 \ "N}$.

The spring constant is then:

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

$= \left(68.6 \setminus \text{N")/(0.63 \ "m"-0.45 \ "m}\right)$

$= \left(68.6 \setminus \text{N")/(0.18 \ "m}\right)$

$\approx 381.1 \setminus \text{N/m}$