If the length of a 63 cm spring increases to 98 cm when a 15 kg weight is hanging from it, what is the spring's constant?

May 14, 2018

$420$ newtons per meter

Explanation:

We use Hooke's law, which states that,

$F = k x$

where:

• $F$ is the force exerted on the spring in newtons

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

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

Here:

• $F = 15 \setminus \text{kg"*9.8 \ "m/s"^2=147 \ "N}$

• x=(98 \ "cm"-63 \ "cm")*(1 \ "m")/(100 \ "cm")=0.35 \ "m"

$\therefore x = \frac{F}{k} = \left(147 \setminus \text{N")/(0.35 \ "m}\right)$

$= 420 \setminus \text{N/m}$