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?

1 Answer
May 14, 2018

420 newtons per meter

Explanation:

We use Hooke's law, which states that,

F=kx

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 \ "kg"*9.8 \ "m/s"^2=147 \ "N"

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

:.x=F/k=(147 \ "N")/(0.35 \ "m")

=420 \ "N/m"