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

Answer:

#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"#