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?

1 Answer
May 5, 2018

Answer:

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=kx#

  • #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 \ "kg"*9.8 \ "m/s"^2=9.8 \ "N"#.

Therefore, the spring constant is:

#k=F/x#

#=(9.8 \ "N")/(86 \ "cm"-45 \ "cm")#

#=(9.8 \ "N")/(41 \ "cm")#

#~~0.24 \ "N/cm"#

#=24 \ "N/m"#