If the length of a #49 cm# spring increases to #73 cm# when a #5 kg# weight is hanging from it, what is the spring's constant?

1 Answer
May 29, 2018

Answer:

About #204.17# kilograms per second squared.

Explanation:

We use Hooke's law, which states that,

#F=kx#

where:

  • #F# is the force in newtons

  • #k# is the spring constant in newtons per meter or kilograms per second squared

  • #x# is the extension in meters

So, we get:

#k=F/x#

#=W/x \ (because "the force here is the weight's weight")#

#=(5 \ "kg"*9.8 \ "m/s"^2)/(73 \ "cm"-49 \ "cm")#

#=(49 \ "kg m/s"^2)/(24 \ "cm")#

#=(49 \ "kg m/s"^2)/(0.24 \ "m") \ (because 1 \ "m"=100 \ "cm")#

#~~204.17 \ "kg/s"^2#