If the length of a #13 cm# spring increases to #45 cm# when a #4 kg# weight is hanging from it, what is the spring's constant?

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Answer 1 · 2018-07-01T14:27:40

#122.5# newtons per meter.

We use Hooke's law, which states that,

#F=kx#

where:

#F# is the force applied in newtons

#k# is the spring's constant

#x# is the extension in meters

Here we have #x=45 \ "cm"-13 \ "cm"=32 \ "cm"=0.32 \ "m"#, #F=4 \ "kg"*9.8 \ "m/s"^2=39.2 \ "N"#.

So, the spring's constant is:

#k=F/x#

#=(39.2 \ "N")/(0.32 \ "m")#

#=122.5 \ "N/m"#