If the length of a #21 cm# spring increases to #37 cm# when a #7 kg# weight is hanging from it, what is the spring's constant?

Question

1650 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-10T12:40:21

I get #428.75 \ "N/m"#.

We use Hooke's law, which states that,

#F=kx#

#k# is the spring constant in newtons per meter

#x# is the extension of the spring in meters

So here, we got:

#x=37 \ "cm"-21 \ "cm"#

#=16 \ "cm"#

#=0.16 \ "m"#

The force is the weight of the weight thingy, which is #7 \ "kg"*9.8 \ "m/s"^2=68.6 \ "N"#.

And so,

#68.6 \ "N"=k*0.16 \ "m"#

#k=(68.6 \ "N")/(0.16 \ "m")#

#=428.75 \ "N/m"#