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?

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Answer 1 · 2018-05-29T13:05:20

About $204.17$ kilograms per second squared.

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:

$k=F/x$

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

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

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

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

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

If the length of a 49 cm49 cm spring increases to 73 cm73 cm when a 5 kg5 kg weight is hanging from it, what is the spring's constant?