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

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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,

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

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

$=428.75 \ "N/m"$

If the length of a 21 cm21cm21 cm spring increases to 37 cm37cm37 cm when a 7 kg7kg7 kg weight is hanging from it, what is the spring's constant?