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

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Answer 1 · 2018-07-01T14:27:40

$122.5$ $122.5$ newtons per meter.

We use Hooke's law, which states that,

$F = k x$ $F=kx$

where:

$F$ $F$ is the force applied in newtons

$k$ $k$ is the spring's constant

$x$ $x$ 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:

$k=F/x$

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

$=122.5 \ "N/m"$

If the length of a 13 cm13cm13 cm spring increases to 45 cm45cm45 cm when a 4 kg4kg4 kg weight is hanging from it, what is the spring's constant?