If the length of a $23 cm$ spring increases to $96 cm$ when a $5 kg$ weight is hanging from it, what is the spring's constant?

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Answer 1 · 2018-03-07T14:30:07

I got $67N/m$

We can use Hooke's Law:
$F_(el)=-kx$
Newton's Second Law...
and the diagram:
enter image source here

At equilibrium:
$F_(el)-W=0$
$-kx-mg=0$
using our data and from our choice of axes (changing in meters):

$-k(-0.73)-(5*9.8)=0$

$k=(5*9.8)/0.73=67N/m$

If the length of a 23 cm23 cm spring increases to 96 cm96 cm when a 5 kg5 kg weight is hanging from it, what is the spring's constant?