If the length of a $23 c m$ $23 cm$ spring increases to $46 c m$ $46 cm$ when a $16 k g$ $16 kg$ weight is hanging from it, what is the spring's constant?

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Answer 1 · 2017-04-26T10:15:17

$682 N m^{- 1}$ $682 N m^(-1)$

Using Hooke's Law, $F = k x$ $F=kx$ ,

$Delta F = k Delta x$
$(Delta m)g = k Delta x$
$(16)(9.81)= k ((46-23)*10^(-2))$
$k= 682Nm^(-1)$

If the length of a 23 cm23cm23 cm spring increases to 46 cm46cm46 cm when a 16 kg16kg16 kg weight is hanging from it, what is the spring's constant?