If the length of a $27 cm$ spring increases to $43 cm$ when a $7 kg$ weight is hanging from it, what is the spring's constant?

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Answer 1 · 2015-12-09T19:38:40

$428,75N//m$

From Hooke's Law, $F=kx$ from which $k=F/x$ , where x is the displacement of the spring from its equilibrium position.

$therefore k=(7xx9,8)/((43-27)xx10^(-2)m)=428,75N//m$

If the length of a 27 cm27 cm spring increases to 43 cm43 cm when a 7 kg7 kg weight is hanging from it, what is the spring's constant?