If the length of a $48$ $48$ $c m$ $cm$ spring increases to $93$ $93$ $c m$ $cm$ when a $6$ $6$ $k g$ $kg$ weight is hanging from it, what is the spring's constant?

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If the length of a $48$ $48$ $c m$ $cm$ spring increases to $93$ $93$ $c m$ $cm$ when a $6$ $6$ $k g$ $kg$ weight is hanging from it, what is the spring's constant?

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Answer 1 · 2016-05-21T23:28:42

First, we need to work in SI units - meters, not centimeters.

The length increases from $0.48$ $0.48$ $m$ $m$ to $0.93$ $0.93$ $m$ $m$ . The force acting is the weight force, $F = m g = 6 \cdot 9.8 = 58.8$ $F=mg=6*9.8=58.8$ $N$ $N$ .

Then $k = \frac{F}{x} = \frac{58.8}{0.93 - 0.48} \approx 130.7$ $k=F/x=58.8/(0.93-0.48)~~130.7$ $N m^{- 1}$ $Nm^-1$ (or $k g s^{- 2}$ $kgs^-2$ ).

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If the length of a $48$ $48$ $c m$ $cm$ spring increases to $93$ $93$ $c m$ $cm$ when a $6$ $6$ $k g$ $kg$ weight is hanging from it, what is the spring's constant?

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