If the length of a $46 c m$ $46 cm$ spring increases to $77 c m$ $77 cm$ when a $16 k g$ $16 kg$ weight is hanging from it, what is the spring's constant?

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If the length of a $46 c m$ $46 cm$ spring increases to $77 c m$ $77 cm$ when a $16 k g$ $16 kg$ weight is hanging from it, what is the spring's constant?

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Answer 1 · 2016-02-17T11:41:23

Hooke's Law can be rearranged as $k = \frac{F}{x}$ $k=F/x$ . In this case $F$ $F$ is the weight force, $m g = 16 \cdot 9.8 = 156.8$ $mg=16*9.8=156.8$ $N$ $N$ . The displacement, $x$ $x$ is $0.77 - 0.46 = 0.31$ $0.77-0.46=0.31$ $m$ $m$ . So $k = \frac{156.8}{0.31} = 505.8$ $k=156.8/0.31=505.8$ $N k g^{- 1}$ $Nkg^-1$ (or $k g^{s} - 2$ $kg^s-2$ ).

Explanation:

Not much more explanation needed, except to note that it was necessary to convert from $c m$ $cm$ to $m$ $m$ , the correct SI unit for length and distance.

The units of the spring constant can be expressed as $N k g^{- 1}$ $Nkg^-1$ or as $k g s^{- 2}$ $kgs^-2$ .

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If the length of a $46 c m$ $46 cm$ spring increases to $77 c m$ $77 cm$ when a $16 k g$ $16 kg$ weight is hanging from it, what is the spring's constant?

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If the length of a 46cm46 cm spring increases to 77cm77 cm when a 16kg16 kg weight is hanging from it, what is the spring's constant?

1 Answer

Explanation:

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If the length of a $46 c m$ $46 cm$ spring increases to $77 c m$ $77 cm$ when a $16 k g$ $16 kg$ weight is hanging from it, what is the spring's constant?