If the length of a $15 c m$ $15 cm$ spring increases to $42 c m$ $42 cm$ when a $4 k g$ $4 kg$ weight is hanging from it, what is the spring's constant?

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If the length of a $15 c m$ $15 cm$ spring increases to $42 c m$ $42 cm$ when a $4 k g$ $4 kg$ weight is hanging from it, what is the spring's constant?

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Answer 1 · 2016-01-19T22:50:12

I found $145.2 \frac{N}{m}$ $145.2N/m$

Explanation:

I tried using Hooke's Law to describe the elastic force $F_{e l}$ $F_(el)$ , and Newton's law to describe the equilibrium of forces:
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According to our choice of axis the displacemeht can be considered negative (below the chosen origin) $y = - h = - 0.27 m$ $y=-h=-0.27m$ ; so we get:
$k = \frac{m g}{- y} = \frac{4 \cdot 9.8}{- (- 0.27)} = 145.2 \frac{N}{m}$ $k=(mg)/(-y)=(4*9.8)/-(-0.27)=145.2N/m$

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If the length of a $15 c m$ $15 cm$ spring increases to $42 c m$ $42 cm$ when a $4 k g$ $4 kg$ weight is hanging from it, what is the spring's constant?

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If the length of a 15 cm15cm15 cm spring increases to 42 cm42cm42 cm when a 4 kg4kg4 kg weight is hanging from it, what is the spring's constant?

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If the length of a $15 c m$ $15 cm$ spring increases to $42 c m$ $42 cm$ when a $4 k g$ $4 kg$ weight is hanging from it, what is the spring's constant?