If the length of a $24 c m$ $24 cm$ spring increases to $59 c m$ $59 cm$ when a $5 k g$ $5 kg$ weight is hanging from it, what is the spring's constant?

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If the length of a $24 c m$ $24 cm$ spring increases to $59 c m$ $59 cm$ when a $5 k g$ $5 kg$ weight is hanging from it, what is the spring's constant?

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Answer 1 · 2015-12-11T02:50:31

I found: $k = 1.96 \frac{N}{m}$ $k=1.96N/m$

Explanation:

Consider the diagram:
enter image source here
Choosing the origin at the extreme of the un-extended spring:
at equilibrium:
$F_{e l} - W = 0$ $F_(el)-W=0$
$- k y = m g$ $-ky=mg$
$- k (- h) = m g$ $-k(-h)=mg$
$- k (- 25) = 5 \cdot 9.8$ $-k(-25)=5*9.8$
$k = 1.96 \frac{N}{m}$ $k=1.96N/m$

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If the length of a $24 c m$ $24 cm$ spring increases to $59 c m$ $59 cm$ when a $5 k g$ $5 kg$ weight is hanging from it, what is the spring's constant?

1 Answer

Explanation:

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Humanities

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If the length of a 24cm24 cm spring increases to 59cm59 cm when a 5kg5 kg weight is hanging from it, what is the spring's constant?

1 Answer

Explanation:

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If the length of a $24 c m$ $24 cm$ spring increases to $59 c m$ $59 cm$ when a $5 k g$ $5 kg$ weight is hanging from it, what is the spring's constant?