If the length of a $26 c m$ $26 cm$ spring increases to $43 c m$ $43 cm$ 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 $26 c m$ $26 cm$ spring increases to $43 c m$ $43 cm$ when a $4 k g$ $4 kg$ weight is hanging from it, what is the spring's constant?

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Answer 1 · 2016-06-23T22:28:18

$k = 230.82 k \frac{g}{m}$ $k = 230.82kg/m$

Explanation:

Hooke's Law states that the spring constant $(k)$ $(k)$ and displacement of the spring are related to the Force applied on the spring by the equation $F = - k x$ $F =-kx$

The Force is determined by the mass times gravity.

$m = 4 k g$ $m = 4kg$
$g = 9.81 \frac{m}{s^{2}}$ $g = 9.81 m/s^2$

$F = (4 k g) (9.81 \frac{m}{s^{2}})$ $F = (4kg)(9.81m/s^2)$

$F = 39.24 N$ $F = 39.24 N$

$k = ?$ $k =?$
$x = - 43 c m + 26 c m = 17 c m = - 0.17 m$ $x = -43cm+26cm = 17cm = -0.17m$

$39.24 N = - k (- 0.17 m)$ $39.24 N = -k(-0.17 m)$

$\frac{39.24 N}{- 0.17 m} = - k$ $(39.24N)/(-0.17m) = -k$

$- 230.82 = - k$ $-230.82 = -k$

$k = 230.82 k \frac{g}{m}$ $k = 230.82kg/m$

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If the length of a $26 c m$ $26 cm$ spring increases to $43 c m$ $43 cm$ when a $4 k g$ $4 kg$ weight is hanging from it, what is the spring's constant?

1 Answer

Explanation:

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If the length of a 26cm26 cm spring increases to 43cm43 cm when a 4kg4 kg weight is hanging from it, what is the spring's constant?

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Explanation:

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If the length of a $26 c m$ $26 cm$ spring increases to $43 c m$ $43 cm$ when a $4 k g$ $4 kg$ weight is hanging from it, what is the spring's constant?