If the length of a $29 c m$ $29 cm$ spring increases to $61 c m$ $61 cm$ when a $6 k g$ $6 kg$ weight is hanging from it, what is the spring's constant?

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If the length of a $29 c m$ $29 cm$ spring increases to $61 c m$ $61 cm$ when a $6 k g$ $6 kg$ weight is hanging from it, what is the spring's constant?

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Answer 1 · 2016-02-05T18:36:43

k = 6xx9.81/(32/100) = 6xx981/32 = 184 N/m#

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Explanation:

See the free body diagram on the image, at equilibrium (0 acceleration) we have :
$sumF = F_k - mg = kDeltax - mg = 0$
$k = mg/(Deltax)$
You are given $m = 6kg; x_i = 29; x_f = 61$
$Deltax = x_f - x_i = 61 - 29 = 32 cm = 32/100$
$k = 6xx9.81/(32/100) = 6xx981/32 = 184 N/m$

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If the length of a $29 c m$ $29 cm$ spring increases to $61 c m$ $61 cm$ when a $6 k g$ $6 kg$ weight is hanging from it, what is the spring's constant?

1 Answer

Explanation:

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If the length of a 29 cm29cm29 cm spring increases to 61 cm61cm61 cm when a 6 kg6kg6 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 $29 c m$ $29 cm$ spring increases to $61 c m$ $61 cm$ when a $6 k g$ $6 kg$ weight is hanging from it, what is the spring's constant?