If the length of a $45 c m$ $45 cm$ spring increases to $86 c m$ $86 cm$ when a $1 k g$ $1 kg$ weight is hanging from it, what is the spring's constant?

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Answer 1 · 2018-05-05T14:40:11

Approximately $24$ $24$ newtons per meter

Well, it looks to me that the spring will increase in length towards the direction of the weight hanging from it, so Hooke's law here will be:

$F = k x$ $F=kx$

$F$ $F$ is the force exerted in newtons

$k$ $k$ is the spring constant

$x$ $x$ is the extension of the spring

The force here is the object's weight, which is $1 \ "kg"*9.8 \ "m/s"^2=9.8 \ "N"$ .

Therefore, the spring constant is:

$k=F/x$

$=(9.8 \ "N")/(86 \ "cm"-45 \ "cm")$

$=(9.8 \ "N")/(41 \ "cm")$

$~~0.24 \ "N/cm"$

$=24 \ "N/m"$

If the length of a 45 cm45cm45 cm spring increases to 86 cm86cm86 cm when a 1 kg1kg1 kg weight is hanging from it, what is the spring's constant?