If the length of a $45 c m$ $45 cm$ spring increases to $63 c m$ $63 cm$ when a $7 k g$ $7 kg$ weight is hanging from it, what is the spring's constant?

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If the length of a $45 c m$ $45 cm$ spring increases to $63 c m$ $63 cm$ when a $7 k g$ $7 kg$ weight is hanging from it, what is the spring's constant?

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Answer 1 · 2018-06-11T00:21:34

Approximately $381.1$ $381.1$ newtons per meter.

Explanation:

We use Hooke's law, which states that,

$F = k x$ $F=kx$

where:

$F$ $F$ is the force applied in newtons

$k$ $k$ is the spring constant in ${kg/s}^{2}$ $"kg/s"^2$ or $N/m$ $"N/m"$

$x$ $x$ is the extension in meters

Here, the force is the weight of the object, which is equal to:

$7 \ "kg"*9.8 \ "m/s"^2=68.6 \ "N"$ .

The spring constant is then:

$k=F/x$

$=(68.6 \ "N")/(0.63 \ "m"-0.45 \ "m")$

$=(68.6 \ "N")/(0.18 \ "m")$

$~~381.1 \ "N/m"$

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If the length of a $45 c m$ $45 cm$ spring increases to $63 c m$ $63 cm$ when a $7 k g$ $7 kg$ weight is hanging from it, what is the spring's constant?

1 Answer

Explanation:

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If the length of a 45 cm45cm45 cm spring increases to 63 cm63cm63 cm when a 7 kg7kg7 kg weight is hanging from it, what is the spring's constant?

1 Answer

Explanation:

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If the length of a $45 c m$ $45 cm$ spring increases to $63 c m$ $63 cm$ when a $7 k g$ $7 kg$ weight is hanging from it, what is the spring's constant?