If the length of a $26 c m$ $26 cm$ spring increases to $46 c m$ $46 cm$ when a $8 k g$ $8 kg$ 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 $46 c m$ $46 cm$ when a $8 k g$ $8 kg$ weight is hanging from it, what is the spring's constant?

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Answer 1 · 2018-03-03T18:18:10

$392.4$ $392.4$ N/m

Explanation:

the spring constant is the direct proportion between:

the force applied on the string

and the extension of the string when the force is applied on it.

they have the relationship

$F = k \cdot e$ $F = k * e$

where

$F$ $F$ is the force in newtons, N

$k$ $k$ is the 'spring constant' in newtons per metre, N/m

and $e$ $e$ is the extension in metres, m.

here, the force is that applied on the $8 k g$ $8kg$ weight.

$W = m g$ $W = mg$

where $W$ $W$ is weight (force due to gravity)

$m$ $m$ is mass

and $g$ $g$ is gravitational field strength.

on Earth, $g \approx 9.81$ $g ~~ 9.81$ newtons per kilogram, N/kg.

the force applied is $8$ $8$ kg $\cdot 9.81$ $* 9.81$ N/kg

which is $78.48$ $78.48$ N.

the extension is from $26$ $26$ cm to $46$ $46$ cm.

$46$ $46$ cm $- 26$ $- 26$ cm $= 20$ $= 20$ cm

$20$ $20$ cm is $\frac{20}{100}$ $20/100$ m, which is $0.2$ $0.2$ m.

$F = k \cdot e$ $F = k * e$

$78.48$ $78.48$ N $= k \cdot 0.2$ $= k * 0.2$ m

$k = (\frac{78.48}{0.2})$ $k = (78.48/0.2)$ N/m

$\frac{78.48}{0.2} = 78.48 \cdot 5 = 392.4$ $78.48/0.2 = 78.48 * 5 = 392.4$

$k = 392.4$ $k = 392.4$ N/m

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If the length of a $26 c m$ $26 cm$ spring increases to $46 c m$ $46 cm$ when a $8 k g$ $8 kg$ weight is hanging from it, what is the spring's constant?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

If the length of a 26cm26 cm spring increases to 46cm46 cm when a 8kg8 kg weight is hanging from it, what is the spring's constant?

1 Answer

Explanation:

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If the length of a $26 c m$ $26 cm$ spring increases to $46 c m$ $46 cm$ when a $8 k g$ $8 kg$ weight is hanging from it, what is the spring's constant?