If the length of a $12 c m$ $12 cm$ spring increases to $20 c m$ $20 cm$ when a $10 k g$ $10 kg$ weight is hanging from it, what is the spring's constant?

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If the length of a $12 c m$ $12 cm$ spring increases to $20 c m$ $20 cm$ when a $10 k g$ $10 kg$ weight is hanging from it, what is the spring's constant?

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Answer 1 · 2015-12-09T15:13:57

$1225 \frac{N}{m}$ $1225 N/m$

Explanation:

A spring constant for a 'simple spring' is usually written as the $k$ $k$ in:
$F = - k x$ $F= -kx$

For force of a $10 k g$ $10 kg$ weight is calculated using gravity:
$F = m g$ $F = mg$

$F = 10 \cdot 9.8 = 98 N$ $F = 10*9.8 = 98 N$

The $x$ $x$ in the equation above is relative to the relaxed length of the spring. If the length of the spring is $12 c m$ $12 cm$ without any mass attached, we call this position zero and we can see that the additional length with the mass attached is $8 c m$ $8 cm$ longer.

We now know $F$ $F$ and $x$ $x$ and we can solve for $k$ $k$ . Notice that the direction of the force is opposite to the direction of gravity, so the sign has changed.

$F = - k x$ $F = -kx$

$- 98 N = - k \cdot 0.08 m$ $-98 N = -k * 0.08 m$

$\frac{98}{0.08} \frac{N}{m} = k = 1225 \frac{N}{m}$ $98/0.08 N/m = k = 1225 N/m$

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If the length of a $12 c m$ $12 cm$ spring increases to $20 c m$ $20 cm$ when a $10 k g$ $10 kg$ 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 12cm12 cm spring increases to 20cm20 cm when a 10kg10 kg weight is hanging from it, what is the spring's constant?

1 Answer

Explanation:

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If the length of a $12 c m$ $12 cm$ spring increases to $20 c m$ $20 cm$ when a $10 k g$ $10 kg$ weight is hanging from it, what is the spring's constant?