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

Dec 9, 2015

#### Answer:

$1225 \frac{N}{m}$

#### Explanation:

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

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

$F = 10 \cdot 9.8 = 98 N$

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

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

$F = - k x$

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

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