# If the length of a 29 cm spring increases to 63 cm when a 15 kg weight is hanging from it, what is the spring's constant?

Dec 19, 2015

#### Answer:

I calculate $k = 432 N {m}^{-} 1$ (3 s.f.)

#### Explanation:

The equation we use to calculate the force constant of a spring is Hooke's Law , $F = - k x$

To make it simpler, I'm going to ignore the negative sign (so $F = k x$). It is only there to show the force is conservative, which means it tries to pull the spring back to its original length when it's extended.

We need to rearrange this equation to make $k$ the subject. This is done by dividing both sides of the equation by $x$ to give us:

$k = \frac{F}{x}$

The value of $F$ is the weight force of the mass. Assuming the acceleration due to gravity $g = 9.8 m {s}^{-} 2$:

$F = m g$
$F = 15 \cdot 9.8$
$F = 147 N$

The value of $x$ is the amount by which the spring's length is extended (or compressed). It's the difference between $29 c m$ and $63 c m$.

$x = 63 - 29$
$x = 34 c m = 0.34 m$

We will use the value in metres, as it's the proper SI unit for distance. Now we can substitute these values into the rearranged equation to find the spring's constant!

$k = \frac{F}{x}$
$k = \frac{147}{0.34}$
$k = 432 N {m}^{-} 1$ (3 s.f.)