# If the length of a 21 cm spring increases to 45 cm when a 2 kg weight is hanging from it, what is the spring's constant?

Jan 8, 2017

I got $k = 81.7 \frac{N}{m}$

#### Explanation:

We can us Hooke's Law to relate the elastic force of a spring of elastic constant $k$ after being elongated of $x$:
${F}_{e l} = - k x$
Also, Newton's Second Law will help us to consider the forces involved through:
$\Sigma \vec{F} = m \vec{a}$

let us see:

So at the equilibrium the elastic force and the weight ($W = m g$) will give zero acceleration:
$- k x - m g = 0$
But the elongation is directed downwards so we write, using numbers:
$- k \left(- 0.24\right) - \left(2 \cdot 9.8\right) = 0$
So:
$k = \frac{2 \cdot 9.8}{0.24} = 81.7 \frac{N}{m}$
I changed the lengths into meters because I am used to this but if you need it in N/cm use $24 c m$ instead of $0.24 m$ in the above.