# If the length of a 13 cm spring increases to 40 cm when a 7 kg weight is hanging from it, what is the spring's constant?

Oct 22, 2016

I got: $k = 254 \frac{N}{m}$

#### Explanation:

Consider that at equilibrium the elastic force ${F}_{e l}$ (given by Hooke's Law: ${F}_{e l} = - k x$) and the weight $W$ must balance to give, into Newton's Second Law, zero acceleration (the sistem is at rest).
We get:
${F}_{e l} + W = m a = 0$
$- k x - m g = 0$
(eleasic force upwards and weight downwards)
The displacement $x$ is considered negative (downwards) so we have:
$- k \left[- \left(0.4 - 0.13\right)\right] - \left(7 \cdot 9.8\right) = 0$
(I changed into meters)
rearranging we get:
$k = 254 \frac{N}{m}$