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?

1 Answer
Oct 22, 2016

Answer:

I got: #k=254N/m#

Explanation:

Consider that at equilibrium the elastic force #F_(el)# (given by Hooke's Law: #F_(el)=-kx#) and the weight #W# must balance to give, into Newton's Second Law, zero acceleration (the sistem is at rest).
We get:
#F_(el)+W=ma=0#
#-kx-mg=0#
(eleasic force upwards and weight downwards)
The displacement #x# is considered negative (downwards) so we have:
#-k[-(0.4-0.13)]-(7*9.8)=0#
(I changed into meters)
rearranging we get:
#k=254N/m#