If the length of a #25 cm# spring increases to #67 cm# when a #15 kg# weight is hanging from it, what is the spring's constant?

1 Answer
Apr 11, 2016

Answer:

#k = 15000/42 = 7500/21 = 2500/7 ("Newtons")/(meters)# or
#k = 357.14# #"Newtons"/"meter"#

Explanation:

#F_(el) = k*x#
#F_(el)# is the force upon the spring, the "elastic" force;
#k# is the spring constant;
#x# is the deformity of the spring;
#F_w# is the force from the weight;
#g# is gravity;
#m# is mass.

Since the length of the spring is #25cm#, the deformity is #67 - 25#, or #42cm#. Using the International Sistem (or S.I.), the deformity must be measured in meters, since the constant(#k#) is measured in Newtons/meter. So # x =0.42# #meters#.

The force applied to the spring is the weight force from the weight hanging from the spring. So #F_w = F_(el)#. Now, we apply the math:
#m*g = k*x#
#15 * 10 = k * 0.42#
#k = 150/(42/100)#
#k = 150/1 * 100/42#
#k = 15000/42 = 7500/21 = 2500/7 ("Newtons")/(meters)# or
#k = 357.14# #"Newtons"/"meter"#