A spring extends in length by #4.0# #cm# when a mass of #1.5# #kg# is hung from it. What is the spring constant of the spring?

1 Answer
Andy Y. · David G.
Jun 7, 2017

#k=(F_x)/(x-x_0)=(14.7" N")/(0.04" m")=367.5# #Nm^-1#

Explanation:

Let these variables represent the following

#F_x=# the force applied to the spring

#k=# the spring force constant

#x=# the distance from equilibrium

#x_0=# the spring equilibrium position

The equation to determine the spring constant, #k# is

#k=(F_x)/(x-x_0)#

We are given the total measure of elongation is #4# #cm#, which is the difference between #x# and #x_0#. That is,

#x-x_0=4# #cm=0.04# #m#

We are also given the mass of the body, #1.5# #kg#, and assuming this measurement is happening on Earth, we have the acceleration due to gravity #9.8# ms#""^(-2)#.

#F_x=9.8# #(ms^-2)xx1.5# #kg=14.7# #N#

Therefore, the final solution is

#k=(F_x)/(x-x_0)=(14.7" N")/(0.04" m")=367.5# #Nm^-1#

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