If the length of a #46 cm# spring increases to #77 cm# when a #16 kg# weight is hanging from it, what is the spring's constant?

1 Answer
Feb 17, 2016

Answer:

Hooke's Law can be rearranged as #k=F/x#. In this case #F# is the weight force, #mg=16*9.8=156.8# #N#. The displacement, #x# is #0.77-0.46=0.31# #m#. So #k=156.8/0.31=505.8# #Nkg^-1# (or #kg^s-2#).

Explanation:

Not much more explanation needed, except to note that it was necessary to convert from #cm# to #m#, the correct SI unit for length and distance.

The units of the spring constant can be expressed as #Nkg^-1# or as #kgs^-2#.