An object with a mass of #6# #kg# is hanging from a spring with a constant of #12# #kgs^-2#. If the spring is stretched by # 2# #m#, what is the net force on the object?

1 Answer
May 5, 2018

Answer:

Hooke's Law says: #F=kx# - the force acting is the product of the spring constant and the displacement of the object from the spring's equilibrium position:

#F=kx=12xx2=24# #N#

Explanation:

(We do not need to know the mass of the object to answer this particular question.)

I have mentioned this in other answers, but I prefer the unit #Nm^-1# rather than #kgs^-2# for expressing the spring constant.

The two units are equivalent, since one newton is one #kgms^-2#.

Given that we are usually talking about forces in #N# and displacements in #m#, the unit #Nm^-1# is just more directly useful.

Just remember that both are directly equivalent and mean the same thing.