An object with a mass of #7 kg# is hanging from a spring with a constant of #2 kgs^-2#. If the spring is stretched by # 9 m#, what is the net force on the object?

1 Answer
Jan 22, 2016

The net force, #F_n#, includes the weight force, #F_w#, acting on the object and the spring force, #F_s#, acting in the opposite direction:

#F_n = F_w + F_s = 68.6 + (-18) = 50.6# #N# downward.

Explanation:

The spring constant can also be expressed in units of #Nm^-1#. Dimensional analysis of units shows this is exactly equivalent to #kgs^-2#, but it makes what is happening much more obvious: if the spring constant is #k# #Nm^-1#, then for every #m# the spring stretches a force of #k# #N# is exerted.

If the spring increases its length by #9m#, the upward force on the object is #9xx2=18# #N#. Call this #-18# #N# since it is in the opposite direction to the weight force.

The downward weight force on the object is #F=mg=7*9.8=68.6# #N#. We will define downward as the positive direction.

Add the weight force, #F_w#, to the spring force, #F_s#, to find the net force, #F_n#:

#F_n = F_w + F_s = 68.6 + (-18) = 50.6# #N# downward.