An object with a mass of #6 kg# is hanging from a spring with a constant of #12 (kg)/s^2#. If the spring is stretched by # 29 m#, what is the net force on the object?

1 Answer
Oct 5, 2017

The net force is 289.2 N upward.

Explanation:

Giving a spring's constant in units of #(kg)/s^2# seems useless to me. But it is equivalent to the more useful units #N/m#. Proof:

From Newton's 2nd Law, we can see that
the Newton, N, is equivalent to #(kg*m)/s^2#.
An algebraic operation shows us that
#s^2# is equivalent to #(kg*m)/N#.

Using that last equivalency on the spring constant given as
#12 (kg)/s^2#,
it can be reworked as follows
#12 (kg)/s^2 = 12 (kg)/((kg*m)/N) = 12 N/m rarr# QED

OK, my rant is over.

If the spring is stretched by 29 m, the spring's force is given by
#12 N/m * 29 m = 348 N#
That force would be upward.

The mass's weight is a force downward. The value of that weight is
#W = m*g = 6 kg*9.8 m/s^2 = 58.8 " kg"*m/s^2 = 58.8 N#

The net force is the vector sum of the spring's force and the mass's weight
#348 N - 58.8 N = 289.2 N#

This mass is apparently in a cycle of bouncing up and down on the spring.

I hope this helps,
Steve