Question

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?

Answer 1

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