# 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?

Oct 5, 2017

The net force is 289.2 N upward.

#### Explanation:

Giving a spring's constant in units of $\frac{k g}{s} ^ 2$ seems useless to me. But it is equivalent to the more useful units $\frac{N}{m}$. Proof:

From Newton's 2nd Law, we can see that
the Newton, N, is equivalent to $\frac{k g \cdot m}{s} ^ 2$.
An algebraic operation shows us that
${s}^{2}$ is equivalent to $\frac{k g \cdot m}{N}$.

Using that last equivalency on the spring constant given as
$12 \frac{k g}{s} ^ 2$,
it can be reworked as follows
$12 \frac{k g}{s} ^ 2 = 12 \frac{k g}{\frac{k g \cdot m}{N}} = 12 \frac{N}{m} \rightarrow$ QED

OK, my rant is over.

If the spring is stretched by 29 m, the spring's force is given by
$12 \frac{N}{m} \cdot 29 m = 348 N$
That force would be upward.

The mass's weight is a force downward. The value of that weight is
$W = m \cdot g = 6 k g \cdot 9.8 \frac{m}{s} ^ 2 = 58.8 \text{ kg} \cdot \frac{m}{s} ^ 2 = 58.8 N$

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