# An object with a mass of 6 kg is hanging from a spring with a constant of 3 (kg)/s^2. If the spring is stretched by 9 m, what is the net force on the object?

Sep 1, 2017

The net force is ${F}_{\text{net}} = 27 N$.

#### Explanation:

My interpretation: The mass is initially hanging, in equilibrium, so ${F}_{\text{net}} = 0$. Then an outside force pulls it down 9m from the initial position. I could say that the mass is now in another equilibrium and then say again that ${F}_{\text{net}} = 0$. But that would be trivial.

Therefore I assume that the outside force is suddenly released and we are asked for net force at that moment.

Hooke's Law gives us that the spring's additional force due to that 9m displacement is
$F = k \cdot x = 3 \frac{k g}{s} ^ 2 \cdot 9 m = 27 \frac{k g \cdot m}{s} ^ 2 = 27 N$
This is the net force on the mass when freed at that position.

Since the external force has been removed, the mass will then start bouncing up and down and net force will vary according to position.

I hope this helps,
Steve