# An object with a mass of 4 kg is hanging from a spring with a constant of 3 (kg)/s^2. If the spring is stretched by  12 m, what is the net force on the object?

##### 2 Answers
Jul 11, 2017

The net force is $= 3.2 N$ acting downwards

#### Explanation:

The mass is $m = 4 k g$

The spring constant is $k = 3 k g {s}^{-} 2$

The extension is $x = 12 m$

Resolving in a vertical direction ${\uparrow}^{+}$

The net force is

$F = k \cdot x - m g$

$= 3 \cdot 12 - 4 \cdot 9.8$

$= - 3.2 N$

Jul 11, 2017

The net force should ideally be zero

#### Explanation:

The object hanging by the spring is acted on by two forces.The gravitational force pulls its down while the spring resists this.

The force exerted by spring is given by Hooke's law as follows ${F}_{s} = - k \cdot x$ where k is referred to as Hooke's constant and x is the displacement.Substituting for values we have ${F}_{s} = - 3 \cdot 12 = - 36 N$

The gravitational force is given by the equation ${F}_{g} = m \cdot g$ which when substituted gives ${F}_{g} = 4 k g \cdot 9.8 \frac{m}{s} ^ 2 = 39.2 N$
In a state of equilibrium ${F}_{s}$ should be equal to ${F}_{g}$.