# 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  8 m, what is the net force on the object?

May 19, 2018

$63.2$ newtons

#### Explanation:

The net force of the spring is basically the force exerted on the spring by Hooke's law and the weight of the object.

Hooke's law states that:

$F = k x$

where:

• $F$ is the force applied in newtons

• $k$ is the spring constant

• $x$ is the extension of the spring

So we get:

${F}_{1} = 3 \setminus \text{kg/s"^2*8 \ "m}$

$= 24 \setminus {\text{kg m/s}}^{2}$

$= 24 \setminus \text{N}$

The weight of the object is:

${F}_{2} = 4 \setminus {\text{kg"*9.8 \ "m/s}}^{2}$

$= 39.2 \setminus \text{N}$

So, the total net force on the object is:

${F}_{\text{net}} = {F}_{1} + {F}_{2}$

$= 24 \setminus \text{N"+39.2 \ "N}$

$= 63.2 \setminus \text{N}$