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  5 m, what is the net force on the object?

Feb 22, 2017

The net force on the object is $= 27.2 N$ (acting downwards)

Explanation:

We resolve in the vertical direction ${\downarrow}_{+}$

The weight acts downwards

${\vec{F}}_{1} = 4 g = 4 \cdot 9.8 = 39.2 N$

According to Hooke's Law

${\vec{F}}_{2} = - k x$

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

The net force is

$\vec{F} = {\vec{F}}_{1} + {\vec{F}}_{2} = 39.2 - 12 = 27.2 N$