Question

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?

Answer 1

The net force is `=3.2N` acting downwards

Explanation:

The mass is `m=4kg`

The spring constant is `k=3kgs^-2`

The extension is `x=12m`

Resolving in a vertical direction `uarr^+`

The net force is

`F=k*x-mg`

`=3*12-4*9.8`

`=-3.2N`

Answer 2

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*x` where k is referred to as Hooke's constant and x is the displacement.Substituting for values we have `F_s=-3*12=-36N`

The gravitational force is given by the equation `F_g=m*g` which when substituted gives `F_g=4 kg * 9.8 m/s^2 =39.2N`
In a state of equilibrium `F_s` should be equal to `F_g`.