Question

An object with a mass of `6` `kg` is hanging from a spring with a constant of `12` `kgs^-2`. If the spring is stretched by `2` `m`, what is the net force on the object?

Answer 1

Hooke's Law says: `F=kx` - the force acting is the product of the spring constant and the displacement of the object from the spring's equilibrium position:

`F=kx=12xx2=24` `N`

Explanation:

(We do not need to know the mass of the object to answer this particular question.)

I have mentioned this in other answers, but I prefer the unit `Nm^-1` rather than `kgs^-2` for expressing the spring constant.

The two units are equivalent, since one newton is one `kgms^-2`.

Given that we are usually talking about forces in `N` and displacements in `m`, the unit `Nm^-1` is just more directly useful.

Just remember that both are directly equivalent and mean the same thing.