# 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?

May 5, 2018

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

$F = k x = 12 \times 2 = 24$ $N$

#### 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 $N {m}^{-} 1$ rather than $k g {s}^{-} 2$ for expressing the spring constant.

The two units are equivalent, since one newton is one $k g m {s}^{-} 2$.

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

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