# An object with a mass of 7 kg is hanging from a spring with a constant of 2 kgs^-2. If the spring is stretched by  9 m, what is the net force on the object?

Jan 22, 2016

The net force, ${F}_{n}$, includes the weight force, ${F}_{w}$, acting on the object and the spring force, ${F}_{s}$, acting in the opposite direction:

${F}_{n} = {F}_{w} + {F}_{s} = 68.6 + \left(- 18\right) = 50.6$ $N$ downward.

#### Explanation:

The spring constant can also be expressed in units of $N {m}^{-} 1$. Dimensional analysis of units shows this is exactly equivalent to $k g {s}^{-} 2$, but it makes what is happening much more obvious: if the spring constant is $k$ $N {m}^{-} 1$, then for every $m$ the spring stretches a force of $k$ $N$ is exerted.

If the spring increases its length by $9 m$, the upward force on the object is $9 \times 2 = 18$ $N$. Call this $- 18$ $N$ since it is in the opposite direction to the weight force.

The downward weight force on the object is $F = m g = 7 \cdot 9.8 = 68.6$ $N$. We will define downward as the positive direction.

Add the weight force, ${F}_{w}$, to the spring force, ${F}_{s}$, to find the net force, ${F}_{n}$:

${F}_{n} = {F}_{w} + {F}_{s} = 68.6 + \left(- 18\right) = 50.6$ $N$ downward.