# An object with a mass of 5 kg is hanging from a spring with a constant of 5 kgs^-2. If the spring is stretched by 3 m, what is the net force on the object?

Jan 31, 2016

The net force on the mass is $49 + \left(- 15\right) = 34$ $N$ in the downward direction.

#### Explanation:

(I've ranted about this in other answers, so won't make too much of a point of it here, but the units for the spring constant $N {m}^{-} 1$ are exactly equivalent to $k g {s}^{-} 2$, and make this kind of question much easier to understand: if the spring constant is $k$ $N {m}^{-} 1$, extending a spring by $1$ $m$ increases the force by $k$ $N$.)

The mass has a downward weight force due to gravity acting on it:

$F = m g = 5 \cdot 9.8 = 49$ $N$

Where $m$ is the mass $\left(k g\right)$ and $g$ is the acceleration due to gravity $\left(m {s}^{-} 2\right)$ or (equivalently) $\left(N k {g}^{-} 1\right)$.

It also has an upward (restoring) force acting on it due to the spring:

$F = k x = 5 \cdot 3 = 15$ $N$

Where $x$ is the distance $\left(m\right)$ and $k$ is the spring constant $\left(k g {s}^{-} 2\right)$ or (equivalently) $\left(N {m}^{-} 1\right)$.

Since these forces are acting in opposite directions, let's call downward the positive direction and upward the negative direction.

The net force on the mass is $49 + \left(- 15\right) = 34$ $N$ in the downward direction.