Question

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?

Answer 1

The net force on the mass is `49+(-15) = 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 `Nm^-1` are exactly equivalent to `kgs^-2`, and make this kind of question much easier to understand: if the spring constant is `k` `Nm^-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=mg = 5*9.8 = 49` `N`

Where `m` is the mass `(kg)` and `g` is the acceleration due to gravity `(ms^-2)` or (equivalently) `(Nkg^-1)`.

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

`F = kx =5*3=15` `N`

Where `x` is the distance `(m)` and `k` is the spring constant `(kgs^-2)` or (equivalently) `(Nm^-1)`.

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+(-15) = 34` `N` in the downward direction.