Question

An object with a mass of `12 kg` is on a surface with a kinetic friction coefficient of `2`. How much force is necessary to accelerate the object horizontally at `14 m/s^2`?

Answer 1

`403` `"N"`

Explanation:

We're asked to find the necessary applied force that must act on an object to make the object accelerate at `14` `"m/s"^2`.

There will be two forces acting on the object:

• an applied force (directed in the positive `x`-direction, although this is arbitrary)

• the kinetic friction force `f_k` (directed in the negative direction, because it will oppose motion)

The net force equation is therefore

`ul(sumF_x = ma_x = overbrace(F_"applied")^"positive" - overbrace(f_k)^"negative"`

We're given that the mass `m = 12` `"kg"`, and that the acceleration `a = 14` `"m/s"^2`, so the net force is equal to

`sumF_x = ma_x = (12color(white)(l)"kg")(14color(white)(l)"m/s"^2) = color(red)(ul(168color(white)(l)"N"`

Updating our equation:

`sumF_x = F_"applied" - f_k = color(red)(168color(white)(l)"N"`

The frictional force `f_k` is given by the equation

`ul(f_k = mu_kn`

The normal force magnitude `n` is equal to its weight, `mg` (because the plane is assumed to be horizontal):

`f_k = mu_kmg = (2)(12color(white)(l)"kg")(9.81color(white)(l)"m/s"^2) = color(green)(ul(235color(white)(l)"N"`

Plugging this in for `f_k`:

`F_"applied" - color(green)(235color(white)(l)"N") = color(red)(168color(white)(l)"N"`

Therefore,

`color(blue)(ulbar(|stackrel(" ")(" "F_"applied" = 403color(white)(l)"N"" ")|)`

The necessary force is thus `color(blue)(403color(white)(l)"newtons"`.