# 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?

Aug 14, 2017

$403$ $\text{N}$

#### Explanation:

We're asked to find the necessary applied force that must act on an object to make the object accelerate at $14$ ${\text{m/s}}^{2}$.

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 $m = 12$ $\text{kg}$, and that the acceleration $a = 14$ ${\text{m/s}}^{2}$, 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:

$\sum {F}_{x} = {F}_{\text{applied" - f_k = color(red)(168color(white)(l)"N}}$

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, $m g$ (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 ${f}_{k}$:

${F}_{\text{applied" - color(green)(235color(white)(l)"N") = color(red)(168color(white)(l)"N}}$

Therefore,

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

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