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?

1 Answer
Aug 14, 2017

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".