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

Answer:

#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"#.