An object with a mass of #12 kg# is on a surface with a kinetic friction coefficient of # 1 #. How much force is necessary to accelerate the object horizontally at # 14 m/s^2#?

1 Answer
Apr 20, 2016

Answer:

#290N["forward"]#

Explanation:

For this problem it is important to note that there are two forces acting on the object in the #x# direction, the applied and friction forces.

Start by listing out the given and required values.

#m=12kgcolor(white)(i)color(teal),color(white)(i)mu_k=1color(white)(i)color(teal),color(white)(i)a=14m/s^2color(white)(i)color(teal),color(white)(i)F_"app"=?#

A free-body diagram can represent the situation as:

http://www.physicsclassroom.com/class/newtlaws/Lesson-2/Drawing-Free-Body-Diagrams

In this case, set the positive directions to be forward and down.

Using #F_("net",y)=ma_y#, add the forces in the #y# direction.

#F_("net",y)=ma_y#

#F_N+Fg=ma_y#

Since the object does not accelerate vertically, the acceleration in the #y# direction is #0m/s^2#.

#F_N+Fg=m(0m/s^2)#

#F_N+Fg=0N#

#F_N=-Fg#

#color(blue)(|bar(ul(color(white)(a/a)color(black)(F_N=-mg)color(white)(a/a)|)))#

After adding the forces in the #y# direction, do the same for the forces in the #x# direction. Rearrange for #F_"app"# to find the force needed to accelerate the object horizontally.

#F_("net",x)=ma_x#

#F_"app"+F_(f,k)=ma_x#

#F_"app"=ma_x-F_(f,k)#

#F_"app"=ma_x-mu_kF_N#

#F_"app"=ma_x-mu_k(color(blue)(-mg))#

#F_"app"=ma_x+mu_kmg#

Substitute your known values.

#F_"app"=(12kg)(14m/s^2)+(1)(12kg)(9.81m/s^2)#

#F_"app"=285.72N#

#F_"app"~~color(green)(|bar(ul(color(white)(a/a)290Ncolor(white)(a/a)|)))#