Question

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`?

Answer 1

`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:

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)|)))`