Question

An object with a mass of `7 kg` is on a surface with a kinetic friction coefficient of `8`. How much force is necessary to accelerate the object horizontally at `21 m/s^2`?

Answer 1

`700N`

Explanation:

Newton's second law of motion states that the sum of the forces acting on an object is equal to its mass multiplied by its acceleration.

Mathematically speaking,

`color(blue)(|bar(ul(color(white)(a/a)F_"net"=macolor(white)(a/a)|)))`

where:
`F_"net"=`net force
`m=`mass `(kg)`
`a=`acceleration `(m/s^2)`

In your case, we will start off by letting the up and forward directions be positive.

Step 1
List out all the forces acting on the object.

`F_N+F_g+F_(app)+F_f=ma`

Since the object is not changing in the `y` direction, or vertically, the normal and gravity forces cancel each other out.

`color(red)cancelcolor(black)(F_N)+color(red)cancelcolor(black)(F_g)+F_(app)+F_f=ma`

So we are left with,

`F_(app)+F_f=ma`

Step 2
Since you are looking for the force necessary to accelerate the object, isolate for `F_(app)`.

`F_(app)=ma-F_f`

Step 3
In the equation, `F_f` can be simplified into

`F_(app)=ma-mu_kF_N`

Note: Since the object is not changing vertically, `F_N=F_g=mg`.

`F_(app)=ma-mu_kmg`

Factoring out `m`,

`F_(app)=m(a-mu_kg)`

Step 4
Plug in your known values.

`F_(app)=(7kg)[21m/s^2-(8)(-9.81m/s^2)]`

Solve.

`F_(app)=696.36N`

Rounding off the answer to one significant figure,

`F_(app)~~color(green)(|bar(ul(color(white)(a/a)color(black)(700N)color(white)(a/a)|)))`