Question

An object with a mass of `5 kg` is on a surface with a kinetic friction coefficient of `4`. How much force is necessary to accelerate the object horizontally at `32 m/s^2`?

Answer 1

Let us use the `2^"nd"` Newton law in order to get this.

Explanation:

According to Newton `2^"nd"` law, the total force made on a body (i.e. the sum of all forces) is proportional to its acceleration, in this way:

`F_"Total"= sum F = m a`

In our question, there are two forces:
- Our own force, which we are going to call `F_"ours"`.
- The friction force, which we are going to call `F_"fr"`.

Both forces must be summed up, although with opposite sign, because friction always acts against movement.

We know that friction force can be obtained by:

`F_"fr" = mu cdot N`

with `mu` being the kinetic friction coefficient, and `N` being the normal force (equals to the weigth of the body in our problem).
So:

`F_"fr" = mu cdot N = mu cdot p = mu cdot (m g) = 4 cdot (5 "kg" cdot 9.8 "m/s"^2) = 196 "N"`

Now that we know this, we can rewrite `2^"nd"` Newton law as:

`sum F = F_"ours" - F_"fr" = m a rightarrow`
`rightarrow F_"ours" = ma + F_"fr" = 5 "kg" cdot 32 "m/s"^2 + 196 "N" = 356 "N"`

So, we must make a force of 356 N in order to move our object with an acceleration of `32 "m/s"^2`.