# 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?

Dec 25, 2015

Let us use the ${2}^{\text{nd}}$ Newton law in order to get this.

#### Explanation:

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

${F}_{\text{Total}} = \sum F = m a$

In our question, there are two forces:
- Our own force, which we are going to call ${F}_{\text{ours}}$.
- The friction force, which we are going to call ${F}_{\text{fr}}$.

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:

${F}_{\text{fr}} = \mu \cdot N$

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

${F}_{\text{fr" = mu cdot N = mu cdot p = mu cdot (m g) = 4 cdot (5 "kg" cdot 9.8 "m/s"^2) = 196 "N}}$

Now that we know this, we can rewrite ${2}^{\text{nd}}$ Newton law as:

$\sum F = {F}_{\text{ours" - F_"fr}} = m a \rightarrow$
$\rightarrow {F}_{\text{ours" = ma + F_"fr" = 5 "kg" cdot 32 "m/s"^2 + 196 "N" = 356 "N}}$

So, we must make a force of 356 N in order to move our object with an acceleration of $32 {\text{m/s}}^{2}$.