# 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?

Jun 16, 2016

$700 N$

#### 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,

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {F}_{\text{net}} = m a \textcolor{w h i t e}{\frac{a}{a}} |}}}$

where:
${F}_{\text{net}} =$net force
$m =$mass $\left(k g\right)$
$a =$acceleration $\left(\frac{m}{s} ^ 2\right)$

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.

${F}_{N} + {F}_{g} + {F}_{a p p} + {F}_{f} = m a$

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

$\textcolor{red}{\cancel{\textcolor{b l a c k}{{F}_{N}}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{{F}_{g}}}} + {F}_{a p p} + {F}_{f} = m a$

So we are left with,

${F}_{a p p} + {F}_{f} = m a$

Step 2
Since you are looking for the force necessary to accelerate the object, isolate for ${F}_{a p p}$.

${F}_{a p p} = m a - {F}_{f}$

Step 3
In the equation, ${F}_{f}$ can be simplified into

${F}_{a p p} = m a - {\mu}_{k} {F}_{N}$

Note: Since the object is not changing vertically, ${F}_{N} = {F}_{g} = m g$.

${F}_{a p p} = m a - {\mu}_{k} m g$

Factoring out $m$,

${F}_{a p p} = m \left(a - {\mu}_{k} g\right)$

Step 4
${F}_{a p p} = \left(7 k g\right) \left[21 \frac{m}{s} ^ 2 - \left(8\right) \left(- 9.81 \frac{m}{s} ^ 2\right)\right]$
${F}_{a p p} = 696.36 N$
${F}_{a p p} \approx \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{700 N} \textcolor{w h i t e}{\frac{a}{a}} |}}}$