# An object with a mass of 7 kg is on a surface with a kinetic friction coefficient of  7 . How much force is necessary to accelerate the object horizontally at 28 m/s^2?

Mar 26, 2016

I found: $676 N$

#### Explanation:

Consider the diagram:

We can use Newton's Second Law: $\Sigma \vec{F} = m \vec{a}$
on the $x$ and $y$ axis as:
$F - f = m a$
$N - W = 0$
where:
$N$ is the normal reaction;
$W$ is the weight ($= m g$);
$f = {\mu}_{k} N$ is kinetic friction;
$a = 28 \frac{m}{s} ^ 2$ is acceleration.
So we get from the second that:
$N = W = m g$
into the first:
$F - {\mu}_{k} \cdot m g = m a$
$F - \left(7 \cdot 7 \cdot 9.8\right) = 7 \cdot 28$
so:
$F = 196 + 480.2 = 676.2 \approx 676 N$