# If an object is moving at 50 m/s over a surface with a kinetic friction coefficient of u_k=8 /g, how far will the object continue to move?

Dec 11, 2015

I found $156 m$

#### Explanation:

Considering that the only force acting on the object will be kinetic friction ${f}_{k}$, it will produce an opposite acceleration slowing down the object.
From Newton's Second Law (horizontally):

$- {f}_{k} = m a$

and so:
$a = - {f}_{k} / m$

Friction will be: ${f}_{k} = {u}_{k} N$
where, in this case, the normal reaction, $N$, will be equal to the weight $W = m g$;
so:
$a = - \frac{{u}_{k} \cancel{m} g}{\cancel{m}} = - \frac{8}{g} \cdot g = - 8 \frac{m}{s} ^ 2$
With this acceleration you can use:
$\textcolor{red}{{v}_{f}^{2} = {v}_{i}^{2} + 2 a d}$
with ${v}_{f} = 0$ (the object stops)
so you get:
${0}^{2} = {50}^{2} - \left(2 \cdot 8 \cdot d\right)$
$d = \frac{2500}{16} = 156 m$