# If an object is moving at 6 m/s over a surface with a kinetic friction coefficient of u_k=20 /g, how far will the object continue to move?

Dec 17, 2015

I found $0.9 m$

#### Explanation:

Your object will move under the acton of Friction that wll produe a negative acceleration and eentually stop it.

To find this acceleration we can use Newton's Second law:

$\Sigma \vec{F} = m \vec{a}$

where the only force will be friction (in the opposite direction to the movement so it will need to enter Newton's Law with a minus sign) given as:

${f}_{k} = {u}_{k} N$

where:

$N =$normal reaction$=$Weight$= m g$ in this case of horizontal movement.

So you have:

$- {f}_{k} = m a$
$- {u}_{k} N = m a$
$- {u}_{k} m g = m a$
$- \frac{20}{\cancel{g}} \cancel{m} \cancel{g} = \cancel{m} a$
so that $a = - 20 \frac{m}{s} ^ 2$
we now use the relationship:
${v}_{f}^{2} = {v}_{i}^{2} + 2 a d$
where ${v}_{f} = 0$ (stops)
$0 = {6}^{2} - 2 \cdot d \cdot 20$
$d = \frac{36}{40} = 0.9 m$