# If an object is moving at 12 m/s over a surface with a kinetic friction coefficient of u_k=24 /g, how far will the object continue to move?

Feb 8, 2016

$x = 3 m e t e r s$

#### Explanation:

${v}^{2} = {v}_{i}^{2} - 2 \cdot a \cdot x$
$v = 0 \text{ when object is stops}$
v_i^²=2*a*x
$k \cdot m \cdot g = m \cdot a$
$\frac{24}{\cancel{g}} \cdot \cancel{m} \cdot \cancel{g} = \cancel{m} \cdot a$
$a = 24 \frac{m}{\sec} ^ 2$
${12}^{2} = 2 \cdot 24 \cdot x$
$144 = 2 \cdot 24 \cdot x$
$144 = 48. x$
$x = \frac{144}{48}$
$x = 3 m e t e r s$