# If a 8 kg object moving at 18 m/s slows to a halt after moving 180 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

Feb 2, 2016

k=0.0917

#### Explanation:

$m :$ mass of object
$g = 9 , 81$
$G = m . g = 8 . 9 , 81 = 78 , 48 N$ "Gravity of object"
$N = - G = - 78 , 48 N$ "reaction to object by surface"
$F = k . N = - k .78 , 48 N$" friction force between surface and object"

$s t e p 1 :$

Object have lost the kinetic energy between the Point A and B.
$\Delta {E}_{k} = \frac{1}{2} m \left({0}^{2} - {18}^{2}\right)$
$\Delta {E}_{k} = \frac{1}{2} 8 \left(0 + 324\right)$
$\Delta {E}_{k} = 4 \cdot 324 = 1296$ Joule

$s t e p 2 :$
$W :$ work doing by friction force
$\Delta {E}_{k} = W$
$1296 = F \cdot \Delta x$
$\Delta x = 180 m e t e r s$
$1296 = k \cdot 78 , 48 \cdot 180$
$1296 = k \cdot 14126 , 4$
$k = 0.0917$ "has no unit"