# If a 1 kg object moving at 25 m/s slows down to a halt after moving 625 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

Feb 29, 2016

${u}_{k} = 0 , 038$

#### Explanation:

${v}_{f}^{2} = {v}_{i}^{2} - 2 \cdot a \cdot x$
${\left(\frac{v}{2}\right)}^{2} = {v}^{2} - 2 \cdot a \cdot 625$
${v}^{2} / 4 = {v}^{2} - 2 \cdot a \cdot 625$
${v}^{2} = 4 {v}^{2} - 8 \cdot a \cdot 625$
$3 {v}^{2} = 8 \cdot a \cdot 625$
$v = 25 \frac{m}{s}$
$3 \cdot {25}^{2} = 8. a .625$
$3 \cdot \cancel{625} = 8 \cdot a \cdot \cancel{625}$
$a = \frac{3}{8} \frac{m}{s} ^ 2 \text{ acceleration of the object}$
${F}_{f} = {u}_{k} \cdot N \text{ the friction force }$
$N = m \cdot g \text{ N:normal force acting contacting surfaces}$
${F}_{f} = m \cdot a \text{ Newton's the second law}$
${u}_{k} \cdot \cancel{m} \cdot g = \cancel{m} \cdot \frac{3}{8}$
${u}_{k} = \frac{3}{8 \cdot 9 , 81}$
${u}_{k} = 0 , 038$