If a 5 kg object moving at 5 m/s slows to a halt after moving 25 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

Jul 7, 2016

${\mu}_{k} = 0.038 \text{ , has no unit}$

Explanation:

${v}_{f}^{2} = {v}_{i}^{2} - 2 \cdot a \cdot \Delta x$
${v}_{f} : \text{represents the final velocity of object}$
${v}_{i} : \text{represents the initial velocity of object}$
$a : \text{represents deceleration of object}$
$\Delta x : \text{represents the displacement of object}$

$\text{Where; "v_i=5 " "m/s " ; " v_f=5/2 " "m/s" ; "Delta x=25" } m$

${\left(\frac{5}{2}\right)}^{2} = {5}^{2} - 2 \cdot a \cdot 25$

$\frac{25}{4} = 25 - 2 \cdot a \cdot 25$

$50 \cdot a = 25 - \frac{25}{4} \text{ ; "50a=75/4" ; } a = \frac{75}{200}$

$a = \frac{3}{8} \text{ } \frac{m}{s} ^ 2$

${F}_{f} = m \cdot a$
${F}_{f} : \text{represents the friction force between contacting surface}$
$N : \text{represents the normal force}$
${\mu}_{k} : \text{represents the coefficient of kinetic friction}$

$N = m \cdot g$

${F}_{f} = {\mu}_{k} \cdot m \cdot g$

${\mu}_{k} \cdot \cancel{m} \cdot g = \cancel{m} \cdot a$

${\mu}_{k} = \frac{a}{g}$

${\mu}_{k} = \frac{\frac{3}{8}}{9.81}$

${\mu}_{k} = \frac{3}{8 \cdot 9.81}$

${\mu}_{k} = \frac{3}{78 , 48}$

${\mu}_{k} = 0.038 \text{ , has no unit}$