# If a 2 kg object moving at 13 m/s slows to a halt after moving 19 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

May 24, 2016

${u}_{k} \cong 0.4$

#### Explanation:

${v}_{f}^{2} = {v}_{i}^{2} - 2 \cdot a \cdot \Delta x$

${v}_{f} = \frac{13}{2} \text{ "m/s" final velocity}$

${v}_{i} = 13 \text{ "m/s " initial velocity}$

$\Delta x = 19 \text{ " m" displacement of object}$

${\left(\frac{13}{2}\right)}^{2} = {13}^{2} - 2 \cdot a \cdot 19$

$2 \cdot a \cdot 19 = 169 - \frac{169}{4}$

$38 \cdot a = \frac{507}{4}$

$a = \frac{597}{38 \cdot 4}$

$a = \frac{597}{152} = 3.93 \frac{m}{s} ^ 2$

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

$a = {F}_{f} / m$

$3.93 = \frac{{u}_{k} \cdot \cancel{m} \cdot g}{\cancel{m}}$

${u}_{k} = \frac{3.93}{9.81}$

$g = 9.81 \frac{N}{k g}$

${u}_{k} \cong 0.4$