# If an object is moving at 22 m/s over a surface with a kinetic friction coefficient of u_k=11 /g, how far will the object continue to move?

Jun 3, 2016

#### Answer:

$\Delta x = 22 \text{ "m" (area under graph)}$

#### Explanation:

${u}_{k} = \frac{11}{g}$
${v}_{i} = 22 \text{ } \frac{m}{s}$

$\text{Friction force: } {F}_{f} = {u}_{k} \cdot m \cdot g$

${F}_{f} = \frac{11}{g} \cdot m \cdot g \text{ ; } {F}_{f} = 11 m$

${F}_{f} = m \cdot a \text{ ; "11*m=m*a" ; "a=11 " } \frac{m}{s} ^ 2$

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

$0 = {v}_{i}^{2} - 2 \cdot a \cdot \Delta x$

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

$\Delta x = {v}_{i}^{2} / \left(2 \cdot a\right)$

$\Delta x = {22}^{2} / \left(2 \cdot 11\right)$

$\Delta x = 22 \text{ "m" (area under graph)}$