# If an object with a mass of 10 kg  is moving on a surface at 45 m/s and slows to a halt after  6 s, what is the friction coefficient of the surface?

Mar 28, 2016

${u}_{k} = 0 , 0051$

#### Explanation:

${v}_{l} = {v}_{i} - a \cdot t$
${v}_{l} : \text{last velocity}$
${v}_{i} : \text{initial velocity}$
$a : \text{acceleration}$
$t : \text{time}$

$v = \frac{v}{2} - a \cdot 6$
$6 \cdot a = v - \frac{v}{2}$
$6 \cdot a = - \frac{v}{2}$
$a = - \frac{v}{12} \text{ "a=-3,75" " m/s^2" 'acceleration of object'}$

${v}_{l}^{2} = {v}_{i}^{2} - 2 \cdot a \cdot \Delta x$
${v}^{2} = {\left(\frac{v}{2}\right)}^{2} - 2.3 , 75 \cdot \Delta x$

${v}^{2} - {v}^{2} / 4 = 7 , 50 \cdot \Delta x$

$\frac{3 \cdot {v}^{2}}{4} = 7 , 50 \cdot \Delta x$
$3 \cdot {v}^{2} = 30 \cdot \Delta x$
${v}^{2} = 10 \cdot \Delta x \text{ ; "45^2=10*Delta x" ; } 2025 = 10 \cdot \Delta x$
$\Delta x = 202 , 5 \text{ m}$
${E}_{k} = \frac{1}{2} \cdot m \cdot {v}^{2} \text{the kinetic energy of object}$
${W}_{f} = {F}_{f} \cdot \Delta x \text{ work doing by friction force}$
${F}_{f} = {u}_{k} \cdot m \cdot g \text{ } {W}_{f} = {u}_{k} \cdot m \cdot g \cdot \Delta x$
${E}_{k} = {W}_{f}$
$\frac{1}{2} \cdot \cancel{m} \cdot {v}^{2} = {u}_{k} \cdot \cancel{m} \cdot g \cdot \Delta x$

${u}_{k} = {v}^{2} / \left(2 \cdot g \cdot \Delta x\right)$

${u}_{k} = {45}^{2} / \left(2 \cdot 9 , 81 \cdot 202 , 5\right) = \frac{2025}{397305}$

${u}_{k} = 0 , 0051$