# An object, previously at rest, slides 4 m down a ramp, with an incline of pi/4 , and then slides horizontally on the floor for another 15 m. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

Mar 6, 2016

${u}_{k} \cong 0 , 158$

#### Explanation:

$\text{Total Energy of object:}$
${E}_{p} = m \cdot g \cdot 4 \cdot \frac{\sqrt{2}}{2} \text{ } {E}_{p} = m \cdot g \cdot 2 \sqrt{2}$
$\text{Work doing by the Friction Force on Ramp}$
${W}_{f} = {u}_{k} \cdot m \cdot g \cdot 4 \cdot \frac{\sqrt{2}}{2} \text{ } {W}_{f} = {u}_{k} \cdot m \cdot g \cdot 2 \sqrt{2}$
$\text{Total Energy on floor}$
$\Delta E = m \cdot g \cdot 2 \sqrt{2} - {u}_{k} \cdot m \cdot g \cdot 2 \sqrt{2}$
$\textcolor{red}{{W}_{f} = \left({u}_{k} \cdot m \cdot g\right) \cdot 15}$
$\textcolor{red}{\text{work doing by The Friction Force on Floor}}$
Delta E=cancel(m*g)*2sqrt2-u_k*cancel(m*g)*2sqrt2=color(red)(u_k*cancel(m*g)*15
$2 \sqrt{2} - {u}_{k} \cdot 2 \sqrt{2} = {u}_{k} \cdot 15$
$2 \sqrt{2} = 15 \cdot {u}_{k} + 2 \sqrt{2} \cdot {u}_{k}$
$2 , 82 = 15 \cdot {u}_{k} + 2 , 82 \cdot {u}_{k}$
$2 , 82 = 17 , 82 \cdot {u}_{k}$
${u}_{k} = \frac{2 , 82}{17 , 82}$
${u}_{k} \cong 0 , 158$