An object, previously at rest, slides 15 m down a ramp, with an incline of pi/3 , and then slides horizontally on the floor for another 12 m. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

Mar 13, 2016

$\mu = \frac{5 \sqrt{3}}{13}$

Explanation: $\text{Total Energy at The Point of A:}$
${E}_{A} = m \cdot g \cdot h$
$\text{Work Doing by The Friction Force Along AB:}$
${W}_{\text{AB}} = \mu \cdot m \cdot g \cdot \cos \left(\frac{\pi}{3}\right) \cdot 15$
$\text{Work Doing by The Friction Force Along BD:}$
${W}_{\text{BD}} = \mu \cdot m \cdot g \cdot 12$
${E}_{A} = {W}_{\text{AB" +W_"BD}}$
$\cancel{m \cdot g} \cdot h = \mu \cdot \cancel{m \cdot g} \cdot \cos \left(\frac{\pi}{3}\right) \cdot 15 + \mu \cdot \cancel{m \cdot g} \cdot 12$
$h = \mu \cdot 15 \cdot \cos \left(\frac{\pi}{3}\right) + \mu \cdot 12 \cdot$
$\alpha = \frac{\pi}{3}$
$h = 15 \cdot \sin \left(\frac{\pi}{3}\right)$
$15 \cdot \sin \left(\frac{\pi}{3}\right) = \mu \left(15 \cdot \cos \left(\frac{\pi}{3}\right) + 12\right)$
$15 \cdot \frac{\sqrt{3}}{2} = \mu \left(15 \cdot \frac{1}{2} + 12\right)$
$7 , 5 \sqrt{3} = \mu \left(\frac{39}{2}\right)$
$15 \sqrt{3} = \mu \cdot 39$
$\mu = \frac{15 \sqrt{3}}{39}$
$\mu = \frac{5 \sqrt{3}}{13}$