An object, previously at rest, slides 5 m down a ramp, with an incline of pi/3 , and then slides horizontally on the floor for another 2 m. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

May 20, 2017

(I really want someone to double check my work since this is one of my first physics answers)

${\mu}_{k} = 0.9622$

Explanation:

Let's use the top of the ramp for initial condition and bottom of the ramp for final condition.

$P {E}_{i} + K {E}_{i} = P {E}_{f} + K {E}_{f} + {W}_{\text{fric}}$

We know that:

$\Delta y = 5 m \left(\sin \left(\frac{\pi}{3}\right)\right) = \frac{5 \sqrt{3}}{2}$
$K {E}_{i} = 0$
$P {E}_{i} = m g \Delta y = \frac{5 \sqrt{3} m g}{2}$
$P {E}_{f} = 0$
${W}_{\text{fric" = 5mu_kF_N = 5mu_kmgcos(pi/3)= 5/2mu_kmgcolor(white)"-}}$

So our equation becomes:

$\frac{5 \sqrt{3} m g}{2} = K {E}_{f} + \frac{5}{2} {\mu}_{k} m g$

We can also have another point ($f 2$) at the end of the 2 meters when the block is at rest.

$P {E}_{f} + K {E}_{f} = P {E}_{\text{f2" + KE_"f2" + W_"fric2}}$

$P {E}_{f} , P {E}_{\text{f2", and KE_"f2}}$ are all 0, and ${W}_{\text{fric2}} = 2 {\mu}_{k} m g$

$K {E}_{f} = 2 {\mu}_{k} m g$

Substituting this into the first equation gives:

$\frac{5 \sqrt{3} m g}{2} = 2 {\mu}_{k} m g + \frac{5}{2} {\mu}_{k} m g$

$\frac{5 \sqrt{3}}{2} = \frac{9}{2} {\mu}_{k}$

${\mu}_{k} = \frac{5 \sqrt{3}}{9} = 0.9622$