# 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

#### Answer:

(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:

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

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

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

Final Answer