# An object, previously at rest, slides 1 m down a ramp, with an incline of pi/3 , and then slides horizontally on the floor for another 3 m. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

Jan 26, 2017

The coefficient of kinetic friction will be $\mu = 0.247$

#### Explanation:

This problem is most easily done by conservation of energy.
Two energy forms are involved:

A change in gravitational potential energy: $m g h$

Frictional heating: $\mu {F}_{N} \Delta {d}_{1} + \mu {F}_{N} \Delta {d}_{2}$

where $\Delta {d}_{1}$ is the distance the object slides along the ramp, and $\Delta {d}_{2}$ is the distance is slides along the horizontal surface.

Before we can continue, we need to be aware of a couple of complications

We must express $h$ (the height the object descends) in terms of $\Delta {d}_{1}$ (its displacement along the ramp).

$h = \Delta {d}_{1} \sin \left(\frac{\pi}{3}\right)$

Also, we must note that the normal force on an incline is not equal to mg, but to $m g \cos \theta$, where $\theta = \frac{\pi}{3}$ in this case. With all that looked after, our equation becomes

$- m g \Delta {d}_{1} \sin \left(\frac{\pi}{3}\right) + \mu m g \cos \left(\frac{\pi}{3}\right) \Delta {d}_{1} + \mu m g \Delta {d}_{2} = 0$

(The first term is negative because the potential energy decreases.)

Notice that we can divide every term by mg, (including the right side of the equation)

So, inserting 1 m for $\Delta {d}_{1}$ and 3m for $\Delta {d}_{2}$ we get:

$- 1 \left(0.866\right) + \left(\mu\right) \left(0.50\right) 1 + \mu \left(3\right) = 0$

$- 0.866 + 0.50 \mu + 3 \mu = 0$

$3.5 \mu = 0.866$

$\mu = 0.247$