# If an object with a mass of 10 kg  is moving on a surface at 4 m/s and slows to a halt after  6 s, what is the friction coefficient of the surface?

Aug 10, 2016

$\mu = \frac{1}{15}$

#### Explanation:

The initial kinetic energy is lost against the friction losses.

$\frac{1}{2} m {v}_{0}^{2} = m g \mu \Delta x$.

Here $\mu$ is the kinetic friction coefficent and $\Delta x$ is the path until repos.

Also the movement along the path $\Delta x$ is dictated by

$\Delta x = {v}_{0} t - \frac{1}{2} \left(\mu g\right) {t}^{2}$

Solving for $\Delta x$ and $\mu$ the system

{ (1/2mv_0^2 = mg mu Delta x), (Delta x = v_0 t - 1/2(mu g)t^2) :}

we obtain

$\Delta x = \frac{{v}_{0} t}{2}$ and
$\mu = {v}_{0} / \left(g t\right)$

Assuming $g = 10$ we have

$\mu = \frac{4}{60} = \frac{1}{15}$