# If a 3 kg object moving at 9 m/s slows down to a halt after moving 27 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

Apr 1, 2017

$\mu = 0.15$

#### Explanation:

The work being done on the object is Work due to friction so the following equation is going to be used:

$\textcolor{w h i t e}{a a a a a a a a a a a a}$Equation (a) ${W}_{f} = \Delta K E$

We can rewrite Equation (a) if we break down both sides step-by-step to become:

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a}$Equation (b)
$\textcolor{w h i t e}{a a a a a a}$$\left(\mu \cdot m g\right) \cdot d \cdot \cos \theta = \left(\frac{1}{2} m {v}_{f}^{2} - \frac{1}{2} m {v}_{i}^{2}\right)$

$\mu = \text{coefficient of kinetic friction}$
$m = \text{mass (kg)}$
$g = \text{acceleration due to gravity} \left(\frac{m}{s} ^ 2\right)$
$d = \text{displacement} \left(m\right)$
$\theta = \text{angle between friction and displacement}$
${v}_{f} = \text{velocity final}$
${v}_{i} = \text{velocity initial}$

Since our object stopped, its final velocity becomes $0$ and therefore $\text{final KE}$ becomes $0$. Friction and displacement are opposite one another so $\cos \left({180}^{\circ}\right) = - 1$. Mass cancels on both sides. Rearrange, plug in, and solve.

$- \mu \cdot g \cdot d = - \frac{1}{2} {v}_{i}^{2}$

$\mu = \frac{0.5 \cdot {9}^{2}}{9.8 \cdot 27} = \frac{40.5}{264.6} = 0.15$