# If a 7/2 kg object moving at 9/4 m/s slows to a halt after moving 3/8 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

Jul 18, 2017

${\mu}_{k} = 0.306$

#### Explanation:

We're asked to find the coefficient of kinetic friction ${\mu}_{k}$ of an object sliding on a surface, with some given information.

While sliding, the vertical forces (gravitational and normal) cancel out, so we'll look at only horizontal motion to find the friction.

We can use kinematics to find the magnitude of the acceleration, given it went from $\frac{9}{4}$ $\text{m/s}$ to rest in $\frac{3}{8}$ $\text{m}$:

${\left({v}_{x}\right)}^{2} = {\left({v}_{0 x}\right)}^{2} + 2 {a}_{x} \left(\Delta x\right)$

We know:

• ${v}_{x} = 0$ (comes to rest)

• ${v}_{0 x} = 2.25$ $\text{m/s}$

• $\Delta x = 0.375$ $\text{m}$

$0 = \left(2.25 \textcolor{w h i t e}{l} \text{m/s")^2 + 2a_x(0.375color(white)(l)"m}\right)$

color(red)(a_x = 3 color(red)("m/s"^2

Now that we know the acceleration, we can use Newton's second law to find the magnitude of the friction force:

sumF_x = ma_x = (7/2color(white)(l)"kg")(color(red)(3color(white)(l)"m/s"^2)) = 10.5 $\text{N}$ $= {f}_{k}$

The coefficient of kinetic friction ${\mu}_{k}$ is given by

${f}_{k} = {\mu}_{k} n$

The normal force $n$ is

$n = m g = \left(\frac{7}{2} \textcolor{w h i t e}{l} {\text{kg")(9.81color(white)(l)"m/s}}^{2}\right) = 34.3$ $\text{N}$

The coefficient of kinetic friction is thus

mu_k = (f_k)/n = (10.5cancel("N"))/(34.3cancel("N")) = color(blue)(0.306