# If a 4 kg object moving at 2/3 m/s slows to a halt after moving 6 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

Sep 4, 2017

${\mu}_{k} = 0.00378$

#### Explanation:

We're asked to find the coefficient of kinetic friction between the object and surface.

I'll solve this problem without using work or energy.

We're given that the object's initial speed was $\frac{2}{3}$ $\text{m/s}$, and it moved $6$ $\text{m}$ before coming to a stop.

We can use the kinematics equation

ul((v_x)^2 = (v_(0x))^2 + 2a_x(Deltax)

to find the acceleration (which should be negative) of the object as it slows to a stop.

• ${v}_{x}$ is the final velocity, which is $0$ because it comes to a stop

• ${v}_{0 x}$ is the initial velocity, given as $\frac{2}{3}$ $\text{m/s}$

• ${a}_{x}$ is the acceleration (what we're trying to find)

• $\Delta x$ is the displacement of the box (given as $6$ $\text{m}$)

Plugging in known values:

$0 = \left(\frac{2}{3} \textcolor{w h i t e}{l} \text{m/s")^2 + 2(a_x)(6color(white)(l)"m}\right)$

color(red)(a_x) = (-4/9color(white)(l)"m"^2"/s"^2)/(2(6color(white)(l)"m")) = color(red)(ul(-0.037color(white)(l)"m/s"^2

The only force acting on the box is the retarding friction force (${f}_{k}$) acting on the box, and so we have

$\sum {F}_{x} = - {f}_{k} = m {a}_{x}$

Recall that the friction force ${f}_{k}$ is equal to

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

Substituting this in:

$m {a}_{x} = - {\mu}_{k} m g$

Or, solving for the desired coefficient of kinetic friction, ${\mu}_{k}$:

color(green)(mu_k) = (-ma_x)/(mg) =color(green)((-a_x)/g

(notice how this is independent of the object's mass)

Thus, we have

color(blue)(mu_k) = (-(-0.037cancel("m/s"^2)))/(9.81cancel("m/s"^2)) = color(blue)(ulbar(|stackrel(" ")(" "0.00378" ")|)