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?

1 Answer
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 2/3 "m/s", and it moved 6 "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_(0x) is the initial velocity, given as 2/3 "m/s"

  • a_x is the acceleration (what we're trying to find)

  • Deltax is the displacement of the box (given as 6 "m")

Plugging in known values:

0 = (2/3color(white)(l)"m/s")^2 + 2(a_x)(6color(white)(l)"m")

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

sumF_x = -f_k = ma_x

Recall that the friction force f_k is equal to

f_k = mu_kn = mu_kmg

Substituting this in:

ma_x = -mu_kmg

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" ")|)