Question

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?

Answer 1

`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" ")|)`