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
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
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 is0 because it comes to a stop -
v_(0x) is the initial velocity, given as2/3 "m/s" -
a_x is the acceleration (what we're trying to find) -
Deltax is the displacement of the box (given as6 "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 (
sumF_x = -f_k = ma_x
Recall that the friction force
f_k = mu_kn = mu_kmg
Substituting this in:
ma_x = -mu_kmg
Or, solving for the desired coefficient of kinetic friction,
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" ")|)