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?

1 Answer
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 9/4 "m/s" to rest in 3/8 "m":

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

We know:

  • v_x = 0 (comes to rest)

  • v_(0x) = 2.25 "m/s"

  • Deltax = 0.375 "m"

0 = (2.25color(white)(l)"m/s")^2 + 2a_x(0.375color(white)(l)"m")

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 "N" = f_k

The coefficient of kinetic friction mu_k is given by

f_k = mu_kn

The normal force n is

n = mg = (7/2color(white)(l)"kg")(9.81color(white)(l)"m/s"^2) = 34.3 "N"

The coefficient of kinetic friction is thus

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