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

Answer:

#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" ")|)#