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
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 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 (
#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" "))#