An object, previously at rest, slides #4 m# down a ramp, with an incline of #pi/3 #, and then slides horizontally on the floor for another #5 m#. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?
WARNING: LONGish ANSWER!
We're asked to find the coefficient of kinetic friction,
We'll split this problem into two parts: the first part is where the object is sliding down the incline, and the second part is where it is sliding across the floor.
The only two forces acting on the object as it slides down the ramp are

The gravitational force (its weight; acting down the ramp)

The kinetic friction force (acting up the ramp because it opposes motion)
The expression for the coefficient of kinetic friction
#mu_k# is#ul(f_k = mu_kn# where

#f_k# is the magnitude of the retarding kinetic friction force acting as it slides down (denoted#f# in the above image) 
#n# is the magnitude of the normal force exerted by the incline, equal to#mgcostheta# (denoted#N# in the above image)The expression for the net horizontal force, which I'll call
#sumF_(1x)# , is#ul(sumF_(1x) = overbrace(mgsintheta)^"gravitational force"  overbrace(color(red)(f_k))^"kinetic friction force"# Since
#color(red)(f_k = mu_kn)# , we have#sumF_(1x) = mgsintheta  color(red)(mu_kn)# And since the normal force
#n = mgcostheta# , we can also write#ul(sumF_(1x) = mgsintheta  color(red)(mu_kmgcostheta)# Or
#ul(sumF_(1x) = mg(sintheta  mu_kcostheta))# #" "# Using Newton's second law, we can find the expression for the acceleration
#a_(1x)# of the object as it slides down the incline:#ul(sumF_(1x) = ma_(1x)# #color(red)(a_(1x)) = (sumF_(1x))/m = (mg(sintheta  mu_kcostheta))/m = color(red)(ul(g(sintheta  mu_kcostheta)# #" "# What we can now do is apply a constantacceleration equation to find the final velocity as it exits the ramp, which we'll call
#v_(1x)# :#ul((v_(1x))^2 = (v_(0x))^2 + 2(a_(1x))(Deltax_"ramp")# where

#v_(0x)# is the initial velocity (which is#0# since it was "previously at rest") 
#a_(1x)# is the acceleration, which we found to be#color(red)(g(sintheta  mu_kcostheta)# 
#Deltax_"ramp"# is the distance it travels down the rampPlugging in these values:
#(v_(1x))^2 = (0)^2 + 2color(red)(g(sintheta  mu_kcostheta))(Deltax_"ramp")# #" "#
#ul(v_(1x) = sqrt(2color(red)(g(sintheta  mu_kcostheta))(Deltax_"ramp"))# This velocity is also the initial velocity of the motion along the floor.
#bb("FLOOR")# As the object slides across the floor, the plane is perfectly horizontal, so the normal force
#n# now equals#n = mg# The only horizontal force acting on the object is the retarding kinetic friction force
#f_k = mu_kn = mu_kmg# (which is different than the first one).
The net horizontal force on the object on the floor, which we'll call
#sumF_(2x)# , is thus#ul(sumF_(2x) = f_k = mu_kmg# (the friction force is negative because it opposes the object's motion)
Using Newton's second law again, we can find the floor acceleration
#a_(2x)# :#color(green)(a_(2x)) = (sumF_(2x))/m = (mu_kmg)/m = color(green)(ul(mu_kg# #" "# We can now use the same constantacceleration equation as before, with only a slight difference in the variables:
#ul((v_(2x))^2 = (v_(1x))^2 + 2(a_(2x))(Deltax_"floor")# where this time

#v_(2x)# is the final velocity, which since it comes to rest will be#0# 
#v_(1x)# is the initial velocity, which we found to be#sqrt(2color(red)(g(sintheta  mu_kcostheta))(Deltax_"ramp")# 
#a_(2x)# is the acceleration, which we found to be#color(green)(mu_kg# 
#Deltax_"floor"# is the distance it travels along the floorPlugging in these values:
#(0)^2 = [sqrt(2color(red)(g(sintheta  mu_kcostheta))(Deltax_"ramp"))color(white)(l)]^2 + 2(color(green)(mu_kg))(Deltax_"floor")# Rearranging gives
#2(color(green)(mu_kg))(Deltax_"floor") = 2color(red)(g(sintheta  mu_kcostheta))(Deltax_"ramp")# #" "# At this point, we're just solving for
#mu_k# , which the following indented portion covers:Divide both sides by
#2g# :#mu_k(Deltax_"floor") = (sintheta  mu_kcostheta)(Deltax_"ramp")# Distribute:
#mu_k(Deltax_"floor") = (Deltax_"ramp")sintheta  (Deltax_"ramp")mu_kcostheta# Now, we can divide all terms by
#mu_k# :#Deltax_"floor" = ((Deltax_"ramp")sintheta)/(mu_k)  (Deltax_"ramp")costheta# Rearrange:
#((Deltax_"ramp")sintheta)/(mu_k) = Deltax_"floor" + (Deltax_"ramp")costheta# Finally, swap
#mu_k# and#Deltax_"floor" + (Deltax_"ramp")costheta# :#color(red)(ulbar(stackrel(" ")(" "mu_k = ((Deltax_"ramp")sintheta)/(Deltax_"floor" + (Deltax_"ramp")costheta)" "))# #" "# The question gives us

#Deltax_"ramp" = 4# #"m"color(white)(al# (distance down ramp) 
#Deltax_"floor" = 5# #"m"color(white)(aa# (distance along floor) 
#theta = pi/3color(white)(aaaaaa.# (angle of inclination)Plugging these in:
#color(blue)(mu_k) = ((4color(white)(l)"m")*sin((pi)/3))/(5color(white)(l)"m"+(4color(white)(l)"m")·cos((pi)/3)) = color(blue)(ulbar(stackrel(" ")(" "0.495" "))# Notice how the coefficient doesn't depend on the mass
#m# or gravitational acceleration#g# if the two surfaces are the same...




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Use the workenergy principle to find the kinetic friction coefficient:
[ \text{Work done by gravity on the ramp} + \text{Work done by kinetic friction on the floor} = \text{Change in kinetic energy} ]
[ mgh(\sin(\pi/3)) + \mu_k \cdot N \cdot d_{\text{floor}} = \frac{1}{2}mv_f^2 ]
Solve for the kinetic friction coefficient ( \mu_k ):
[ \mu_k = \frac{mgh(\sin(\pi/3)) + \frac{1}{2}mv_f^2}{N \cdot d_{\text{floor}}} ]
Substitute the values and compute the result.
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