An object, previously at rest, slides #1 m# down a ramp, with an incline of #(3pi)/8 #, and then slides horizontally on the floor for another #8 m#. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

Question

An object, previously at rest, slides #1  m# down a ramp, with an incline of #(3pi)/8 #, and then slides horizontally on the floor for another #8 m#. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

Lily Adams · Accepted Answer

#mu_k = 0.110#

WARNING: LONG-ish ANSWER!

We're asked to find the coefficient of kinetic friction, #mu_k#, between the object and the ramp.

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.

#-----------bb("INCLINE")-----------#

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 constant-acceleration 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 ramp
    
    Plugging 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 constant-acceleration 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 floor
      
      Plugging 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" = 1# #"m"color(white)(al# (distance down ramp)
      
      #Deltax_"floor" = 8# #"m"color(white)(aa# (distance along floor)
      
      #theta = (3pi)/8color(white)(aaaaaa.# (angle of inclination)
      
      Plugging these in:
      
      #color(blue)(mu_k) = ((1color(white)(l)"m")*sin((3pi)/8))/(8color(white)(l)"m"+(1color(white)(l)"m")·cos((3pi)/8)) = color(blue)(ulbar(|stackrel(" ")(" "0.110" ")|)#
      
      Notice how the coefficient doesn't depend on the mass #m# or gravitational acceleration #g# if the two surfaces are the same...

Nicholas Agrusa · Answer

undefined To find the coefficient of kinetic friction (μ) for the material, we can use the following equation:

μ = tan(θ)

Where θ is the angle of inclination of the ramp.

Given that the incline of the ramp is (3π)/8, we can substitute this value into the equation:

μ = tan((3π)/8)

Using a calculator, we find:

μ ≈ 0.661

Therefore, the material's kinetic friction coefficient is approximately 0.661.

Nicholas Agrusa · Answer

undefined To solve for the material's kinetic friction coefficient, we can use the principles of conservation of energy. The object initially possesses gravitational potential energy due to its height on the ramp, and as it moves down the ramp and along the floor, this potential energy is converted into kinetic energy. The work done against friction on the floor also converts some of this energy into thermal energy.

First, let's find the change in gravitational potential energy as the object moves down the ramp. The height of the ramp can be calculated using trigonometry:

$$ h = d \sin(\theta) $$

Where:
- $ d = 1 $ m (the distance moved down the ramp)
- $ \theta = \frac{3\pi}{8} $ radians (the incline of the ramp)

Then, we can calculate the change in potential energy as:

$$ \Delta U = mgh $$

Where:
- $ m $ is the mass of the object
- $ g $ is the acceleration due to gravity ($ 9.81 \, \text{m/s}^2 $)

Next, we find the kinetic energy of the object as it slides along the floor. Since there is no change in height, the gravitational potential energy at this point is zero, and all the energy is in the form of kinetic energy:

$$ KE = \frac{1}{2}mv^2 $$

Where:
- $ v $ is the final velocity of the object

Now, we can equate the change in potential energy to the change in kinetic energy to solve for the final velocity $ v $. We'll also consider the work done against friction on the horizontal floor:

$$ \Delta U = KE + W_{\text{friction}} $$

Given that the work done against friction $ W_{\text{friction}} = f_k \cdot d' $, where $ d' $ is the horizontal distance traveled and $ f_k $ is the kinetic friction force, we can find $ f_k $ using $ f_k = \mu_k \cdot N $, where $ \mu_k $ is the coefficient of kinetic friction and $ N $ is the normal force.

Once we have $ f_k $, we can calculate $ \mu_k $ using $ \mu_k = \frac{f_k}{mg} $.

By solving these equations, we can find the coefficient of kinetic friction for the material.