An object, previously at rest, slides #9 m# down a ramp, with an incline of #(pi)/3 #, and then slides horizontally on the floor for another #12 m#. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

Question

Layla Ahmed · Accepted Answer

The frictional coefficient of the surface is given by #\mu = xx#

Frictional coefficients have no units.

We are not told the mass of the object, but we can just call it #m#, and I have a feeling it will cancel out anyway. Two forces due to the weight force of gravity on the object's mass are acting while the object is on the slope: a force parallel to the slope, #F_"//" = mgsin heta#, and a force perpendicular to the slope, #F_|-- = mgcos heta#. A third force is also acting, the frictional force, given by #F_"frict" = \muF_|--# where #\mu# is the coefficient of friction. The frictional force acts in the opposite direction to #F_"//"#. When the object is moving across the flat floor, only the frictional force is acting. In this case, # heta = 0^o# and #cos heta = 1#, so the frictional force (it will be different from that when the object on the ramp although the frictional coefficient is the same) is given by #F_"fric" = \mumg#. While traveling, the object accelerates on the inclined plane and decelerates on the floor. It starts and ends its journey at rest. Its deceleration in the floor will be given by #f=F_"fric"/m = (\mumg)/m = \mug# As I suspected, the mass of the object cancels out. We can find its velocity, #v# at the bottom of the ramp, at least in terms of #\mu#, by solving #v^2 = u^2 + 2as# (some people write the last term as '#2ad#') where #v#, the final velocity on the floor, is #0# #ms^-1# and #u#, the initial velocity on the floor, is what we want to find. I won't go through all of the solution here for space reasons, but substituting in the #12# #m# traveled I get #u = # #ms^-1#. \sigh} Will try again tomorrow to finish it up; I'm tired and having trouble with this one.

Olivia Askew · Answer

undefined To solve for the material's kinetic friction coefficient, we need to analyze the motion of the object and apply principles of energy conservation.

The total work done on the object equals the change in its kinetic energy. This can be expressed as the work done against friction on the ramp plus the work done against friction on the horizontal floor.

The work done against friction on the ramp can be calculated using the gravitational potential energy the object loses as it descends the ramp. The work done against friction on the horizontal floor is equal to the frictional force multiplied by the distance.

Let's denote the kinetic friction coefficient as $ \mu_k $.

The gravitational potential energy lost on the ramp is converted into kinetic energy and work against friction:

$ mgh = \frac{1}{2}mv_r^2 + f_r d_r $,

where $ m $ is the mass of the object, $ g $ is the acceleration due to gravity, $ h $ is the vertical height of the ramp, $ v_r $ is the velocity of the object on the ramp, $ f_r $ is the frictional force on the ramp, and $ d_r $ is the distance down the ramp.

The work done on the horizontal floor is:

$ f_h d_h $,

where $ f_h $ is the frictional force on the horizontal floor, and $ d_h $ is the horizontal distance.

Given that the object slides down the ramp, its vertical height is $ h = 9 \sin(\pi/3) $. The velocity on the ramp, $ v_r $, can be found using the equations of motion. Since the object started from rest, we can use $ v_r^2 = 2gh $.

Now, since the material is the same, the coefficient of friction is consistent throughout, so $ f_r = f_h = \mu_k mg $.

By substituting the known values and solving for $ \mu_k $, we can find the kinetic friction coefficient.