An object, previously at rest, slides 4 m4m down a ramp, with an incline of pi/12 π12, and then slides horizontally on the floor for another 4 m4m. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

2 Answers
Jul 10, 2017

The kinetic friction coefficient is 0.130.13.

Explanation:

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Jul 10, 2017

mu_k=0.13μk=0.13

Explanation:

Another approach, making use of the work-energy theorem...

Since the object begins from rest at the top of the ramp and comes to rest on the floor below, we know that all of the gravitational potential energy originally possessed by the object has been lost to friction by the time it comes to a stop. Therefore, we know that the work done by friction, or the energy lost to friction, is equal to the object's gravitational potential energy at the top of the ramp.

U_g=mghUg=mgh

mgh=W_fmgh=Wf

mgh=vecf_k*dmgh=fkd

It is necessary to break up the right side into a sum of the work done by friction while the object is on the ramp as well as when it is sliding across the floor. Let's call the distance down the ramp dd, and the distance across the floor rr.

mgh=vecf_kd+vecf_krmgh=fkd+fkr

mgh=mu_kvecnd+mu_kvecnrmgh=μknd+μknr

We can find hh, the height of the ramp, using basic trigonometry. We have an angle of pi/12π12 and a hypotenuse of 44, so:

sin(pi/12)=h/4sin(π12)=h4

h=4sin(pi/12)h=4sin(π12)

Now let's take inventory of the forces acting on the object while on the ramp and the floor.

Ramp:

sumvecF_x=F_gsin(theta)-f_k=maFx=Fgsin(θ)fk=ma

sumvecF_y=n-F_gcos(theta)=0Fy=nFgcos(θ)=0

Note that there is no net perpendicular (y) force and therefore no acceleration in that direction (dynamic equilibrium)

We can solve for n:

n=F_gcos(theta)n=Fgcos(θ)

=mgcos(theta)=mgcos(θ)

Floor:

sumvecF_x=-f_k=maFx=fk=ma

sumvecF_y=n-F_g=0Fy=nFg=0

Again, we can solve for n, and we find that n=mgn=mg.

Substituting back in...

mgh=mu_kmgcos(theta)d+mu_kmgrmgh=μkmgcos(θ)d+μkmgr

cancelcolor(purple)(m)cancelcolor(blue)(g)h=mu_kcancelcolor(purple)(m)cancelcolor(blue)(g)cos(theta)d+mu_kcancelcolor(purple)(m)cancelcolor(blue)(g)r

h=mu_kcos(theta)d+mu_kr

Solve for mu_k

h=mu_k(cos(theta)d+r)

mu_k=h/(cos(theta)d+r)

And now, substituting in our known values:

mu_k=(4sin(pi/12))/(4cos(pi/12)+4)

mu_k~~0.13

I give a more detailed explanation of a similar problem here if you would like elaboration on any of the above steps or concepts.