# An object, previously at rest, slides #15 m# down a ramp, with an incline of #pi/3 #, and then slides horizontally on the floor for another #10 m#. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

##### 1 Answer

#### Answer:

#### Explanation:

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 is 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.**

Gravitational potential energy:

#U_g=mgh#

Work done by kinetic friction:

#W_f=f_k*d#

Therefore, we have:

#=>mgh=W_f#

#=>color(blue)(mgh=f_k*d)#

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

#mgh=W_(f" total")#

#mgh=W_(f " ramp")+W_(f" floor")#

#=>color(blue)(mgh=f_k*r+f_k*d)#

**We have the following information:**

#|->r=15"m"# #|->theta=pi/3# #|->d=10"m"#

**Diagram:**

where

#vecn# is the normal force,#vecf_k# is the force of kinetic friction, and#F_g# is the force of gravity, decomposed into its parallel (x, horizontal) and perpendicular (y, vertical) components.

We can find

#sin(theta)="opposite"/"hypotenuse"#

#=>sin(pi/3)=h/15#

#=>h=15sin(pi/3)#

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

**Ramp:**

#color(blue)(sumF_x=F_gsin(theta)-f_k=ma_x)#

#color(blue)(sumF_y=n-F_gcos(theta)=0)#

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

We can solve for n:

#n=F_gcos(theta)#

#=>=mgcos(theta)#

**Floor:**

#color(darkblue)(sumF_x=-f_k=ma_x)#

#color(darkblue)(sumF_y=n-F_g=0)#

Again, we can solve for n, and we find that

Substituting back in...

#mgh=mu_kmgcos(theta)r+mu_kmgd#

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

#h=mu_kcos(theta)r+mu_kd#

Solve for

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

#color(blue)(mu_k=h/(cos(theta)r+d))#

And now, substituting in our known values:

#mu_k=(15sin(pi/3))/(15cos(pi/3)+10)#

#mu_k~~0.742#