A box with an initial speed of #3 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #8/5 # and an incline of #pi /8 #. How far along the ramp will the box go?

We're asked to find how far the box will travel up the ramp, given its initial speed, the coefficient of kinetic friction, and the angle of inclination.

I will solve this problem using only Newton's laws and kinematics (i.e. without using work/energy).

I will take the positive #x#-direction as *up the ramp.*

When the box reaches its maximum distance, the instantaneous velocity will be #0#. We are ultimately going to use the constant-acceleration equation

#ul((v_x)^2 = (v_(0x))^2 + 2a_x(Deltax)#

where

• #v_x# is the instantaneous velocity (which is #0#)

• #v_(0x)# is the initial velocity

• #a_x# is the (constant) acceleration

• #Deltax# is the distance it travels (what we're trying to find)

Since the velocity #v_x = 0#, we can also write this equation as

#0 = (v_(0x))^2 + 2a_x(Deltax)#

We also figure that the acceleration will be negative because it slows down and comes to a brief stop at its maximum height, so we then have

#0 = (v_(0x))^2 + 2(-a_x)(Deltax)#

And we can move it to the other side:

#ul(2a_x(Deltax) = (v_(0x))^2#

Rearranging for the distance traveled #Deltax#:

#ulbar(|stackrel(" ")(" "Deltax = ((v_(0x))^2)/(2a_x)" ")|)#

We already know the initial velocity, so we need to find the acceleration of the box.

#" "#

Let's use Newton's second law of motion to find the acceleration, which is

#ul(sumF_x = ma_x#

where

• #sumF_x# is the net force acting on the box

• #m# is the mass of the box

• #a_x# is the acceleration of the box (what we're trying to find)

The only forces acting on the box are

• the gravitational force (acting down the ramp), equal to #mgsintheta#

• the friction force (acting down the ramp), equal to #mu_kn#

And so we have our net force equation:

#sumF_x = mgsintheta + mu_kn#

The normal force #n# exerted by the incline is equal to

#n = color(purple)(mgcostheta)#

So we can plug this in to the net force equation above:

#ul(sumF_x = mgsintheta + mu_kcolor(purple)(mgcostheta)#

Or

#ul(sumF_x = mg(sintheta + mu_kcostheta)#

Now, we can plug this in for #sumF_x# in the Newton's second law equation:

#sumF_x = ma_x#

#ma_x = mg(sintheta + mu_kcostheta)#

We can cancel the mass #m# by dividing both sides by #m#, leaving us with

#color(green)(ul(a_x = g(sintheta + mu_kcostheta)#

#" "#

Now that we have found an expression for the acceleration, let's plug it into the equation

#Deltax = ((v_(0x))^2)/(2a_x)#

And we get

#color(red)(ulbar(|stackrel(" ")(" "Deltax = ((v_(0x))^2)/(2g(sintheta + mu_kcostheta))" ")|)#

We're given in the problem

• #v_(0x) = 3# #"m/s"#

• #theta = (pi)/8#

• #mu_k = 8/5#

• and the gravitational acceleration #g = 9.81# #"m/s"#

Plugging these in:

#color(blue)(Deltax) = ((3color(white)(l)"m/s")^2)/(2(9.81color(white)(l)"m/s"^2)(sin[(pi)/8] + 8/5cos[(pi)/8])) = color(blue)(ulbar(|stackrel(" ")(" "0.247color(white)(l)"m"" ")|)#