A box with an initial speed of #3 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #5/7 # and an incline of #( 5 pi )/12 #. How far along the ramp will the box go?

1 Answer
Aug 14, 2017

Answer:

#"distance" = 0.399# #"m"#

Explanation:

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 = (5pi)/12#

  • #mu_k = 5/7#

  • 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[(5pi)/12] + 5/7cos[(pi)/12])) = color(blue)(ulbar(|stackrel(" ")(" "0.399color(white)(l)"m"" ")|)#