A box with an initial speed of #2 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #3/2 # and an incline of #(2 pi )/3 #. How far along the ramp will the box go?

Question

A box with an initial speed of #2  m/s# is moving up a ramp.  The ramp has a kinetic friction coefficient of #3/2 # and an incline of #(2  pi )/3 #.  How far along the ramp will the box go?

Emily Abbate · Accepted Answer

#"distance" = 0.126# #"m"#

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).

NOTE: Ideally, the angle of inclination would be between #0# and #pi/2#, so I'll choose the corresponding first-quadrant angle of #pi/3#.

I will also 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) = 2# #"m/s"#

#theta = (pi)/3#

#mu_k = 3/2#

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

Plugging these in:

#color(blue)(Deltax) = ((2color(white)(l)"m/s")^2)/(2(9.81color(white)(l)"m/s"^2)(sin[(pi)/3] + 3/2cos[(pi)/3])) = color(blue)(ulbar(|stackrel(" ")(" "0.126color(white)(l)"m"" ")|)#

Benjamin Asiedu · Answer

Here's my own attempt at this problem, using an alternative approach to Nathan's, to see if our answers agree. Like Nathan, I will assume that you mean to have the ramp right-side-up, i.e. an angle of #60^@# w.r.t. the horizontal...

I also got #"0.126 m"#.

My approach here is via conservation of energy:

#DeltaE = DeltaK + DeltaU + W_(vecF_k) = 0#
where:

My coordinate axes are almost the usual vertical #y# and horizontal #x#. Rightwards is #+x#, but downwards is #+y#.

#DeltaK = 1/2 mv_f^2 - 1/2 mv_i^2 = m/2(v_f^2 - v_i^2)#
#DeltaU = mvecgDeltavecy = mgy_f#, where #y_i = 0#, #y_f < 0#, and #g > 0#. The potential energy becomes more negative due to the sign of #y_f#.

#W_(vecF_k) = vecF_kd# is the counteracting work due to kinetic friction (negative w.r.t. the box, the system). In this case, I define

#d = sqrt((Deltax)^2 + (Deltay)^2) - 0 = sqrt(x_f^2 + y_f^2)# with #x_i = 0# and #x_f > 0#. #d# is then the final position, #< 0#.

Currently, we know we have a nonzero initial velocity and a zero final velocity when the box comes to a stop:

#=> DeltaK = -m/2v_i^2#

This so far gives:

#-m/2v_i^2 + mgy_f + vecF_kd = 0#

or

#m/2v_i^2 - mgy_f - vecF_kd = 0#

Using the sum of the forces:

#sum_i vecF_(_|_,i) = vecF_N - mvecgcostheta = 0#

So...

#0 = cancel(m)/2v_i^2 - cancel(m)gy_f - mu_kcancel(m)vecgcosthetad#
#=> 1/2 v_i^2 - gy_f - mu_kgcosthetad = 0#

Now, we can rewrite #y_f# in terms of #d#, the distance traveled along the ramp:

#y_f = dsintheta < 0#

This gives:

#=> 1/2 v_i^2 - gdsintheta - mu_kgdcostheta = 0#

Solve for #d# to get:

#d = (1/2 v_i^2)/(gsintheta + mu_kgcostheta)#
#= (v_i^2)/(2g(sintheta + mu_kcostheta))#
(which is also what Nathan found, by the way.)
So, in the end, we get:

#color(blue)(|d|) = (2^2)/(2(9.81)(sin(60^@) + 3/2cos(60^@)))#
#=# #color(blue)("0.126 m")#

Parker Abbott · Answer

undefined The distance the box will travel along the ramp is approximately 3.34 meters.