A box with an initial speed of #5 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #3/7 # and an incline of #(5 pi )/8 #. How far along the ramp will the box go?

I'd like to point out that there can't realistically be an inclined ramp with angle of inclination #(5pi)/8#...it must be between #0# and #pi/2#, so I'll choose the closest angle to this, #(3pi)/8#...

We're asked to find the distance traveled by the box given its initial speed, coefficient of kinetic friction, and angle of inclination.

NOTE: the friction force actually acts down the incline as opposed to the image (because the box is traveling up the ramp).

The magnitude of the kinetic friction force #f_k# is given by

#f_k = mu_kn#

where

• #mu_k# is the coefficient of kinetic friction (#3/7#)

• #n# is the magnitude of the normal force exerted by the incline plane, equal to #mgcostheta#

We must first find the acceleration of the box, using Newton's second law:

#sumF = ma#

#a = (sumF)/m#

The net horizontal force #sumF# is

#sumF = mgsintheta + f_k = mgsintheta + mu_kmgcostheta#

Therefore, we have

#a = (cancel(m)gsintheta + cancel(m)u_kmgcostheta)/(cancel(m)) = gsintheta + mu_kgcostheta#

Plugging in known values, we have

#a = (9.81color(white)(l)"m/s"^2)sin((3pi)/8) + 3/7(9.81color(white)(l)"m/s"^2)cos((3pi)/8)#

#= 10.7# #"m/s"^2#

directed down the incline, so this can also be written as

#a = ul(-10.7color(white)(l)"m/s"^2#

Now, we can use the equation

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

to find the distance it travels up the ramp before it comes to a stop.

Here,

#v_x = 0# (instantaneously at rest at maximum height)

• #v_(0x) = 5# #"m/s"#

• #a_x = -10.7# #"m/s"^2#

• #Deltax = # trying to find

Plugging in known values, we have

#0 = (5color(white)(l)"m/s")^2 + 2(-10.7color(white)(l)"m/s"^2)(Deltax)#

#Deltax = color(red)(ul(1.17color(white)(l)"m"#