A box with an initial speed of #8 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #7/4 # and an incline of #(2 pi )/3 #. How far along the ramp will the box go?

Since there were differing answers here, I thought I'd settle it:)

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

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 (#7/4#)

• #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((pi)/3) + 7/4(9.81color(white)(l)"m/s"^2)cos((pi)/3)#

#= 17.1# #"m/s"^2#

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

#a = ul(-17.1# #"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) = 8# #"m/s"#

• #a_x = -17.1# #"m/s"^2#

• #Deltax = # trying to find

Plugging in known values, we have

#0 = (8color(white)(l)"m/s")^2 + 2(-17.1color(white)(l)"m/s"^2)(Deltax)#

#Deltax = color(red)(ul(1.87color(white)(l)"m"#

Here's explanation.👇