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

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Answer 1 · 2017-01-29T17:14:00

The box will travel #0.135m# along the ramp.

This problem is most easily done by conservation of energy.
Three energy forms are involved:
A change in kinetic energy: #1/2mv_f^2-1/2mv_i^2#

A change in gravitational potential energy: #mgh#

Frictional heating: #muF_NDelta d#

Before we can continue, we need to be aware of a couple of complications

We must express #h# (the height the object rises) in terms of #Deltad# (its displacement along the ramp).

#h=Deltadsintheta#

Also, we must note that the normal force on an incline is not equal to mg, but to #mgcostheta#, where #theta=pi/4# in this case.

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With all that looked after, our equation becomes

#1/2mv_f^2-1/2mv_i^2+mgDeltadsin(pi/4)+mumgcos(pi/4)Delta d=0#

Notice that we can divide every term by the mass m, (including the right side of the equation)

Also, #v_f=0# , and inserting the values for #mu#, #v_i# and g, we get:

#-1/2(2)^2+(9.8)Deltad(0.707)+(8/7)(9.8)(0.707)Delta d=0#

#-2.0+6.93Deltad+7.92Deltad=0#

#14.85Deltad=2.0#

#Deltad=0.135m#