A box with an initial speed of #3 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #5/2 # and an incline of #pi /4 #. How far along the ramp will the box go?

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Answer 1 · 2017-07-09T17:58:13

The distance is #=0.19m#

Taking the direction up and parallel to the plane as positive #↗^+#

The coefficient of kinetic friction is #mu_k=F_r/N#

Then the net force on the object is

#F=-F_r-Wsintheta#

#=-F_r-mgsintheta#

#=-mu_kN-mgsintheta#

#=mmu_kgcostheta-mgsintheta#

#F=m*a#

Where #a# is the acceleration
So

#ma=-mu_kgcostheta-mgsintheta#

#a=-g(mu_kcostheta+sintheta)#

The coefficient of kinetic friction is #mu_k=5/2#

The incline of the ramp is #theta=1/4pi#

#a=-9.8*(5/2cos(1/4pi)+sin(1/4pi))#

#=-24.3ms^-2#

The negative sign indicates a deceleration

We apply the equation of motion

#v^2=u^2+2as#

#u=3ms^-1#

#v=0#

#a=-24.3ms^-2#

#s=(v^2-u^2)/(2a)#

#=(0-9)/(-2*24.3)#

#=0.19m#