A box with an initial speed of #6 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #9/5 # and an incline of #( 5 pi )/12 #. How far along the ramp will the box go?

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Answer 1 · 2017-07-29T13:22:13

The distance is #=1.28m#

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

#ma=-mu_kgcostheta-mgsintheta#

#a=-g(mu_kcostheta+sintheta)#

The coefficient of kinetic friction is #mu_k=9/5#

The incline of the ramp is #theta=5/12pi#

#a=-9.8*(9/5cos(5/12pi)+sin(5/12pi))#

#=-14.03ms^-2#

The negative sign indicates a deceleration

We apply the equation of motion

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

#u=6ms^-1#

#v=0#

#a=-11.5ms^-2#

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

#=(0-36)/(-2*14.03)#

#=1.28m#