A box with an initial speed of #3 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #5/6 # and an incline of #( pi )/3 #. How far along the ramp will the box go?

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Answer 1 · 2017-06-27T07:35:36

The distance is #=0.36m#

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=5/6#

The incline of the ramp is #theta=1/3pi#

#a=-9.8*(5/6cos(1/3pi)+sin(1/3pi))#

#=-12.57ms^-2#

The negative sign indicates a deceleration

We apply the equation of motion

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

#u=3ms^-1#

#v=0#

#a=-12.7ms^-2#

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

#=(0-9)/(-2*12.57)#

#=0.36m#