A box with an initial speed of #7 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #5/4 # and an incline of #pi /6 #. How far along the ramp will the box go?

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Answer 1 · 2017-08-13T09:05:07

The distance is #=1.58m#

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=5/4#

The incline of the ramp is #theta=1/6pi#

#a=-9.8*(5/4cos(1/6pi)+sin(1/6pi))#

#=-15.51ms^-2#

The negative sign indicates a deceleration

We apply the equation of motion

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

#u=7ms^-1#

#v=0#

#a=-15.51ms^-2#

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

#=(0-49)/(-2*15.51)#

#=1.58m#