A box with an initial speed of #8 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #3/4 # and an incline of #(2 pi )/3 #. How far along the ramp will the box go?

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Answer 1 · 2018-01-17T14:07:17

The distance is #=6.65m#

Resolving in 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=3/4#

The acceleration due to gravity is #g=9.8ms^-2#

The incline of the ramp is #theta=2/3pi#

#a=-9.8*(3/4cos(2/3pi)+sin(2/3pi))#

#=-4.81ms^-2#

The negative sign indicates a deceleration

We apply the equation of motion

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

#u=8ms^-1#

#v=0#

#a=-4.81ms^-2#

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

#=(0-64)/(-2*4.81)#

#=6.65m#