A box with an initial speed of #7 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #8/3 # and an incline of #pi /8 #. How far along the ramp will the box go?

Question

1237 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-18T18:38:46

The distance is #=0.88m#

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=8/3#

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

The incline of the ramp is #theta=1/8pi#

#a=-9.8*(8/3cos(1/8pi)+sin(1/8pi))#

#=-27.89ms^-2#

The negative sign indicates a deceleration

We apply the equation of motion

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

#u=7ms^-1#

#v=0#

#a=-27.89ms^-2#

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

#=(0-49)/(-2*27.89)#

#=0.88m#