A box with an initial speed of #9 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #5/3 # and an incline of #(3 pi )/8 #. How far along the ramp will the box go?

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Answer 1 · 2017-04-09T08:02:25

The distance is #=2.65m#

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Solving in the direction of the plane #↗^+#

#mu_k=F_k/N#

#F_k=mu_kN#

#N=mgcostheta#

#F_k=mu_kmgcostheta#

The component of the weight is

#=mgsintheta#

Applying Newton' second Law

#-mu_kmgcostheta-mgsintheta=ma#

Therefore,

#a=-u_kgcostheta-gsintheta#

#=-5/3*g*cos(3/8pi)-g*sin(3/8pi)#

#=-15.3ms^-2#

The initial velocity is #u=9ms^-1#

We apply the equation of motion

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

#0=9^2-2*7.43s#

#s=81/(2*15.3)#

#=2.65m#