A box with an initial speed of #5 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #1/2 # and an incline of #(3 pi )/8 #. How far along the ramp will the box go?

1 Answer
Apr 12, 2017

Answer:

The distance is #=1.14m#

Explanation:

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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#

#=-1/2*g*cos(3/8pi)-g*sin(3/8pi)#

#=-10.93ms^-2#

The initial velocity is #u=5ms^-1#

We apply the equation of motion

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

#0=5^2-2*10.93s#

#s=25/(2*10.93)#

#=1.14m#