A box with an initial speed of #5# m/s is moving up a ramp. The ramp has a kinetic friction coefficient of #2/5 # and an incline of #pi /12 #. How far along the ramp will the box go?

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Answer 1 · 2017-04-01T17:02:02

The distance is #=1.98m#

The coefficient of kinetic friction is

#mu_k=F_r/N#

Let the mass of the object be #=m kg#

The incline is #=theta#

Therefore,

#F_r=mu_k*N=mu_k*mg costheta#

The net force on the box is

#=-mu_kmgcostheta-mgsintheta#

According to Newtons' second Law

#F=ma#

Where the acceleration is #=ah#

#ma=-mu_kmgcostheta-mgsintheta#

#a=-2/5gcos(pi/12)-gsin(pi/12)#

#a=-6.32ms^-2#

We apply the equation of motion

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

#v=0#

#u=5ms^-1#

So,

#2as=u^2#

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

#=25/(2*6.32)=1.98m#