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A box with an initial speed of #7 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #3/4 # and an incline of #pi /8 #. How far along the ramp will the box go?

Answer 1

Nathan Ashcroft

The distance is #=2.32m#

Taking the direction up and parallel to the plane as positive #↗^+#

The coefficient of kinetic friction is #mu_k=F_r/N#

Consequently, the object's net force is

#F=-F_r-Wsintheta#

#=-F_r-mgsintheta#

#=-mu_kN-mgsintheta#

#=mmu_kgcostheta-mgsintheta#

According to Newton's Second Law

#F=m*a#

Where #a# is the acceleration

#ma=-mu_kgcostheta-mgsintheta#

#a=-g(mu_kcostheta+sintheta)#

The coefficient of kinetic friction is #mu_k=3/4#

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

#a=-9.8*(3/4cos(1/8pi)+sin(1/8pi))#

#=-10.54ms^-2#

The negative sign indicates a deceleration

We apply the equation of motion

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

#u=7ms^-1#

#v=0#

#a=-10.54ms^-2#

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

#=(0-49)/(-2*10.54)#

#=2.32m#

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Answer 2

William Audy

The distance along the ramp the box will go is approximately 9.8 meters.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.