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A box with an initial speed of #3 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #2/5 # and an incline of #pi /4 #. How far along the ramp will the box go?

Answer 1

Isabella Alvarez

The distance is #=0.46m#

Resolving in 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#

Newton's Second Law states

#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=2/5#

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

The incline of the ramp is #theta=1/4pi#

#a=-9.8*(2/5cos(1/4pi)+sin(1/4pi))#

#=-9.7ms^-2#

A deceleration is indicated by the negative sign.

We utilize the equation of motion.

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

#u=3ms^-1#

#v=0#

#a=-9.7ms^-2#

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

#=(0-9)/(-2*9.7)#

#=0.46m#

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Answer 2

William Audy

The distance along the ramp the box will go is approximately 5.65 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.