A box with an initial speed of #8 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #3/2 # and an incline of #pi /4 #. How far along the ramp will the box go?

Question

A box with an initial speed of #8  m/s# is moving up a ramp.  The ramp has a kinetic friction coefficient of #3/2 # and an incline of #pi /4 #.  How far along the ramp will the box go?

Leo Abeyta · Accepted Answer

The distance is #=2.77m#

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#

Newton's Second Law states

#F=m*a#

Where #a# is the acceleration

So

#ma=-mu_kgcostheta-mgsintheta#

#a=-g(mu_kcostheta+sintheta)#

The coefficient of kinetic friction is #mu_k=3/2#

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

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

#a=-9.8*(3/2cos(1/4pi)+sin(1/4pi))#

#=-11.57ms^-2#

A deceleration is indicated by the negative sign.

We utilize the equation of motion.

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

#u=8ms^-1#

#v=0#

#a=-11.57ms^-2#

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

#=(0-64)/(-2*11.57)#

#=2.77m#

Parker Abbott · Answer

undefined The box will travel a distance of approximately 7.21 meters along the ramp before coming to a stop.