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

Question

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

Lily Adams · Accepted Answer

The distance is #=0.19m#

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=5/2#

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

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

#=-24.3ms^-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=-24.3ms^-2#

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

#=(0-9)/(-2*24.3)#

#=0.19m#

Xavier Adame · Answer

undefined To find how far the box will go along the ramp, we can use the work-energy principle. The work done by the friction force will be equal to the change in kinetic energy of the box. The frictional force can be found using the formula $ f_{friction} = \mu_k \times f_{normal} $, where $ f_{normal} $ is the normal force. The normal force can be found using the component of the gravitational force perpendicular to the ramp. The kinetic energy change can be found using the formula $ \Delta KE = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 $, where $ v_i $ and $ v_f $ are the initial and final velocities, respectively. Finally, the work done by friction can be calculated using the formula $ W_{friction} = f_{friction} \times d $, where $ d $ is the distance along the ramp. Rearranging the formulas, we can solve for $ d $. After substituting the given values and solving the equation, we find that the box will travel approximately $ 6.63 $ meters along the ramp.