A box with an initial speed of #7 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #8/3 # and an incline of #pi /12 #. How far along the ramp will the box go?

Question

A box with an initial speed of #7  m/s# is moving up a ramp.  The ramp has a kinetic friction coefficient of #8/3 # and an incline of #pi /12 #.  How far along the ramp will the box go?

Charles Aeschliman · Accepted Answer

The distance is #=0.88m#

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

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

Then the net force on the object is

#F=-F_r-Wsintheta#

#=-F_r-mgsintheta#

#=-mu_kN-mgsintheta#

#=mmu_kgcostheta-mgsintheta#

According to Newton's Second Law of Motion

#F=m*a#

Where #a# is the acceleration of the box

So

#ma=-mu_kgcostheta-mgsintheta#

#a=-g(mu_kcostheta+sintheta)#

The coefficient of kinetic friction is #mu_k=8/3#

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

The incline of the ramp is #theta=1/12pi#

The acceleration is #a=-9.8*(8/3cos(1/12pi)+sin(1/12pi))#

#=-27.78ms^-2#

The negative sign indicates a deceleration

Apply the equation of motion

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

The initial velocity is #u=7ms^-1#

The final velocity is #v=0#

The acceleration is #a=-27.78ms^-2#

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

#=(0-49)/(-2*27.78)#

#=0.88m#

Sebastian Acosta · Answer

undefined To find the distance along the ramp (d), you can use the following equation:

$$ d = \frac{{v_0^2}}{{2g}} \left(1 - \sqrt{1 - \frac{{2g \cdot \mu_k \cdot \theta}}{{v_0^2}}}\right) $$

Where:
$ v_0 = 7 \, \text{m/s} $ (initial speed),
$ g = 9.8 \, \text{m/s}^2 $ (acceleration due to gravity),
$ \mu_k = \frac{8}{3} $ (kinetic friction coefficient),
$ \theta = \frac{\pi}{12} $ (incline angle).

Plug in the values to find the distance along the ramp.