A box with an initial speed of #2 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #5/3 # and an incline of #(3 pi )/8 #. How far along the ramp will the box go?

Question

A box with an initial speed of #2  m/s# is moving up a ramp.  The ramp has a kinetic friction coefficient of #5/3 # and an incline of #(3  pi )/8 #.  How far along the ramp will the box go?

Charles Aeschliman · Accepted Answer

The distance is #=0.13m#

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 of Motion states

#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=5/3#

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

The incline of the ramp is #theta=3/8pi#

The acceleration is #a=-9.8*(5/3cos(3/8pi)+sin(3/8pi))#

#=-15.30ms^-2#

A deceleration is indicated by the negative sign.

Utilize the equation of motion.

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

The initial velocity is #u=2ms^-1#

The final velocity is #v=0#

The acceleration is #a=-15.30ms^-2#

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

#=(0-4)/(-2*15.30)#

#=0.13m#

Isabella Alvarez · Answer

undefined To find the distance, use the kinematic equation:

$$d = \frac{v_i^2}{2g} \left(\sin\theta - \mu_k\cos\theta\right)$$

Where:
$v_i = 2 \, \text{m/s}$ (initial speed),
$\theta = \frac{3\pi}{8}$ (ramp incline angle),
$\mu_k = \frac{5}{3}$ (kinetic friction coefficient),
$g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity).

Substitute the values into the equation to find $d$.