A box with an initial speed of #9 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #5/6 # and an incline of #(3 pi )/8 #. How far along the ramp will the box go?

Question

A box with an initial speed of #9  m/s# is moving up a ramp.  The ramp has a kinetic friction coefficient of #5/6 # and an incline of #(3  pi )/8 #.  How far along the ramp will the box go?

Elena Albarran · Accepted Answer

The distance is #=3.33m#

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=5/6#

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*(5/6cos(3/8pi)+sin(3/8pi))#

#=-12.18ms^-2#

A deceleration is indicated by the negative sign.

Utilize the equation of motion.

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

The initial velocity is #u=9ms^-1#

The final velocity is #v=0#

The acceleration is #a=-6.32ms^-2#

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

#=(0-81)/(-2*12.18)#

#=3.33m#

Elena Albarran · Answer

undefined To find how far along the ramp the box will go, you can use the equations of motion. First, calculate the acceleration of the box along the ramp using the gravitational force component parallel to the ramp and subtracting the frictional force. Then, use the kinematic equation to find the distance traveled. Given the incline angle, you can decompose the gravitational force into components parallel and perpendicular to the ramp. The parallel component provides the acceleration. Use the equation:

$$ a = g \cdot \sin(\theta) - \mu_k \cdot g \cdot \cos(\theta) $$

where:
- $ a $ = acceleration
- $ g $ = acceleration due to gravity (approximately $ 9.8 \, \text{m/s}^2 $)
- $ \theta $ = incline angle
- $ \mu_k $ = coefficient of kinetic friction

Then, use the equation of motion:

$$ d = v_i \cdot t + \frac{1}{2} \cdot a \cdot t^2 $$

where:
- $ d $ = distance traveled
- $ v_i $ = initial velocity
- $ t $ = time

Given:
- $ v_i = 9 \, \text{m/s} $
- $ \theta = \frac{3\pi}{8} $
- $ \mu_k = \frac{5}{6} $

Substitute the values and solve for $ d $.