A box with an initial speed of #4 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #5/6 # and an incline of #pi /4 #. How far along the ramp will the box go?

Question

A box with an initial speed of #4  m/s# is moving up a ramp.  The ramp has a kinetic friction coefficient of #5/6 # and an incline of #pi /4 #.  How far along the ramp will the box go?

Sebastian Acosta · Accepted Answer

The distance is #=0.63ms^-1#

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=5/6#

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

#a=-9.8*(5/6cos(1/4pi)+sin(1/4pi))#

#=-12.7ms^-2#

A deceleration is indicated by the negative sign.

We utilize the equation of motion.

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

#u=4ms^-1#

#v=0#

#a=-12.7ms^-2#

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

#=(0-16)/(-2*12.7)#

#=0.63m#

Nora Arzate · Answer

undefined To find the distance the box travels up the ramp, you can use the formula for the work done by friction:

$$ W_{friction} = \mu_k \cdot m \cdot g \cdot d $$

Where:
$ \mu_k $ = coefficient of kinetic friction (5/6 in this case)
$ m $ = mass of the box
$ g $ = acceleration due to gravity (approximately 9.8 m/s²)
$ d $ = distance traveled up the ramp

You can also use the formula for gravitational force along the ramp:

$$ F_{gravity} = m \cdot g \cdot \sin(\theta) $$

Where:
$ \theta $ = angle of the incline (π/4 in this case)

Since the box is moving up the ramp, the force of friction opposes the gravitational force. Therefore:

$$ W_{friction} = F_{gravity} $$

Solve for $ d $:

$$ \mu_k \cdot m \cdot g \cdot d = m \cdot g \cdot \sin(\theta) $$

$$ d = \frac{\sin(\theta)}{\mu_k} $$

Substitute the given values:

$$ d = \frac{\sin(\frac{\pi}{4})}{\frac{5}{6}} $$

$$ d = \frac{\frac{\sqrt{2}}{2}}{\frac{5}{6}} $$

$$ d = \frac{6\sqrt{2}}{10} $$

$$ d = \frac{3\sqrt{2}}{5} $$

So, the box will travel $ \frac{3\sqrt{2}}{5} $ meters along the ramp.