A box with an initial speed of #1 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 #1  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?

Lily Adams · Accepted Answer

The distance is #=0.04m#

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=1/4pi#

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

#=-12.7ms^-2#

A deceleration is indicated by the negative sign.

Utilize the equation of motion.

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

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

The final velocity is #v=0#

The acceleration is #a=-12.7ms^-2#

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

#=(0-1)/(-2*12.7)#

#=0.04m#

Nathan Ashcroft · Answer

undefined To find the distance the box travels along the ramp, you can use the equations of motion along the incline. The key equation to use here is the equation for the distance traveled along the incline, which is given by:

$$ d = \frac{{v_i^2}}{{2g}} \left( \frac{{\mu_k + \tan(\theta)}}{{1 - \mu_k \cdot \tan(\theta)}} \right) $$

Where:
- $ d $ is the distance traveled along the incline.
- $ v_i $ is the initial velocity of the box.
- $ g $ is the acceleration due to gravity.
- $ \mu_k $ is the coefficient of kinetic friction.
- $ \theta $ is the angle of incline.

Substitute the given values into the equation and solve for $ d $:

$$ d = \frac{{(1 \, \text{m/s})^2}}{{2 \cdot 9.8 \, \text{m/s}^2}} \left( \frac{{\frac{5}{6} + \tan\left(\frac{\pi}{4}\right)}}{{1 - \frac{5}{6} \cdot \tan\left(\frac{\pi}{4}\right)}} \right) $$

$$ d = \frac{1}{19.6} \left( \frac{{\frac{5}{6} + 1}}{{1 - \frac{5}{6}}} \right) $$

$$ d = \frac{1}{19.6} \left( \frac{{\frac{11}{6}}}{{\frac{1}{6}}} \right) $$

$$ d = \frac{1}{19.6} \cdot 11 = \frac{11}{19.6} \, \text{m} $$

So, the box will travel approximately $ \frac{11}{19.6} $ meters along the ramp.