A box with an initial speed of #5# m/s is moving up a ramp. The ramp has a kinetic friction coefficient of #2/5 # and an incline of #pi /12 #. How far along the ramp will the box go?

Question

A box with an initial speed of #5# m/s is moving up a ramp.  The ramp has a kinetic friction coefficient of #2/5 # and an incline of #pi /12 #.  How far along the ramp will the box go?

Benjamin Asiedu · Accepted Answer

The distance is #=1.98m#

The coefficient of kinetic friction is

#mu_k=F_r/N#

Let the mass of the object be #=m kg#

The incline is #=theta#

Therefore,

#F_r=mu_k*N=mu_k*mg costheta#

The net force on the box is

#=-mu_kmgcostheta-mgsintheta#

According to Newtons' second Law

#F=ma#

Where the acceleration is #=ah#

#ma=-mu_kmgcostheta-mgsintheta#

#a=-2/5gcos(pi/12)-gsin(pi/12)#

#a=-6.32ms^-2#

We apply the equation of motion

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

#v=0#

#u=5ms^-1#

So,

#2as=u^2#

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

#=25/(2*6.32)=1.98m#

Axel Acevedo · Answer

undefined To find how far the box will go along the ramp, you can use the equations of motion along the incline. The distance the box travels can be found using the equation:

$$ d = \frac{v_0^2}{2g} \left( \frac{\mu_k}{\sin(\theta)} + \cos(\theta) \right) $$

where:
- $ d $ is the distance traveled along the incline,
- $ v_0 $ is the initial velocity of the box (5 m/s in this case),
- $ g $ is the acceleration due to gravity (approximately $ 9.81 \, \text{m/s}^2 $),
- $ \mu_k $ is the coefficient of kinetic friction (2/5 in this case),
- $ \theta $ is the angle of the incline (pi/12 radians in this case).

Substitute the given values into the equation to find the distance traveled by the box along the ramp.