A box with an initial speed of #8 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #4/5 # and an incline of #( pi )/6 #. How far along the ramp will the box go?

Question

A box with an initial speed of #8  m/s# is moving up a ramp.  The ramp has a kinetic friction coefficient of #4/5 # and an incline of #(  pi )/6 #.  How far along the ramp will the box go?

Amelia Adams · Accepted Answer

The box will move #~~2.7"m"# up the ramp.

I used a combination of kinematics and Newton's second law.

We have the following information:

#|->v_i=8"m"//"s"#
#|->mu_k=4/5#
#|->theta=pi/6->30^o#
#|->g=9.81"m"//"s"^2#

Diagram:

#color(darkblue)(F_(y"net")=sumF_y=n-(F_G)_y=0)#

Note no acceleration in the vertical direction (dynamic equilibrium).

Therefore, we have #color(darkblue)(n=(F_G)_y)#.

The force of kinetic friction is given by

#color(darkblue)(f_k=mu_kn)#

We can find the perpendicular component of gravity and consequently the magnitude of the normal force using trigonometry. We can see from the above diagram that the angle between the force of gravity and the vertical is the same as the angle of incline of the ramp.

#cos(theta)="adjacent"/"hypotenuse"#
#=>cos(theta)=(F_G)_y/(F_G)#
#=>color(darkblue)((F_G)_y=F_Gcos(theta))#

Similarly, we find that #color(darkblue)((F_G)_x=mgsin(theta))#.

As #F_G=mg#, we have:

#color(darkblue)(n=mgcos(theta))#

Therefore:

#color(darkblue)(f_k=mu_k*mgcos(theta))#

Substituting this and the equation we derived for the force of gravity parallel above into our statement for the net force parallel:

#mu_kcancelcolor(skyblue)(m)gcos(theta)+cancelcolor(skyblue)(m)gsin(theta)=cancelcolor(skyblue)(m)a_x#
#=>color(blue)(a_x=g(mu_kcos(theta)+sin(theta)))#

We can now use this kinematic equation to find the distance traveled up the ramp:

#v_f^2=v_i^2+2aDeltas#

Solving for #Deltas#:

#Deltas=(cancel(v_f^2)-v_i^2)/(2a)#

Substituting in the equation we derived above for the acceleration:

#color(blue)(Deltas=-v_i^2/(2g(mu_kcos(theta)+sin(theta)))#

Now we can substitute in our known values:

#Deltas=-(8"m"//"s")^2/(2(9.81"m"//"s"^2)(4/5cos(30^o)+sin(30^o))#
#=>color(blue)(~~-2.7"m")#

Because we defined up the ramp as negative, getting a negative value for displacement tells us that our final position was less than our initial position, which makes sense as we set at #s_i=0"m"#.

Note: this question can also be solved using the work-energy theorem, but it can be very easy to misplace negative signs and alter your answer since addition is involved in the denominator. Here is a link to an example of a very similar problem solved using the W-E theorem.

Adam Arnold · Answer

undefined The distance the box travels along the ramp can be calculated using the equation:

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

Where:
- $ d $ = distance along the ramp
- $ v_0 $ = initial speed of the box (8 m/s)
- $ g $ = acceleration due to gravity (9.8 m/s²)
- $ \mu_k $ = kinetic friction coefficient (4/5)
- $ \theta $ = angle of incline ($ \pi/6 $ radians)

Plugging in the values:

$$ d = \frac{{(8 \, \text{m/s})^2}}{{2 \times 9.8 \, \text{m/s}^2}} \left( \frac{4}{5} \cos\left(\frac{\pi}{6}\right) + \sin\left(\frac{\pi}{6}\right) \right) $$

$$ d \approx \frac{{64}}{{19.6}} \left( \frac{4}{5} \times 0.866 + 0.5 \right) $$

$$ d \approx \frac{{64}}{{19.6}} \left( 0.693 + 0.5 \right) $$

$$ d \approx \frac{{64}}{{19.6}} \times 1.193 $$

$$ d \approx 3.257 \, \text{m} $$

So, the box will travel approximately 3.257 meters along the ramp.