# A box with an initial speed of #3 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #3/4 # and an incline of #pi /8 #. How far along the ramp will the box go?

##### 1 Answer

#### Answer:

#### Explanation:

This question can be solved using a combination of **kinematics** and **Newton's second law,** or through the **work-energy** theorem. I will provide both solutions.

**We have the following information:**

#|->v_i=3"m"//"s"# #|->mu_k=3/4# #|->theta=pi/8# #|->g=9.81"m"//"s"^2#

**Diagram:**

where

#vecn# is the normal force,#vecf_k# is the force of kinetic friction, and#vecF_G# is the force of gravity, decomposed into its parallel and perpendicular components

**I wil define**.*up*the ramp as the positive direction

#color(darkblue)("Method 1: Newton's Second Law and Kinematics")# **

We can set up statements of the net force parallel (x) and perpendicular (y).

#color(darkblue)(F_(x"net")=sumF_x=-f_k-(F_G)_x=ma_x)#

#color(darkblue)(F_(y"net")=sumF_y=n-(F_G)_y=0)# Note no acceleration in the vertical direction (dynamic equilibrium).

Therefore, we have

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

As

#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=(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=(3"m"//"s")^2/(2(9.81"m"//"s"^2)(3/4cos(pi/8)+sin(pi/8))#

#=>color(blue)(~~0.42"m")#