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Object Traveling Up an Inclined Plane

  1. Mar 13, 2008 #1
    [SOLVED] Object Traveling Up an Inclined Plane

    1. The problem statement, all variables and given/known data

    In a circus performance, a monkey is strapped to a sled and both are given an initial speed of 2.0 m/s up a 19.0° inclined track. The combined mass of monkey and sled is 16 kg, and the coefficient of kinetic friction between sled and incline is 0.20. How far up the incline do the monkey and sled move?

    2. Relevant equations

    Sigma(F) = ma

    3. The attempt at a solution

    I drew a free body diagram and found the normal force using Sigma(Fy)=ma.
    Normal Force - Weight y = 0
    Normal Force = Weight y
    Normal Force = (weight)(cos 19)
    Normal Force = 148.257 N

    I can now find the Force of Friction:
    Force of friction = coefficient of friction * Normal Force
    Force of Friction = (.20)(148.257 N)
    Force of Friction = 29.65 N (In the negative x direction.)

    Now I'm stuck. I'm not sure how to apply Newton's laws since there is no force in the positive x direction. Any help is appreciated. Thanks.
  2. jcsd
  3. Mar 13, 2008 #2


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    Homework Helper

    Well you could use energy considerations to find how far it travels up the plane. (Conservation of energy)
  4. Mar 13, 2008 #3
    I was thinking about using (mgh + .5mv^2)initial = (mgh + .5mv^2)final, but I thought this is only true for conservative forces.
  5. Mar 13, 2008 #4
    I believe I've solved it using energy. Here's what I did (My main equation is in bold):

    Work (nonconservative) = (KE(f) + PE(f)) - (KE(i) + PE(i))
    Work = F*Delta x

    Substituting the latter into the former and getting rid of things that will be zero, I get:
    Force*Delta x = (mgh(f)) - (.5mv(i)^2)

    Now, using simple trigonometric relationships, I see that Sin theta = h/x. This gets plugged into my main equation, so:
    Force*Delta x = (mg(x*sin theta)(f)) - (.5mv(i)^2)

    I also know that my only nonconservative force is my friction force, so:
    -Force friction*Delta x = (mg(x*sin theta)(f)) - (.5mv(i)^2)

    I now have one equation and one unknown, and can solve. Plug in everything:
    -29.651 N*Delta x = (16 kg)(9.8 m/s^2)(x*Sin(19)) - (.5)(16 kg)(2 m/s)^2

    Simplified and units removed for clarification:
    -29.651*Delta x = 51.04x - 32
    -80.691x = -32
    x = -32/-80.691

    x = .397 m

    Does this look right to you all?
  6. Mar 15, 2008 #5
    Can anybody comment on whether or not this looks correct?
  7. Mar 15, 2008 #6
    This does look correct and is a clever way of solving the problem. More commonly, because the combined force against motion up the inclined plane is always constant (friction force you found + force from gravity), you can apply one of the four kinematic equations for constant acceleration.

    Your original question was:

    When doing these problems, you usually change coordinate axes from the ordinary vertical is y and horizontal is x, to one where the direction of motion (up the plane) is x, and the normal force, which is perpendicular to that motion, is y.

    You have done this by using cosine to find how much of the total force from gravity contributes to the normal force and used that to find the force opposing motion up the plane from friction. You also need to use sine to find the force from gravity that is directly opposing movement up the plane.

    Then, try to move on from there.
  8. Mar 16, 2008 #7
    I see. Didn't even think to use kinematics.
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