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Newton's Laws - Incline. Do I have the right idea?

  1. Oct 7, 2016 #1
    I am hoping that I can have someone look over my answers. I have provided what I think is necessary to be appropriate posting, including the questions and attempts at all my answers. I would just like some feedback that I am correct in my answers. Thank you!

    This is an overview that all the questions are based off of: An incline with theta 24 degrees, length of incline is 50m, base is 45.7 m, and height is 20.3 meters.

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

    Assuming a block at the top starts from rest (ignoring friction), how fast would it be moving when it reaches the bottom of the ramp? i.e. after sliding a distance of 50m.

    2. Relevant equations

    3. The attempt at a solution
    v=sqrt2gh = 19.94 m/s

    Question 2)
    Now include sliding friction. Draw a free-body diagram for the block when it is partway down the ramp. Label all the forces

    No relevant Equations needed

    Attempt at solution: Since I can't upload a photo, I'll describe my free body diagram: dot in middle represents block, downward arrow represents gravity, upward and to the right represents normal force, and upward perpendicular to normal force (to the left) represents friction. Am I missing any forces?

    Question 3)
    Using the free body diagram and Newton's 2nd law to derive the acceleration down the ramp (assume coefficient of kinetic fricion uk) Show reasoning/derivation.

    Attempt at solution: Solve for Normal, N=mgcostheta
    Sum of forces down the ramp are positive = ma=mgsintheta-friction
    Friction is negative because it opposes the motion. Friction =uN Friction = umgcostheta
    So, ma=mgsintheta-umgcostheta
    a=g(sintheta) - uk (costheta)

    Question 4) In the show watched in class (I don't think anyone needs to watch this to help), Adam suggests the second block will go faster down the incline because it weighs more. Based on your result in #2, do you agree?

    Relevant Equations:
    v=sqrt(2gh) Conservation of Energy

    Attempt at solution: No, I don't agree. This is tricky. If I assume no air resistance (which since this hasn't been mentioned in the assignment, I'll assume no air resistence), then mass won't matter. If air resistance is considered, then yes - he would move faster.

    Question 5) Suppose the block starts at rest, slides a distance of 50m to the bottom of the ramp, ending at a known (measured) speed v. Which expression tells us the coefficient of kinetic friction? Show the algebra.

    My options for this one are:

    A) tan(theta)
    B) tan(theta) + v^2/2gL
    C) tan(theta) - v^2/2gl
    D) tan(theta) + v^2/2gl(cos(theta))
    E) tan(theta) - v^2/2gl(cos(theta))

    Relevant Equations:


    Attempt at a solution:

    I believe the answer for this one is A because i think i remember it being taught in class that way. However, I'm not sure how to derive the answer. I know I have to start with the kinetic friction formula which is Fr=uN with N being the force of gravity? This is the one that I need more help on. I hesitate to choose that answer because my intuition tells me that this equation should include velocity.

    Thanks for all the help, everyone!

  2. jcsd
  3. Oct 7, 2016 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Good, but make sure you don't leave out that factor of 'g' in the second term.

    Your intuition is correct: you'll need to use the given velocity. Hint: You just derived a formula that relates the acceleration and the coefficient of friction. Use it! Hint 2: Use kinematics to derive another equation for acceleration.
  4. Oct 7, 2016 #3
    Doc Al, Thank you for your very helpful quick reply. I redid Question 5 and got E) for my answer.
    I have one more question pertaining to this problem set:
    1. The problem statement, all variables and given/known data
    We are told the block reaches a max speed of about 30 mi/h at the bottom of the ramp. Use question 5 to calculate the coefficient of kinetic friction down the incline.

    2. Relevant equations
    tan(theta) - v^2/2gL(costheta)

    3. The attempt at a solution
    Using dimensional analysis, I converted 30 mi/h to 13.4 m/s.
    tan(24) - 13.4^2/2(9.8)(50)(cos24)
    .445 - 13.4^2/895.27
    .445 - .201 = .244
  5. Oct 7, 2016 #4

    Doc Al

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    Staff: Mentor

    Excellent! :smile:
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