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Energy and Circular Motion Problem

  1. Dec 1, 2008 #1
    1. The problem statement, all variables and given/known data

    This is a simple, yet rather convoluted problem, so please, bear with me.

    An 80 Kg man on a sled is being chased by a lion down a 60 meter slope (μk = .13) with an incline of 30°. At the end of the slope, there is a 100 meter cliff. Also, there is a 20-meter-long rope at the end of the slope, which is tied to an overhanging tree branch. The rope is angled at 50° below the horizontal. The man grabs the rope and swings away to escape the lion. Let the point at the end of the slope = Point A, and let the point at which the rope is perpendicular to the ground = Point B.

    Find:
    1. The man's Potential Energy.
    2. The Work of Friction.
    3. The Velocity at Point A.
    4. The Kinetic Energy at Point A.
    5. The Potential Energy from Point A to Point B.
    6. The total Kinetic Energy at Point B.
    7. The Velocity at Point B.
    8. The Tension in the Rope at Point B.
    9. The angle of the rope when it reaches its highest point.

    10. Assume that the lion is located at Point A when the man grabs the rope. Where does the lion land after falling off the cliff?

    *Use 10 m/s² for gravity.


    2. Relevant equations

    Too many to name, but I'll try.

    PE = mgy
    W = μk*n*d
    KE = (1/2)m(v²)
    F(centripetal) = (mv²)/r
    Tension = F(centripetal) + ω

    And other basic equations. I'm just having a difficult time factoring each of them in.


    3. The attempt at a solution

    Well, firstly, finding the initial PE is fairly simple: PE = (80 Kg)(10 m/s²)(60sin30°) = 24000 J (#1).

    Then, I assume I would use the equation W(Friction) = μk*n*d, so W(Friction) = (.13)(800cos30°)(60m) = 692.82 J (#2).

    I would then procede to take the difference between the two, so PE - W(Friction) = 24000 - 692.82 = 23307.18 J, which would be the Kinetic Energy at Point B (#4).

    By then making that equal to the equation for KE, I would would use 23307.18 = (1/2)(80)(v²), and solving for v, which would equal 24.14 m/s (#3).

    Now, the fifth part is worded ambiguously to me. So, from here, I'm pretty much lost. Knowing my teacher, I doubt he is asking for a representation of the PE for all points between A and B. I suspect it is merely supposed to mean what the PE is at Point A. However, if that is true, I am unsure as to how to go about factoring the initial velocity into the equation (Our curriculum is all over the map, and I can't seem to find it in my textbook). I also have an inkling that I made a mistake in the first part of the equation, so I would appreciate a double-check.

    Please, if you could, help me. Thank you
     
    Last edited: Dec 1, 2008
  2. jcsd
  3. Dec 1, 2008 #2

    LowlyPion

    User Avatar
    Homework Helper

    Welcome to PF.

    I'd say that part 5 is wanting to know what potential energy is at the 50° angle that the rope is at relative to where it will be of interest at Point B in the later parts.

    To determine the answers for 6 and so forth you would take the initial Kinetic and add the additional potential energy that he would have gained by the time he reaches Point B.

    KEb = PE + KEa
     
  4. Dec 1, 2008 #3
    Thanks for the reply, the insight, and the welcome.

    So, if I'm understanding what you're saying, then #5 is simply asking me to find the Potential Energy that exists at Point A, without taking into account the remaining Potential Energy (albeit in a different direction) that will exist beyond Point B. If that is so, then PE = (80)(10)(20 - 20cos40°) = 3743.29 J.

    And here I was mulling over that same conservation of energy equation, but couldn't figure out what to substitute for PEFinal. Well, that clarifies things a bit.

    So, adding the Kinetic Energy of A to the newly found Potential Energy, I find KEB = 24000 + 3743.29 = 27743.29 J (#6).

    Then, I would be able to take that Kinetic Energy and set it equal to (1/2)mv², so 27743.29 = (1/2)(80)v², so v at Point B would equal 26.34 m/s (#7).

    So, for the tension, I would utilize T = Fc + w. T = ([80*26.34]/20) + 800 = 905.36 N (#8).

    Now, I've caught a slight snag. To find the highest location of the rope, I thought I could use angular velocity/centripetal acceleration, but each time I run through it, the answers are either illogically large or small. I'm fairly sure there's an easier way to do it, so could someone get me back on track?

    The final number 10 is easy, but there's too many equations to type here, so I'll just disregard it.

    However, I still can't shake the feeling that something is still wrong with my calculations. Could someone perhaps check my work?
     
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