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Math Help - Some Questions concerning Work, Energy and motion...

  1. #1
    PFX
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    Some Questions concerning Work, Energy and motion...

    1. A cyclist starts from rest and freewheels down a hill inclined at arcsin 1/20 to the horizontal. After traveling 80 m the road becomes horizontal and the cyclist travels another 80 m before coming to rest without using his brakes. Given that the combined mass of the cyclist and his machine is 80 kg, model the cyclist and his machine as a single particle and hence find the resistive force, assumed constant throughout the motion.

    2. A boy and his skateboard have a mass of 50 kg. He descends a slope inclined at 12(degrees) to the horizontal from rest. At the bottom, the ground becomes horizontal for 10 m before rising at 8(degrees) to the horizontal. The boy travels 30 m up the incline before coming to rest again. He is subject to a constant resistance of 20 N throughout the motion. By modeling the boy and his skateboard as a single particle find the distance the boy traveled down the slope.

    3. A book slides down a rough desk lid which is inclined at 30(degrees) to the horizontal and falls to the ground which is 0.8 m below the edge of the lid. The coefficient of friction between the book and the lid is 1/4. Given that the book starts from rest 0.4 m from the bottom of the lid which is hinged is at the edge of the desk, find the speed with which the book hits the floor.

    4. A rough plane is inclined at arcsin 3/5 to the horizontal. A package of mass 40 kg is released from rest and travels 15 m while increasing speed to 12 m/s. Find, using energy considerations, the coefficient of friction between the package and the plane.


    Im revising for my Mechanics 2 A2 Level exam so ill be really grateful.... Thankyou....
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    Quote Originally Posted by PFX View Post
    1. A cyclist starts from rest and freewheels down a hill inclined at arcsin 1/20 to the horizontal. After traveling 80 m the road becomes horizontal and the cyclist travels another 80 m before coming to rest without using his brakes. Given that the combined mass of the cyclist and his machine is 80 kg, model the cyclist and his machine as a single particle and hence find the resistive force, assumed constant throughout the motion.
    These are all basically on the same concept so I'll help you out with this one and ask you to try the rest again. If you still need help, please let us know in this thread.

    The weird thing about this problem is the constant frictional force. Typically we would have a constant coefficient of friction. Strange.

    The picture I have is that the slope is angling downward to the right. I am choosing a coordinate system for this part to have +x down the slope and +y perpendicular (and upward) to the slope. The Free Body Diagram shows three forces on the cyclist: A normal force (n) in the +y direction, a weight (w) acting straight down, and a kinetic friction force (f) acting up the slope.

    On the horizontal surface we have a similar situation. I am taking +x to the right and +y upward. The FBD has a normal force (N) acting in the +y direction, a friction force (f) the same magnitude as before in the -x direction, and a weight (w) acting straight downward. (Note that the normal force here is different than the normal force on the slope.)

    It will be simpler here to use the work-energy theorem rather than Newton's 2nd Law. (Typically we would use the latter approach.)
    W_{nc} = E - E_0

    Picking the starting point to be at the top of the slope and the ending point to be where the cyclist stops is probably the easiest way since we don't care what happens in between. I am choosing the 0 point for the gravitational potential energy to be at the level where the cyclist is on the horizontal part of the motion.

    The friction is constant for both parts so the nonconservative work done by it is simply
    W_{nc} = -(80~m + 80~m)f = -160f

    Thus
    -160f = \left ( \frac{1}{2}mv^2 + mgh \right) - \left ( \frac{1}{2}mv_0^2 + mgh_0 \right )

    -160f = -mgh_0 = -mg(80)sin(asn(1/20))

    You can do the rest from here.

    -Dan
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  3. #3
    PFX
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    @ topsquark

    thanx !!! ur answer really helped me a lot. i was worried cuz xam is in a couple of days and i couldnt figure this out...

    one last thing. could u help me out with the second problem. i solved the third and the fourth ones but cant seem to solve the second one even though its fairly similar to the first one ...
    Last edited by PFX; May 19th 2008 at 03:13 AM.
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