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Potential Energy Questions - I'm lost, help!

  1. Mar 7, 2006 #1
    1. A daredevil on a motorcycle leaves the end of a ramp with a speed of 35 m/s. If his speed is 33 m/s when he reaches the peak of the path, what is the maximum height that he reaches? Ignore friction and air resistance.

    I don't know how to approach this problem or set it up? Do I use vectors?

    2. Three objects with masses m1 = 5.0 kg, m2 = 10 kg, and m3 = 15 kg, respectively, are attached by strings over frictionless pulleys. The horizontal surface is frictionless and the system is released from rest. Using energy concepts, find the speed of m3 after it moves down a distance of 4.0 m.

    There is a diagram showing m2 in the middle of the table and m1 & m3 hanging off of opposite ends.
  2. jcsd
  3. Mar 7, 2006 #2


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    1. When he leaves the ramp he has KE only at the peak of his motion he has KE and PE. The total energy must remain constant.
  4. Mar 7, 2006 #3
    What about the vectors? Do I set this up with X and Y components? I understand and agree with what you said but I'm lost.
  5. Mar 7, 2006 #4


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    Potential Energies are scalars (single numbers), not vectors.
    You have to know the functional forms ("formulas"), but no component work.
    (Most of us prefer Energy approaches, because they're MUCH easier to use!)

    gravitational PE formula is m g h , so you need a vertical coordinate.
    ... KE formula is (1/2) m v^2
  6. Mar 8, 2006 #5


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    1. Since the daredevil is subjected to gravity alone we know that his total mechanical energy will be conserved. That is
    [tex]E\ =\ K\ +\ U\ =\ constant[/tex]
    [tex]E_{bottom}\ =\ E_{top}[/tex]

    2. Here we have the conservation of mechanical energy of the system of masses again, since the only force doing work along the line of motion of the masses is gravity. since the masses are connected together they will all move with the same speed. Another way of expressing
    [tex]E\ =\ K\ +\ U\ =\ constant[/tex]
    is to say that the change in [itex]E[/itex] of the system at two different stages will be zero. That is
    [tex]E_2\ -\ E_1\ =\ 0[/tex]
    which can be written as
    which after a little bit of arrangement comes to
    [tex]\Delta K\ +\ \Delta U\ =\ 0[/tex]
    that is the change in kinetic energy of the system plus the change in potential energy of the system will be zero. Or stated another way, what the system loses in potential energy it gains in kinetic energy or vice versa.
    Last edited: Mar 8, 2006
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