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Work problem

  1. Mar 13, 2010 #1
    1. The problem statement, all variables and given/known data
    67. How fast must a cyclist climb a 6 degrees hill to maintain a power output of 0.25 horse power? Neglect work done by friction and assume the mass of cyclist plus bicycle is 70 kg.

    A. 2.6 m/s

    2. Relevant equations
    W = mgh = Fdsin[tex]\theta[/tex]= [tex]\frac{1}{2}[/tex]mv[tex]^{2}[/tex]


    3. The attempt at a solution
    (0.25 hp * 735.5 W * 3600 J) = 70 * 9.8 * h * sin(6) = [tex]\frac{1}{2}[/tex]*70v[tex]^{2}[/tex]

    661950=71.707h =35v[tex]^{2}[/tex]

    I don't get how the answer should be only 2.6 m/s.
     
    Last edited: Mar 14, 2010
  2. jcsd
  3. Mar 13, 2010 #2
    I think your equations may be confusing you.

    For instance what is h? h or (d) should be the distance that the cyclist is traveling.
    so h=d sin (6) make sense?

    now [tex] W = -\Delta U[/tex] so if you call your initial position h=0 your final potential is mgh

    now can you use d to solve for velocity?
     
  4. Mar 14, 2010 #3
    but how do I find the velocity to maintain 0.25 hp?

    661950 = 35v[tex]^{2}[/tex]
    v[tex]^{2}[/tex] = 18912
    v = 137.524 m/s

    this is not what I should get. I need to get 2.6 m/s.
     
  5. Mar 14, 2010 #4
    This may help you out

    [tex]P=\frac{dW}{dt}=\frac{F\cdot ds}{dt} = F \cdot v[/tex]

    So you should figure out the force he needs to cancel out gravity, and from there get the velocity. Be careful of units, and it should be a straight shot.
     
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