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Power output of cyclist up slope at costant speed

  1. Sep 25, 2011 #1
    1. The problem statement, all variables and given/known data

    A bicyclist coasts down a 7.0 degree slope at a steady speed of 5.0 m/s. Assuming a total mass of 75 kg (bicycle plus rider), what must the cyclist's power output be to pedal up the same slope at the same speed?

    2. Relevant equations
    [tex]P=Fvcos\theta[/tex]
    [tex]K=\frac{1}{2}mv^2[/tex]
    [tex]F_g=mg[/tex]



    3. The attempt at a solution

    I need to find the net force but since there is presumably no acceleration (constant velocity of 5.0 m/s) the only force is the force to overcome gravity and maintain a constant speed of 5.0 m/s. Correct?

    I'm missing a formula somewhere I think. How do I find the force? Once I have that I can determine the Power using the first formula.
     
  2. jcsd
  3. Sep 25, 2011 #2

    Borek

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    Staff: Mentor

    Don't worry about force. When he climbs his potential energy changes.
     
  4. Sep 25, 2011 #3
    Potential energy isn't covered until the next chapter. Also, I don't know the height, so how would I find PE?
     
  5. Sep 25, 2011 #4

    Borek

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    Staff: Mentor

    Speed and angle give you height change per second.
     
  6. Sep 25, 2011 #5
    What about

    [tex]F_biker=mg\sin\theta => (75)(9.8)(\sin7)=9.0*10^1[/tex]

    [tex]P=(90)(5.0)(\cos7)=4.47*10^2 => 4.5*10^2 = 450W[/tex]
     
  7. Sep 25, 2011 #6
    Why do you have cos(7)? It doesn't change the answer much but I don't think it should be there.
     
  8. Sep 25, 2011 #7
    I thought you needed it because theta is the angle between the force vector and the velocity vector.

    No?

    It changes the answer a lot without it. 90*50 = 4.5*10^3 I got 4.5*10^2
     
  9. Sep 26, 2011 #8

    Borek

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    Staff: Mentor

    Check your math - cos(7) is 0.993, it can't change the answer tenfold. You have already taken care of the angle calculating vertical component of the force, assuming no friction vertical component is the only one doing work.
     
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