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Power of a bicycle accelerating up a slope.

  1. Jun 27, 2016 #1
    • No template because originally posted in wrong thread
    Hello everyone :alien:
    I have been assigned to create a physics problem involving bicycles. I thought up the following, but I had questions regarding if it was properly made / worked out.

    A cyclist is accelerating constantly up a slope at 1 m/s ^-2 The slope is at a 15 degree angle and is 30 m long. It takes the cyclist 1 minute to reach the end of the slope. The combined mass of the cyclist and the bicycle is 80 kg. How much power is needed to make this possible ? (Neglect air resistance and drag)

    I was thinking of using the formula: F = mg sin θ + ma to calculate the force needed to go up the slope, and then incorporating the found value of F into the formula: P = work / time = ( F x d ) / t to find the amount power needed to complete this task. The working out looks like this:

    P = ( ( mg sin θ + ma ) x d ) / t
    P = ( (( 80 x 9.81 ) sin ( 15 ) + ( 80 x 1 )) 30 ) / 60
    P = 141.6 W

    Is this correct ? Are there any variables of e.g. vectors, forces, etc. that I have forgotten to include in this calculation? or am I over complicating things ?

    Thank you very much !
     
  2. jcsd
  3. Jun 27, 2016 #2
    How do you know that after 1 minute, the bicycle only covers 30 m? I get 1800 m (almost 2 km). The work is also calculated incorrectly. The work is equal to the change in potential energy plus the change in kinetic energy. Also, the power is not constant. The work divided by the time is equal to the average power.
     
  4. Jun 27, 2016 #3
    Thank you for making me notice that ! At the time, I was just trying to make up some figures off the top of my head, but now I realise I had to use the formula:
    d = 1/2 a t^2 to figure out the distance traveled at that acceleration.
    I also appreciate the correction on the work done formula. If I'm correct, it should look like this now :

    W = Δ Ep + Δ Ek
    ____________________________________________________________________________________________

    ΔEp = mg Δh
    = ( 80 ) ( 9.81 ) Δ( length of X = 465.9 m - see right angled triangle )
    = 319915.8 J

    vtimxrajzbuard4hsj78_jpg_150×200_pixels.png
    SOH
    X = sin(15) x 1800 = 465.9 m
    ___________________________________________________________________________________________

    ΔEk = Ek f - Ek i
    = (1/2 m (v final)^2) - (1/2 m (v initial)^2)
    v initial = 0
    v final = (v initial) + at = (0) + (1)(60) = 60 m.s^-1
    = (1/2 (80) (60)^2) - (1/2 (80) (0)^2)
    = 144000 J
    ___________________________________________________________________________________________
    W = (319915.8) + (144000)
    = 463,915 .8 J

    Average Power = work / time


    Av P = ( 463,915.8) / 60
    = 7731.9 W
    ____________________________________________________________________________________

    However, upon seeing the final answer I cant help but feel like I miscalculated something ( I suspect an error when calculating the change in kinetic energy, or maybe the scenario simply isn't realistic ) because the magnitude of the answer - I feel - is too big for a cyclist.

    Thank you for your feedback !
     
  5. Jun 28, 2016 #4

    billy_joule

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    Science Advisor

    The scenario isn't realistic, cyclists don't cycle at over 200km/hr (60m/s) and can't accelerate at a constant rate. They are close enough to constant power device, like a car. A quick google says an elite cyclist can maintain ~500W over an hour, ~1,200W in short bursts.
     
  6. Jun 28, 2016 #5
    Thank you !
     
  7. Jun 28, 2016 #6

    CWatters

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    Science Advisor
    Homework Helper

    Perhaps simplify the problem by making it constant velocity rather than constant acceleration? or would that be too easy?

    Perhaps turn the problem on it's head. The max power a cyclist can generate is 300W, how fast can he go up the slope?

    How about a problem involving a unicyclist and working out the angle he must lean forwards or something like that?
     
  8. Jun 28, 2016 #7
    You can still use the problem as stated, but you have some input numbers that you can play with to make the problem more realistic: the angle, the time, the acceleration. Play around with these until you get a more realistic power value.
     
  9. Jun 28, 2016 #8
    Thats a very good idea! Thanks for helping me out.
     
  10. Jun 28, 2016 #9
    Thank you! Ill toy around with the equation a bit
     
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