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A cyclist exerts 80N of energy pedalling. How much Work is done?

  1. Oct 28, 2013 #1
    1. The problem statement, all variables and given/known data
    Cyclist pedals downward stroke with 80N of energy. The diameter of the circle traced by his pedals is 36 cm, how much work is done each stroke?

    2. Relevant equations

    W=F•d

    3. The attempt at a solution

    I figured that the force exerted downwards means half of the circle traced, which would mean that it would be 80 N per 18 centimetres. It was not so. does centripetal stuff have anything to do with this?
     
  2. jcsd
  3. Oct 28, 2013 #2

    rock.freak667

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    Homework Helper

    The downward thing is just describing the stroke so I believe the problem is telling you that with one stroke, the rider produces 80 N of force forward, so if the diameter is 36 cm, the linear distance in one stroke would be 1 circumference.
     
  4. Oct 28, 2013 #3

    NascentOxygen

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    Staff: Mentor

    If you have ridden a conventional bicycle, you'll have noticed how while standing on the pedals you can force your foot DOWNWARDS for almost the whole depth of the circle traced by the pedals. You stand vertically, so your leg is exerting the main component force vertically downwards. (You do need to exert a small force tangentially to encourage the pedals to continue turning at the top & bottom of the path, but that tangential force is slight and does insignificant work in relation to that of the main downwards thrust.)

    Work done = force • distance through which that force acts

    (I've retained that dot you used, as it is very appropriate.)
     
  5. Oct 29, 2013 #4
    I thank thee profusely, and I am honoured that my dot is appropriate. I bashed random buttons on my keyboard till it came out.
     
  6. Oct 29, 2013 #5
    You can use the equation for work: W=F * D where D is the distance.

    Substitute for D: D = circumference = pi * d, where d is the 0.36 meter diameter. Now the rest of the calculations should be self explanatory.
     
  7. Oct 29, 2013 #6

    NascentOxygen

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    Staff: Mentor

    As I indicated in bold, I do not believe circumference is the applicable distance in this problem.
     
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