Work Bicycle Problem: Calculate Work Done in Each Stroke

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In calculating the work done by a cyclist pedaling uphill, a downward force of 500N is applied during each stroke with a pedal diameter of 38cm. The confusion arises from the relationship between force and distance, as the cyclist initially considers using the circumference for calculations. However, the correct approach involves recognizing that work is done only during the downward stroke, where the displacement corresponds to the diameter rather than the circumference. The work done is thus calculated as Force multiplied by Diameter. Understanding this concept clarifies the calculation despite the cyclist's initial uncertainty about torque.
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Homework Statement



In pedaling a bicycle uphill, a cyclist exerts a downward force of 500N during each stroke. If the diameter of the circle traced by each pedal is 38cm, calculate how much work is done in each stroke.

Homework Equations



??

The Attempt at a Solution



I'm at a loss, I thought that I could multiply the F * circumference but that didn't work. I don't know where to start. Can anyone help me?

Thanks.
 
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Hi timficker, welcome to PF. Think "torque" and "work done by a torque". What are the relevant equations?
 
Well, I figured it out finally. We haven't come to torque yet, in a couple weeks. Although I got the answer it was only due to guessing. I don't really understand how it could be true.

But the answer came from F*diameter. I don't know how this could be because I thought the work was the force times the distance traveled which in this case would be the circumference.

Thanks for your help.
 
timficker said:
Well, I figured it out finally. We haven't come to torque yet, in a couple weeks. Although I got the answer it was only due to guessing. I don't really understand how it could be true.

But the answer came from F*diameter. I don't know how this could be because I thought the work was the force times the distance traveled which in this case would be the circumference.

Thanks for your help.
The result can be derived quite easily if you knew about torques. Since you don't, here is a different way to look at it. Work is done when you push down on the pedal, i.e. during the downward stroke only. The displacement is the diameter and the force is F down in the same direction as the displacement. Therefore the work done is Force*Diameter.

Don't forget that it is only the component of the force parallel to the displacement that does work and in this case the force is always down.
 
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