How do nonconservative forces affect work done in a round trip?

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SUMMARY

The discussion centers on calculating the work done by a resistive force during a round trip made by a bicyclist. The bicyclist travels 5.0 km east and then returns 5.0 km west, encountering a resistive force of 3.0 N in opposite directions. The correct calculation for work done is based on the definition of a Joule, which is the work done by a force of 1 Newton over a distance of 1 meter. The total work done by the resistive force during the round trip is -30000 J, as the distance must be converted from kilometers to meters.

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  • Knowledge of unit conversions, specifically between kilometers and meters
  • Familiarity with the concept of resistive forces
  • Basic proficiency in calculating work using the formula W = Fd
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  • Study the work-energy theorem in physics
  • Learn about unit conversions in physics, focusing on Joules and Newtons
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Hi!

I've been needing help with a question in my textbook. It seems so stupidly easy, but I can't get it. (Ironically enough, I understood all the others).. here's the question:

A bicyclist rides 5.0km due east, while the resistive foce from the air has a magnitude of 3.0N and points due west. The rider then turns around and rides 5.0km due west, back to her starting point. The resistive force from the air on the return trip has a magnitude of 3.0 N and points due east. Find the work done by the resisitve force during the round trip.

Here's what I did:
W = Fd
W = (3.0N)(5.0km) = -15J
since its a round trip, I doubled it and get -30J, but the book says that the answer is -30000J

Any help would be greatly appreciated!
 
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You have a "units" problem. A "Joule" is defined as the work done by a force of 1 Newton acting over a distance of 1 meter- not 1 kilometer.

How much work is done by a force of 3 N acting over a (total) distance of 1000 m?
 
oh.. a UNITS problem.. I feel very stupid.. sorry for bothering you with that, but thanks! :)
 

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