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

AI Thread Summary
The discussion revolves around calculating the work done by a resistive force during a round trip on a bicycle. The bicyclist rides 5.0 km east and then returns 5.0 km west, with a resistive force of 3.0 N acting in opposite directions on each leg of the trip. The initial calculation mistakenly used kilometers instead of meters, leading to an incorrect answer of -30 J instead of the correct -30000 J. The key takeaway is that work is calculated in Joules, which requires distances to be in meters. The confusion highlights the importance of using the correct units in physics calculations.
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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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