Calculating Net Force on a Cyclist: Newton's Law Q Explained

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The discussion revolves around calculating the net force acting on a cyclist who is pedaling against opposing forces such as friction and air resistance. The subject area pertains to Newton's laws of motion and force analysis.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the calculation of net force by considering the forward force exerted by the cyclist and the opposing forces of friction and air resistance. There are differing opinions on the resulting net force, with some suggesting it could be 200N or 0N.

Discussion Status

The discussion includes various interpretations of the net force calculation. Some participants provide insights into the directionality of forces and the concept of net force, while others express confusion about the correct approach. Guidance has been offered regarding the addition of forces with consideration of their signs.

Contextual Notes

Participants question the implications of the cyclist's speed and the definitions of opposing forces in the context of net force calculations. There is an acknowledgment of the cyclist's ongoing motion and the conditions under which that motion would change.

pencilcase
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A cyclist is pedaling so that a force of 100N pushes her forward while there are 50N of friction and 50 N of air resistance opposing her movement. What is the net force on the cyclist?
 
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pencilcase said:
A cyclist is pedaling so that a force of 100N pushes her forward while there are 50N of friction and 50 N of air resistance opposing her movement. What is the net force on the cyclist?

Sounds pretty easy to me. What do you think? That's the important part.
 
i think its either 200n or 0n :/
 
pencilcase said:
i think its either 200n or 0n :/

That's good. So which? 100N forward. I think 'opposing her movement' should tell you which direction the other two are, yes? 'Net force' means you add the forces considering the sign of each one.
 
so its 200 n
 
pencilcase said:
So there is no net force on her?

Right. If she's moving 20km/hr she'll keep on moving 20km/hr until she stops working and slows down or works harder and speeds up.
 
pencilcase said:
so its 200 n

I liked your first answer better. If you call forward positive then the other backward forces are negative. 100N+(-50N)+(-50N)=0N.
 
ooooooh ok thanksss!
 

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