Finding the coefficient of friction

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SUMMARY

The coefficient of friction between the race-car driver's tires and the road can be determined using the acceleration of 4.0 m/s² without spinning the tires. The correct approach involves recognizing that static friction applies until the tires begin to slip. The calculation shows that the static coefficient of friction (u_s) is 0.41, which aligns with the derived formula u_s = a/g, where g is the acceleration due to gravity. The confusion arises from the distinction between static and kinetic friction, emphasizing the importance of understanding these concepts in physics.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Knowledge of static and kinetic friction forces
  • Familiarity with basic physics equations involving acceleration and force
  • Concept of gravitational acceleration (g = 9.81 m/s²)
NEXT STEPS
  • Study the differences between static and kinetic friction in detail
  • Learn how to apply Newton's laws to various physical scenarios
  • Explore real-world applications of friction in automotive engineering
  • Investigate how tire materials affect the coefficient of friction
USEFUL FOR

Students studying physics, automotive engineers, and anyone interested in the mechanics of friction and acceleration in vehicles.

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Homework Statement



A race-car driver discovers that he can accelerate at 4.0 m/s2 without spinning his tires, but if he tries to accelerate more rapidly, he always "burns rubber." What is the coefficient of friction between his tires and the road?

Answer behind the book:
0.62

Homework Equations



Kinetic friction force = ukN
Static friction force <= usN

The Attempt at a Solution



I'm really lost with this one. I tried starting with Newton's second law and end up at:
ukmg = ma
ukg = a
uk = a/g = 0.41

Which is not the answer behind the book. Is this considered static friction? I'd think not because the tires are slipping relative to the ground. Anyone can put me on the right path?
 
Physics news on Phys.org
It is static friction up to the point where the tires start to slip. So I get u_s = 0.41. I don't understand the given answer, unless some info is missing.
 

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