Coefficient of Friction [Needing Help]

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SUMMARY

The discussion centers on calculating the average forward force required for a skater weighing 500 Newtons to maintain a constant velocity on ice with a kinetic coefficient of friction of 0.01. The formula used is f = (Uk)(N), where f represents the frictional force, Uk is the kinetic coefficient of friction, and N is the normal force. The calculation yields a frictional force of 5 Newtons, confirming that the initial calculation is correct. The participants agree on the accuracy of the solution and the underlying principles.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the concept of friction and its coefficients
  • Basic algebra for solving equations
  • Knowledge of forces acting on objects in motion
NEXT STEPS
  • Study the principles of kinetic and static friction in physics
  • Explore real-world applications of friction in sports and engineering
  • Learn about the effects of different surfaces on frictional forces
  • Investigate advanced topics such as friction in fluid dynamics
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Students studying physics, particularly those focusing on mechanics, as well as educators and anyone interested in the practical applications of friction in sports and motion analysis.

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Homework Statement


If the kinetic coefficient of friction between ice skates and ice is 0.01, what is the average forward force necessary for a 500 Nt skater to maintain a constant forward velocity?


Homework Equations


f = (Uk)(N)


The Attempt at a Solution



f = (Uk) (N)
f = (0.01) (500 Nt)
f = 5 Nt

Is this right? Just seems too simple. Can someone explain and walk me through this problem if I'm wrong.

Also I realized from a diagram I did in my head Fapp - f = 0
Fapp = f
Fapp = (Uk) (N)...right?
 
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Your calculation to get 5 N looks correct to me!
 
Alright thanks a bunch.
 

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