Question about coefficient of kinetic friction

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SUMMARY

The discussion centers on calculating the coefficient of kinetic friction between a hockey puck and ice, given an initial speed of 5.0 m/s and a stopping distance of 20 m. The user identifies the equation for kinetic friction, f(sub k) = (mu)(normal force), and realizes that the mass of the puck is not necessary for the calculation. By applying Newton's Second Law, F = ma, the user concludes that equating the friction force to the deceleration allows for the determination of the coefficient of kinetic friction.

PREREQUISITES
  • Understanding of Newton's Second Law (F = ma)
  • Familiarity with the concept of kinetic friction
  • Basic knowledge of motion equations
  • Ability to manipulate algebraic equations
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  • Study the derivation of the coefficient of kinetic friction from motion equations
  • Learn how to apply Newton's laws in friction scenarios
  • Explore the relationship between mass, force, and acceleration in physics
  • Investigate real-world applications of kinetic friction in sports physics
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Physics students, educators, and anyone interested in understanding the principles of motion and friction in sports contexts.

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A hockey player hits a puck with her stick, giving the puck an initial speed of 5.0 m/s. If the puck slows uniformly and comes to rest in a distance of 20 m, what is the coefficient of kinetic friction between the ice and the puck?

I know that this question must be rather easy, and that my mind has missed some logical leap in order to complete it.

Given the equation for kinetic friction (I can't format, so I'm not sure how to write this...)

f(sub k)=(mu)(normal force)

How can I solve for the coefficient (f sub k) without the mass? It seems that without the mass I am stuck.
 
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You have a force (friction) you need to find the acceleration. What happens when you equate your friction force to Newton's Third, F=ma?
 
Wonderful things happen.

I wonder why I didn't think of that. -blush-

Thank you!
 

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