Coeffiecient of Kinetic Friction problem? (Shuffleboard and Distance traveled)

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SUMMARY

The discussion centers on calculating the distance a shuffleboard disk travels before stopping, given an initial speed of 5.8 m/s and a coefficient of kinetic friction of 0.31. The relevant kinematic equation used is vf² = vi² + 2aΔd, where vf is the final velocity (0 m/s), vi is the initial velocity (5.8 m/s), and a is the acceleration due to friction. Participants clarify the correct application of the equation to find the stopping distance, emphasizing the importance of substituting known values accurately.

PREREQUISITES
  • Understanding of kinematic equations, specifically vf² = vi² + 2aΔd
  • Knowledge of the coefficient of kinetic friction and its role in motion
  • Basic algebra skills for solving equations
  • Familiarity with concepts of initial and final velocity
NEXT STEPS
  • Calculate acceleration using the formula a = -μk * g, where g is the acceleration due to gravity
  • Practice similar problems involving kinetic friction and motion
  • Explore the effects of different coefficients of friction on stopping distances
  • Learn about energy conservation in relation to friction and motion
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and motion, as well as educators looking for practical examples of kinematic equations in real-world scenarios.

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Coeffiecient of Kinetic Friction problem? (Shuffleboard and Distance traveled)

Homework Statement



A shuffleboard disk is accelerated to a speed of 5.8m/s and released. If the coefficient of kinetic friction between the disk and the concrete court is 0.31, how far does the disk go before it comes to a stop? The courts are 15.8m long.

Known info:
vi = 5.8m/s
vf = 0m/s
\muk = 0.31


Homework Equations



I have no idea!

The Attempt at a Solution



So, I originally tried vf2 = vi2+ 2a\Deltad, but that would make no sense because I used 15.8 as the distance but then the problem would be solved...
 
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Welcome to PF!

glindawantsme said:
A shuffleboard disk is accelerated to a speed of 5.8m/s and released. If the coefficient of kinetic friction between the disk and the concrete court is 0.31, how far does the disk go before it comes to a stop? The courts are 15.8m long.
…
So, I originally tried vf2 = vi2+ 2a\Deltad, but that would make no sense because I used 15.8 as the distance but then the problem would be solved...

Hi glindawantsme! Welcome to PF! :smile:

(have a delta: ∆ :smile:)

Yes, that is the right equation …

So what is vf? what is vi? and what is a?

Put them all in, and find ∆d. :smile:
 

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