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

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A shuffleboard disk is released at a speed of 5.8 m/s with a coefficient of kinetic friction of 0.31. The problem involves calculating the distance the disk travels before stopping on a 15.8 m court. The equation vf² = vi² + 2aΔd is suggested for solving the problem, 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 discuss substituting the known values into the equation to find Δd. The conversation emphasizes understanding the variables and applying the correct physics principles to solve the problem.
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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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