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I am Lyosha [343]
3 years ago
15

A shuffleboard disk is accelerated to a speed of 5.8 m/s and released. If the coefficient of kinetic friction between the disk a

nd the concrete court is 0.31, how far does the disk go before it comes to a stop?
Physics
1 answer:
MatroZZZ [7]3 years ago
3 0

Answer:

5.53 m

Explanation:

v_{o} = initial speed of  shuffleboard disk = 5.8 ms^{-1}

v_{f} = final speed of  shuffleboard disk = 0 ms^{-1}

\mu = Coefficient of kinetic friction between the disk and concrete court = 0.31

acceleration due to friction is given as

a = - \mu g\\a = - (0.31) (9.8)\\a = - 3.04 ms^{-2}

d = distance traveled by disk before it stops

using the kinematics equation that fits the above the data, we have

v_{f}^{2} = v_{o}^{2} + 2 a d\\0^{2} = 5.8^{2} + 2 (- 3.04) d\\d = 5.53 m

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Explanation:

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Substituting these values into equation 1

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Substituting the values of a and b into equation 2

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Answer:

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