Question

A shuffleboard disk is accelerated to a speed of 5.8 m/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?

Answer 1

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

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