Asked by Karl
A shuffleboard disk is accelerated to a speed of 5.6 m/s and released. If the coefficient of kinetic friction between the disk and the concrete court is 0.29, how far does the disk go before it comes to a stop? The courts are 15.7 m long.
Answers
Answered by
Scott
the frictional force is __ .29 g
the time to stop is __ 5.6 / (.29 g)
the average velocity is __ (5.6 + 0) / 2
the distance traveled is the average velocity divided by the time
the time to stop is __ 5.6 / (.29 g)
the average velocity is __ (5.6 + 0) / 2
the distance traveled is the average velocity divided by the time
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