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11 October, 20:47

A 7.0kg ball is given an initial velocity of 14m/s and then is rolled along the floor. A friction force of 2.5N opposes its motion. For what length of time will it roll before stopping.

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  1. 11 October, 21:40
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    A=F/m

    =2.5/7.0

    =0.36 m/s^2

    The ball eventually stops, so the final velocity is 0 m/s.

    t = (v-u) / a

    = (14-0) / 0.36

    =38.9 s
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