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22 July, 00:45

A woman is driving her truck with speed 45.0 mi/h on a horizontal stretch of road. (a) When the road is wet, the coefficient of static friction between the road and the tires is 0.105. Find the minimum stopping distance (in m). m (b) When the road is dry, μs = 0.602. Find the minimum stopping distance (in m).

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  1. 22 July, 01:49
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    45 mi / h = 45 x 1.6 x 1000 / (60 x 60) m / s

    = 20 m / s

    Maximum frictional force possible on wet road

    = μs x mg where μs is coefficient of static friction and m is mass of body

    Applying work energy theorem

    work done by friction = kinetic energy of truck

    μs x mg x d = 1/2 m v ² where v is velocity of body and d is stopping distance

    2xμs x g x d = v²

    2 x. 105 x 9.8 x d = 20 x 20

    d = 194.36 m

    b)

    In this case

    μs = 0.602

    Inserting this value in the relation above

    2xμs x g x d = v²

    2 x. 602 x 9.8 x d = 20 x 20

    d = 33.9 m.
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