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16 October, 20:29

A 50g mass is placed on a straight air track sloping at an angle of 45° to the horizontal. Calculate in metre per second, the acceleration of the load as it slides down and also the distance it will be moved from rest in two seconds

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  1. 16 October, 20:39
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    a = 0.347 m/s^2 and S = 0.694 m

    Explanation:

    on a slope which is friction less acceleration is m g sin Ф whereФ is angle of inclined plane with horizontal

    here m = 50g = 0.050 kg, g = 9.81 m/s^2, Ф = 45°

    a = 0.050 * 9.81 * sin45°

    a = 0.347 m/s^2

    now

    S = Ut + (1/2) at^2

    where S is displacement U is initial velocity, a is acceleration and t is time

    S = (0) (2) + (1/2) (0.347) (2) ^2

    S = 0 + (0.5) (0.347) (4)

    S = 0.694 m
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