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28 March, 02:50

Two forces F1 and F2 act on a 5.00 kg object. Taking F1=20.0N and F2=15.00N, find the acceleration of the object for the configurations:

a) mass m with F1 acting in the positive x direction and F2 acting perpendicular in the positive y direction

b) mass m with F1 acting in the positive x direction and F2 acting on the object at 60 degrees above the horizontal.

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  1. 28 March, 02:53
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    A) mass m with F1 acting in the positive x direction and F2 acting perpendicular in the positive y direction

    m = 5.00 kg

    F1=20.0N ... x direction

    F2=15.00N ... y direction

    Net force ^2 = F1^2 + F2^2 = (20N) ^2 + (15n) ^2 = 625N^2 = >

    Net force = √625 = 25N

    F = m*a = > a = F/m = 25.0 N / 5.00 kg = 5 m/s^2

    Answer: 5.00 m/s^2

    b) mass m with F1 acting in the positive x direction and F2 acting on the object at 60 degrees above the horizontal.

    m = 5.00 kg

    F1=20.0N ... x direction

    F2=15.00N ... 60 degress above x direction

    Components of F2

    F2, x = F2*cos (60) = 15N / 2 = 7.5N

    F2, y = F2*sin (60) = 15N * 0.866 = 12.99 N ≈ 13 N

    Total force in x = F1 + F2, x = 20.0 N + 7.5 N = 27.5 N

    Total force in y = F2, y = 13.0 N

    Net force^2 = (27.5N) ^2 + (13.0N) ^2 = 925.25 N^2 = Net force = √ (925.25N^2) =

    = 30.42N

    a = F / m = 30.42 N / 5.00 kg = 6.08 m/s^2

    Answer: 6.08 m/s^2
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