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9 December, 09:38

A time-varying net force acting on a 2.6 kg particle causes the object to have a displacement given by x = a + b t + d t2 + e t3, where a = 2 m, b = 1.4 m/s, d = - 1.8 m/s 2, and e = 1.2 m/s 3, with x in meters and t in seconds. Find the work done on the particle in the first 3.3 s of motion. Answer in units of J.

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  1. 9 December, 11:47
    0
    W = 1579.94J

    Explanation:

    x = a + bt + dt² + et³

    First, we find acceleration:

    v = dx/dt = b + 2dt + 3et²

    a = dv/dt = 2d + 6et

    d = - 1.8 m/s², e = 1.2 m/s³

    a = 2 * (-1.8) + 6*1.2*t

    a = - 3.6 + 7.2t

    Force, F is given as:

    F = m * a

    F = 2.6 * (-3.6 + 7.2t)

    F = - 9.36 + 18.72t

    When t = 3.3 secs:

    F = - 9.36 + (18.72*3.3)

    F = 52.42N

    x = a + bt + dt² + et³

    Inputting values of a, b, d, e:

    x = 2 + (1.4*3.3) + (-1.8*3.3²) + (1.2*3.3³)

    x = 2 + 4.62 - 19.602 + 43.1244

    x = 30.1424m

    Therefore, Work done, W is

    W = F * x

    W = 52.42 * 30.1424

    W = 1579.94J
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