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24 October, 11:33

An initially stationary 4.3 kg object accelerates horizontally and uniformly to a speed of 11 m/s in 3.4 s. (a) In that 3.4 s interval, how much work is done on the object by the force accelerating it

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  1. 24 October, 12:17
    0
    Answer: W=260.174J

    Explanation: since the object is stationary, it initial velocity U = 0

    final velocity V = 11 m/s, time t = 3.4s, distance S = ?, acceleration a = ? work done W = ? force F = ?

    W = FS

    a = v-u/t = 11-0/3.4 = 3.2353m/s^2

    to calculate the distance, let look at one of the equations of motion

    V^2=U^2+2as hence s = V^2-U^2/2a = 11*11/2*3.235 = 18.7017m

    But force F = MA (mass*acceleration)

    F = 4.3*3.2353 = 13.91179N

    therefore work done W = 13.91179*18.7017 = 260.174J
  2. 24 October, 14:37
    0
    The work done on the object by the force accelerating it is 520.31 J

    Explanation:

    Given;

    mass of an object = 4.3 kg

    horizontal velocity of the object, v = 11 m/s

    time of acceleration, t = 3.4 s

    work done is given as the product of force and distance

    Work done = Fd

    horizontal distance traveled by the object within 3.4 s, is calculated as follows;

    X = Vt + ¹/₂gt², gravity has little or no influence on horizontal displacement, thus g = 0

    X = Vt

    X = 11*3.4 = 37.4 m

    Force on the object, F = ma = m (v/t) = 4.3 (11/3.4) = 13.912 N

    work done = Fd = 13.912 x 37.4 = 520.31 J

    Therefore, the work done on the object by the force accelerating it is 520.31 J
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