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1 March, 09:18

A hoop (I = MR2) of mass 2.0 kg and radius 0.50 m is rolling at a center-of-mass speed of 15 m/s. An external force does 750 J of work on the hoop. What is the new speed of the center of mass of the hoop?

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  1. 1 March, 09:42
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    Answer: 24.49 m/s

    Explanation:

    Given

    Mass of hoop, m = 2 kg

    Radius of hoop, r = 0.5 m

    Speed of rolling, v = 15 m/s

    Work done, = 750 J

    If we assume that the hoop is not skidding, then it's Kinetic Energy of rotation is 1/2mv²

    also the Kinetic Energy of translation is 1/2mv², thus the total Kinetic Energy is

    KE = KE (r) + KE (t)

    KE = 1/2mv² + 1/2mv²

    KE = mv²

    Recall, the change in kinetic energy = work done = 750 J

    mv² - mv•² = 750

    m (v² - v•²) = 750

    v² - v•² = 750/m

    v² - 15² = 750/2

    v² = 375 + 225

    v² = 600

    v = 24.49 m/s

    Therefore, the new speed is 24.49 m/s
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