Ask Question
18 May, 15:30

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?

+3
Answers (1)
  1. 18 May, 15:54
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “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 ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers