We wrap a light, nonstretching cable around a 10.0 kg kg solid cylinder with diameter of 38.0 cm cm . The cylinder rotates with negligible f

Question

We wrap a light, nonstretching cable around a 10.0 kg kg solid cylinder with diameter of 38.0 cm cm . The cylinder rotates with negligible friction about a stationary horizontal axis. We tie the free end of the cable to a 14.0 kg kg block and release the block from rest. As the block falls, the cable unwinds without stretching or slipping.How far will the mass have to descend to give the cylinder 510 J of kinetic energy?

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Thu Cúc 5 years 2021-08-13T21:52:35+00:00 1 Answers 29 views 0

Answers ( )

  1. niczorrrr
    0
    2021-08-13T21:54:09+00:00
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    Answer: 14.16

    Explanation:

    Given

    d = 38cm

    r = d/2 = 38/2 = 19cm = 0.19m

    K.E = 510J

    m = 10kg

    I = 1/2mr²

    I = 1/2*10*0.19²

    I = 0.18kgm²

    When it has 510J of Kinetic Energy then,

    510J = 1/2Iω²

    ω² = 1020/I

    ω² = 1020/0.18

    ω² = 5666.67

    ω = √5666.67 = 75.28 rad/s

    Velocity is the block, v = ωr

    V = 75.28 * 0.19

    V = 14.30m/s

    The “effective mass” M of the system is

    M = (14.0 + ½*10.0) kg = 19.0 kg

    The motive force would be

    F = ma

    F = 14 * 9.8

    F = 137.2N

    so that the acceleration would be

    a = F/m

    a = 137.2/19

    a = 7.22m/s²

    Finally, using equation of motion.

    V² = u² + 2as

    14.3² = 0 + 2*7.22*s

    204.49 = 14.44s

    s = 204.49/14.44

    s = 14.16m

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