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30 May, 06:03

A 30.0 kg wheel, essentially a thin hoop with radius 0.620 m, is rotating at 202 rev/min. It must be brought to a stop in 26.0 s. (a) How much work must be done to stop it? (b) What is the required average power? Give absolute values for both parts.

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  1. 30 May, 08:01
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    (a) 2579.99 J.

    (b) 99.23 W

    Explanation:

    (a)

    The work done by the wheel = Kinetic energy of the wheel.

    W = 1/2mv² ... Equation 1

    Where W = work done, m = mass of the wheel, v = velocity of the wheel.

    But,

    v = rω ... Equation 2

    Where r = radius of the wheel, ω = angular velocity

    Substitute equation 2 into equation 1

    W = 1/2mr²ω² ... Equation 3

    Given: m = 30 kg, r = 0.620 m, ω = 202 rev/min = (202*0.10472) rad/s = 21.153 rad/s

    Substitute into equation 3

    W = 1/2 (30) (0.62²) (21.153²)

    W = 2579.99 J.

    (b)

    Power = work done/time

    P = W/t ... Equation 4

    where P = power, W = work done, t = time

    Given: W = 2579.99 J, t = 26 s

    Substitute into equation 4

    P = 2579.99/26

    P = 99.23 W.
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