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31 August, 21:50

A rubber ball with mass 0.20 kg is dropped vertically from a height of 1.5 m above a floor. The ball bounces off of the floor, and during the bounce 0.60 J of energy is dissipated. What is the maximum height of the ball after the bounce

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  1. 31 August, 22:15
    0
    The maximum height of the ball after the bounce is 1.2 m

    Explanation:

    Potential Energy = mass * Height * acceleration of gravity

    PE=mgh

    = 0.2 x 9.8 x 1.5

    P. E = 2.94 J

    During bounce of ball 0.60 J of energy is lost. So

    2.94 - 0.6 = 2.34 J

    now new energy is 2.34

    New P. E = mgh

    2.34 = 0.2 x 9.8 x h

    h = 2.34 / 0.2 x 9.8

    h = 1.2 m

    The maximum height of the ball after the bounce is 1.2 m
  2. 31 August, 22:16
    0
    Given:

    Mass = 0.2 kg

    Height = 1.5 m

    Potential energy of the drop, PE = m * g * h

    = 0.2 x 9.81 x 1.5

    = 2.94 J

    After the drop, 0.6 J of energy is dissipated, so amount of energy left

    = 2.94 - 0.6

    = 2.34 J

    This energy is eqal to thenew potential energy which is:

    m * g * h = 2.34

    0.2 * 9.81 * h = 2.34

    = 2.34/1.962

    = 1.19 m.
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