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11 August, 20:09

A 12.0-g plastic ball is dropped from a height of 2.50 m. Just as it strikes the floor, it is moving at a speed of 3.20 m/s. How much mechanical energy did the ball lose during its fall?

0.233 J

0.256 J

0.128 J

0.512 J

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  1. 11 August, 20:34
    0
    0·233 J

    Explanation:

    Given

    Mass of the ball = 0·012 kg

    Initially the ball is at a height of 2·5 m

    As initially the ball is dropped, it's initial velocity will be equal to 0

    Therefore initially it has zero kinetic energy and has only potential energy

    ∴ Initially total mechanical energy of the ball = potential energy of the ball

    Initial potential energy of the ball = m * g * h

    where

    m is the mass of the ball

    g is the acceleration due to gravity

    h is the height of the ball

    ∴ Potential energy = 0·012 * 9·8 * 2·5 = 0·294 J

    Velocity of the ball after striking the floor = 3·2 m/s

    After striking the floor, the total mechanical energy = kinetic energy just after striking the floor

    Kinetic energy = 0·5 * m * v²

    where m is the mass of the ball

    v is the velocity of the ball

    ∴ Kinetic energy of the ball = 0·5 * 0·012 * 3·2² = 0·061 J

    Mechanical energy that is lost = 0·294 - 0·061 = 0·233 J

    ∴ Mechanical energy that the ball lost during its fall = 0·233 J
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