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8 October, 13:15

While leaning out a window that is 6.4 m above the ground, you drop a 0.60-kg basketball to a friend at ground level. Your friend catches the ball at a height of 1.4 m above the ground. Determine the following.

(a) the amount of work done by the force of gravity on the ball

(b) the gravitational potential energy of the ball-earth system, relative to the ground when it is released

(c) the gravitational potential energy of the ball-earth system, relative to the ground when it is caught

(d) the ratio of the change (PEf - PE0) in the gravitational potential energy of the ball-earth system to the work done on the ball by the force of gravity

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  1. 8 October, 15:18
    0
    Given that,

    Height of window (h) = 6.4m

    Mass of basketball M=0.6kg

    If the ball is cough at a height of 1.4m

    Let g=9.81m/s²

    a. Work done by gravity is given by

    weight = mg

    Work done by the weight of the body is weight of the body times height of the body from the ground

    Wg=w•h

    Wg=mgh

    The work is done to where the ball was caught is height i. e h=6.4-1.4

    h=5m

    Wg=0.6*9.81*5

    Wg=29.43J

    b. Gravitation potential energy relative to the ground is give as

    Wp=mgh

    This is the potential energy over the whole height i. e h=6.4m

    Wp=0.6*9.81*6.4

    Wp=37.67J

    c. Gravitational potential energy relative to when the ball was caught

    The height when the ball was caught is 1.4m

    Therefore, potential energy is given as

    Wp=mgh

    Wp=0.6*9.81*1.4

    Wp=8.24J

    d. Ratio of change in potential energy to the work done

    ∆P. E = P. E (final) - P. E (initial)

    ∆P. E = 8.24-37.67

    ∆P. E = - 29.43

    The ratio will be ∆P. E/Wg

    Ratio = - 29.43/29.43

    Ratio = - 1

    Which is right

    Because work is conservative and

    W=-∆U

    Then

    W/∆U=-1
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