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1 December, 01:39

A person throws a 0.5 kg ball up in the air in 1.1 s. During this process the position of the ball changes from 0.5 m to 1.5 m (before being released) and the speed of the ball changes from 0 m/s to 7 m/s. What is the work (in J) done by the force of gravity on the ball while the ball is being thrown?

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  1. 1 December, 02:45
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    A ball is thrown up and it works against gravity

    Mass = 0.5kg

    Time 1.1sec

    Change of position ∆h=0.5 to 1.5m

    ∆h=1.5-0.5

    ∆h=1m

    Work done by gravity is given as=W*d

    W=mg, let g, = 9.8m/s²

    W=0.5*9.8

    W=4.9N

    Then the upward motion

    Initial velocity 7m/s

    Final velocity 0m/s

    So this final velocity will be the initial velocity of the upward motion

    We need to get the height, the ball traveled

    S=ut-1/2gt² motion against gravity

    u=7m/s

    S=7*1.1-0.5*9.8*1.1²

    S=1.771m

    The height the ball reached is 1.771m

    The total height of the ball from start is ∆h + S

    H=1.771+1

    H=2.771m

    Work done done by gravity = W*H

    Workdone = 4.9 * 2.771

    Work done=13.58Joules

    But the work done from when the ball is release up to reach the maximum height is given as

    Work done = W*s

    Work done = 4.9 * 1.771

    Work done = 8.68J

    This is the work done by gravity during the thrown alone, when the ball left the hand
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