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8 December, 00:23

A shot-putter puts a shot (weight = 71.5 N) that leaves his hand at a distance of 1.84 m above the ground. (a) Find the work done by the gravitational force when the shot has risen to a height of 2.50 m above the ground. Include the correct sign for the work. (b) Determine the change (PEf - PE0) in the gravitational potential energy of the shot.

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  1. 8 December, 03:24
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    a) W = - 47.19 J

    b) 47.19 J

    Explanation:

    Given that

    Weight, mg = 71.5 N

    y = 1.84 m

    H = 2.5

    a)

    The distance above the hand, h = 2.5 - 1.84 m

    h = 0.66 m

    We know that gravitaional force act in the downward direction but the displacement is in upward direction that is why work done will be negative.

    W = - m g h

    W = - 71.5 x 0.66 J

    W = - 47.19 J

    b)

    The potential energy at initial position = m g y

    The potential energy at final position = m g H

    So change in the potential energy = m g H - m g y

    = mg (H - y)

    =71.5 (2.5 - 1.84) = 47.19 J

    Therefore change in the potential energy = 47.19 J
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