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12 March, 15:16

A bag of sand originally weighing 144 lb is lifted at a constant rate. As it rises, sand also leaks out at a constant rate. At the instant when the bag has been lifted to a height of 18 feet, exactly half of the original amount of sand remains. (Neglect the weight of the bag and the lifting equipment.) How much work was done lifting the sand this far?

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  1. 12 March, 15:17
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    Step-by-step explanation:

    The bag of sand wave is

    W=144lb at x=0

    The bag of sand is lifted at a constant rate, i. e the it is not accelerating, so a=0m/s²

    Let W be the mass of the sand,

    Sand leaks out at a constant rate

    dW/dx = C

    At a height=18ft, half of the original mass.

    i. e x=18ft, M=72lb

    We are asked to find the work at 18ft

    Work is given as

    Work=∫F•dx,

    Where F is the force and also the weight of the sand, F=W

    The Weight of the bag is a linear function,

    dW/dx = C

    Then, using variable separation

    dW=Cdx

    Integrating both sides

    ∫dW=∫Cdt

    W=Cx + B

    So, at x=0, W=144

    144=B

    B=144

    W=Cx+144

    Also at x=18, W=72lb

    72=C*18+144

    72-144=18C

    -72=18C

    C=-4

    Then, the weight function becomes

    W=-4x+144

    W=144-4x

    Then applying work formula

    Work=∫W•dx

    Work=∫ (144-4x) dx. x=0 to x=18

    Work = 144x-4x²/2. x=0 to x=18

    Work=144x-2x² x=0 to x=18

    Work=144 (18) - 2 (18²) - 0 - 0

    Work = 2592-648

    Work = 1944 ft lbs

    The work done in lifting the sand to 18ft is 1944 ft lbs
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