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Math Help - integral physics application problems

  1. #1
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    integral physics application problems

    We are now applying calculus to physics problems and I am having trouble coming up with the correct integrals.

    An anchor weighing 100 lb in water is attached to a chain weighing 3 lb per foot in water. Find the work done to haul the anchor and chain to the surface of the water from a depth of 25 ft.

    I know I need an integral from 0 to 25 but what do I need to multiply my distance term by????


    A rectangular swimming pool 50 ft long by 20 ft wide and 10 ft deep is filled with water to a depth of 9 ft. Use an integral to find the work required to pump all the water out over the top.

    This one I thought would just be the integral from 0 to 10 of

    50 x 20 x 9 x density of water per cubic foot times y dy but I am not getting the correct answer.

    Thank you for making this site so helpful and helping not pull all of my hair out!!!!!
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  2. #2
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    A rectangular swimming pool 50 ft long by 20 ft wide and 10 ft deep is filled with water to a depth of 9 ft. Use an integral to find the work required to pump all the water out over the top.

    This one I thought would just be the integral from 0 to 10 of

    50 x 20 x 9 x density of water per cubic foot times y dy but I am not getting the correct answer.

    Thank you for making this site so helpful and helping not pull all of my hair out!!!!!
    Since the pool has 9 feet of water, but is 10 feet deep, we can use the

    62.4(50)(20)\int_{0}^{9}(10-y)dy

    62,400\int_{0}^{9}(10-y)dy
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  3. #3
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    An anchor weighing 100 lb in water is attached to a chain weighing 3 lb per foot in water. Find the work done to haul the anchor and chain to the surface of the water from a depth of 25 ft.

    I know I need an integral from 0 to 25 but what do I need to multiply my distance term by????
    Hello Frostking,

    Lets write a linear function for the weight(force) exerted by the chain and anchor.

    Facts we know:
    Before we pull any of the chain up it the total weight is
    100 +3(25)=175 \\\ f(0)=175

    When the anchor is at the surface we only have the weight of the anchor
    f(25)=100

    We can now write the equation of the line between these two points to get

    f(x)=-3x+175

    Now we just need to integrate

    \int_{0}^{25}(-3x+175)dx

    I hope this helps.
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  4. #4
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    We are now applying calculus to physics problems and I am having trouble coming up with the correct integrals.

    An anchor weighing 100 lb in water is attached to a chain weighing 3 lb per foot in water. Find the work done to haul the anchor and chain to the surface of the water from a depth of 25 ft.

    I know I need an integral from 0 to 25 but what do I need to multiply my distance term by????
    \int_{0}^{25}(100+3(25-y))dy
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