# Thread: integral physics application problems

1. ## 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!!!!!

2. 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

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

$62,400\int_{0}^{9}(10-y)dy$

3. 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.

4. 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????
$\int_{0}^{25}(100+3(25-y))dy$