# Thread: Work needed to pump water from inverted cone

1. ## Work needed to pump water from inverted cone

Hi, I know

the gist of how to do these problems but I am getting the wrong answers.

A tank in the shape of an inverted right circular cone has height 6 meters and radius 3 meters. It is filled with 5 meters of hot chocolate. Find the work required to empty the tank by pumping the hot chocolate over the top of the tank. The density of hot chocolate is

volume= pi r^2
r=3, y=6 so r=.5y
distance is from 0 to 5 meters
density is given as 1010

So W= density * volume * distance

W= 1010 [pi(.5y)^2]dy(6-y)
w=intregal from 0 to 5 (6-y)1010pi(1/4)y^2dy
=(1010pi/4) intregal from 0 to 5, 6y^3-y^3dy

and so on... I got 74367.388 J but it's wrong. any help will be appreciated. Thanks

2. ## Re: Work needed to pump water from inverted cone

$\displaystyle W = \delta g \pi \int_0^5 \left(\frac{y}{2}\right)^2 \cdot (6-y) \, dy$

where $\displaystyle g = 9.8 \, m/s^2$

so ... multiply your result by 9.8

3. ## Re: Work needed to pump water from inverted cone

Thank you!

Also I have a really quick question, if instead of giving you density of a liquid, it give you the weight. Example. the weight of the liquid is 60 lbs/ft^3

What does that mean? I know F=ma=Density * Volume*acceleration so does that weight of a liquid= density*volume? Thanks

4. ## Re: Work needed to pump water from inverted cone

Originally Posted by makeupgirl1107
Thank you!

Also I have a really quick question, if instead of giving you density of a liquid, it give you the weight. Example. the weight of the liquid is 60 lbs/ft^3

What does that mean? I know F=ma=Density * Volume*acceleration so does that weight of a liquid= density*volume? Thanks
that would be the weight density ... $\displaystyle \delta g$

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