# Work problem

• Nov 18th 2006, 08:20 PM
synnexster
Work problem
find how much work it takes to empty a full hemispherical water reservoir of radius 5m by pumping the water to a height of 4 m above the top of the reservoir. water weighs 9800 N/m^3.

I need to find the volume first so i think it's v=integral (pi*r^2)

so V = pi * integral (25-y^2) dy from 0 to 5
which is 250*pi/3
mass = density * volume = 9800n/m^3 * 250*pi/3 = 2.45*10^6*pi / 3
force = mass * 9.8m/s^2 (gravity) = 2.401*pi*10^7 /3
then work = force*x = [2.401*pi*10^7 /3] *4m

am i doing this right?
• Nov 19th 2006, 01:45 AM
CaptainBlack
Quote:

Originally Posted by synnexster
find how much work it takes to empty a full hemispherical water reservoir of radius 5m by pumping the water to a height of 4 m above the top of the reservoir. water weighs 9800 N/m^3.

I need to find the volume first so i think it's v=integral (pi*r^2)

so V = pi * integral (25-y^2) dy from 0 to 5
which is 250*pi/3
mass = density * volume = 9800n/m^3 * 250*pi/3 = 2.45*10^6*pi / 3
force = mass * 9.8m/s^2 (gravity) = 2.401*pi*10^7 /3
then work = force*x = [2.401*pi*10^7 /3] *4m

am i doing this right?

Consider a disk of thickness $ds$ $s$ metres below the brim of the tank. this has volume:

$
dV = \pi (5^2-s^2) ds
$

which has mass:

$
dM=\rho dV=1000 dV
$

the $1000$ is the density of water which is $1000 kg/m^3$.

The work done to lift this to a height of $4m$ above the brim of the tank is:

$
dW=dM\ g\ (s+4)
$

that is the mass times g times the height lifted; as this is the change in potential energy of the water.

So the total work to empty the tank is:

$
W=\int_0^5 (s+4)\ g\ 1000\ \pi\ (5^2-s^2) ds=\int_0^5 (s+4)\ 9800 \ \pi\ (5^2-s^2) ds
$

(where we take $g\approx 9.8 m/s^2$.)

RonL