Question

An aquarium 4 m long, 1 m wide, and 1 m deep is full of water. Find the work needed to pump half of the water out of the aquarium. (Use 9.8 m/s2 for g and the fact that the density of water is 1000 kg/m3.)

Answer 1

Answer: 4900 J

Explanation:

Let's begin by explaining that the Work $W$ done by a Force $F$ refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a distance $y$.  When the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:  

$W=F.y$ (1)

In the case of this rectangular aquarium (see figure attached), we know its initial volume $V_{i}$ (which is the same volume occupied by water, because the tank is full) is:

$V_{i}=4m.1m.1m=4m^{3}$ (2)

In addition, we know the force needed to pump a small amount of water is:

$F=m_{water}.g$ (3)

On the other hand, we know the density of the water $\rho_{water}$ is given by the following equation:

$\rho_{water}=\frac{m_{water}}{V}$ (4)

Where $m_{water}$ is the mass of water and $V=4m.1m.dy$ is the volume of a thin "sheet" of water.

Finding $m_{water}$:

$m_{water}=\rho_{water}.V$  (5)

Substituting (5) in (3):

$F=\rho_{water}.V.g$   (6)

And substituting (6) in (1):

$W=\rho_{water}.V.g.y$ (7)

Now, we are asked to find the work needed to pump half of the water out of the aquarium. So, if the aquarium is $1m$ deep, the half is $0.5m$:

$W=(1000\frac{kg}{m^{3}})(4m.1m.dy)(9.8\frac{m}{s^{2}})y$ (8)

$W=39200\frac{kg.m^{6}}{s^{2}}ydy$ (9)

Well, in order to solve this, we have to write the definite integral from y=0 mto y=0.5m:

$W=39200\frac{kg.m^{6}}{s^{2}}\int\limits^{0.5}_0 {y} \, dy$ (10)

Knowing $\int\limits^b_a {f(y)} \, dy= F(b)-F(a)$:

$W=39200\frac{kg.m^{6}}{s^{2}}(\frac{(0.5m)^{2}}{2}-\frac{(0m)^{2}}{2})$ (11)

$W=4900kg\frac{m^{2}}{s^{2}}$ (12)

$W=4900J$ >>>>This is the work needed to pump half of the water out of the aquarium