Physics, 28.06.2019 03:30 Jana1517

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.)

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Work is the product of force and displacement. The amount of work needed to pump half of the water out of the aquarium is 4900 J.

What is work done?

Work done can be defined as the amount of force needed to displace an object from one location to another.

$W = F \cdot ds$

Given to us

Volume of the aquarium, V = 4 x 1 x s = 4m³

Height of the aquarium, s = 1 m

Acceleration due to gravity, g = 9.8 m/s²

Density of water, ρ = 1000 kg/m³

We know about work done, it is given as,

$W = F \cdot ds$

We also know that force can be written as,

$F = m \cdot a$

Also, the mass can be written as,

$m = \rho \times V$

Therefore, work can be written as,

$W = F\cdot ds\\\\W = m \cdot a \cdot ds\\\\W = (\rho \times V \times a )ds\\\\W = \int (\rho \times V \times a )ds$

As we need to pump half of the water, therefore, from below the tank to half the distance

$W = \int_0^{\frac{1}{2}} (\rho \times V \times a )ds\\\\W = \int_0^{\frac{1}{2}} (\rho \times (4 \times 1 \times s) \times a )ds\\\\W = (\rho \times 4 \times a )[s^2]_0^{\frac{1}{2}\\\\$

Substitute all the values,

$W = 1000\times 4 \times g \times[(\dfrac{0.5}{2}^2) -0^2]\\\\W = 1000\times 4 \times 9.8 \times 0.125\\\\W = 4900\ J$

Hence, the amount of work needed to pump half of the water out of the aquarium is 4900 J.

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An aquarium 4 m long, 1 m wide, and 1 m deep is full of water. find the work needed to pump half of

Answer from: ennasawesome

4900 J

Explanation:

Let's begin by explaining that the Work done by a Force 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 . 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 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

Answer from: Quest

higher potential

explanation:

because it is in a high speed so it must move into a of high potential

Answer from: Quest

god i hope this !

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