Question

An aquarium 2 m long, 1 m wide, and 1 m deep is full of water. Find the work (in J) needed to pump half of the water out of the aquarium. Use the fact that the density of water is 1000kg/m3.

Answer 1

To pump out half the water from an aquarium that is 2 m long, 1 m wide, and 1 m deep, you need to perform 4900 J of work. This calculation takes into account the volume of water, the density of water, and gravitational force.

Step-by-step explanation:

To calculate the work needed to pump half of the water out of the aquarium, we have to consider the volume of water in the aquarium, the density of water, and the gravitational force that we need to overcome. The volume of the aquarium is 2 m long by 1 m wide by 1 m deep, which equals 2 cubic meters. Since we need to pump out half of the water, the volume of water to be pumped is 1 cubic meter.

Given that the density of water is 1000 kg/m³, the mass of 1 cubic meter of water is 1000 kg. To find the work done against gravity, we use the formula Work = force × distance. The force here is the weight of the water, which is mass multiplied by the acceleration due to gravity (9.8 m/s²). Therefore, the force is 1000 kg × 9.8 m/s² = 9800 N (newtons).

The distance through which this force must act is the average height at which the water is pumped out. Since the tank is 1 m deep, and we're only pumping out the top half, the average height is 0.5 m. Thus, the work done to pump out half the water is 9800 N × 0.5 m = 4900 J (joules).

Answer 2

`W=2452.5 J`

Step-by-step explanation:

The work is define as the integral of the force times distance. So we have:

`W=\int^(a)_(b)Fxdx`

Now, we can write the force in terms of density.

`F=m*g=\rho Vg`

V is the volume (V=2*1*1=2 m³)

So the work will be:

`W=\int^(0.5)_(0) 2*1000*9.81*xdx=\int^(0.5)_(0) 2*1000*9.81*xdx=\int^(0.5)_(0)19620xdx=19620((x^(2))/(2))|^(0.5)_(0)`

The limit of integration is between 0 and 0.5 because we want to pump half of the water out of the aquarium.

`W=19620((x^(2))/(2))|^(0.5)_(0)=19620((0.5^(2))/(2))`

`W=2452.5 J`

I hope it helps you!