A tank is full of water. Find the work W required to pump the water out of the spout. (Use 9.8 for g and 3.14 for π. If you enter your answer in scientific notation, round the decimal value to two decimal places. Use equivalent rounding if you do not enter your answer in scientific notation.)
r = 3
h = 1.5
Assume all measurements are in metres (since g is given as 9.8 m/s^2)
The centre of gravity of the full tank is at 1.5/2=0.75 from the bottom.
Work done is equal to the potential energy of the water at the spout (top) minus the potential of the full tank.
Volume of tank, V=(πr^2 h)
Mass of water = V*ρ
where ρ=density of water = 1000 kg/m^3
PE of water in tank, PE1
=ρV*1.5g/2
PE of water at spout, PE2
=ρV*1.5g
Therefore
work done
=PE2-PE1
=ρV(1.5-0.75)g
=0.75ρg V
To find the work required to pump the water out of the spout, we can use the formula:
\(W = \rho \cdot V \cdot g \cdot h\)
Where:
\(W\) is the work required,
\(\rho\) is the density of water,
\(V\) is the volume of the water in the tank,
\(g\) is the acceleration due to gravity, and
\(h\) is the height of the water column
First, let's find the volume of the water in the tank. The tank is in the shape of a cylinder, so we can use the formula for the volume of a cylinder:
\(V = \pi \cdot r^2 \cdot h\)
Given that:
\(r = 3\) (radius of the tank)
\(h = 1.5\) (height of the water column)
Substituting these values into the formula, we get:
\(V = 3.14 \cdot 3^2 \cdot 1.5\)
Simplifying the equation:
\(V = 3.14 \cdot 9 \cdot 1.5\)
\(V = 42.39\) (rounded to two decimal places)
Next, substituting the given values and the calculated volume into the work formula:
\(W = \rho \cdot V \cdot g \cdot h\)
We need the density of water to continue.
Can you please provide the density of water?
To find the work required to pump the water out of the spout, we need to calculate the gravitational potential energy of the water in the tank. The work is equal to the change in gravitational potential energy.
The gravitational potential energy is given by the formula:
PE = mgh
Where:
m = mass of the water
g = acceleration due to gravity
h = height of the water
To find the mass of the water, we can use the formula for the volume of a cylinder:
V = πr^2h
Where:
V = volume
r = radius of the base
h = height
Substituting the given values:
r = 3
h = 1.5
V = π(3^2)(1.5)
V = 3.14(9)(1.5)
V ≈ 42.39 cubic units
The mass of the water can be found by multiplying the volume by the density of water:
m = Vρ
Where:
ρ = density of water
The density of water is approximately 1000 kg/m^3. Since the units of volume are cubic units, we need to convert the density to the appropriate units:
ρ = 1000 kg/m^3 * (1 m/100 cm)^3 = 1000 kg/cm^3
m = 42.39 cubic units * 1000 kg/cm^3
m ≈ 42,390 kg
Now, we can calculate the gravitational potential energy:
PE = mgh
PE = (42,390 kg)(9.8 m/s^2)(1.5 cm)
PE ≈ 613,392 J
Therefore, the work required to pump the water out of the spout is approximately 613,392 Joules.