Answer: To find the work required to pump the water to the top of the tank, we need to calculate the change in potential energy of the water as it is moved from the halfway point (y = 2 meters) to the top of the tank (y = 4 meters).
Given data:
Square base side length (s) = 2 meters
Height of the tank (h) = 4 meters
Density of water (ρ) = 1000 kg/m³
Acceleration due to gravity (g) = 9.8 m/s²
Let's proceed with the calculation:
Volume of water in the tank (V_water):
The tank is half-filled with water, so the volume of water is half the total volume of the tank.
Volume of water = (1/2) * Base area * Height
V_water = (1/2) * (s²) * h
V_water = (1/2) * (2 m)² * 4 m
V_water = 4 m³
Mass of the water (m_water):
Mass = Volume * Density
m_water = V_water * ρ
m_water = 4 m³ * 1000 kg/m³
m_water = 4000 kg
Change in height (Δy):
The change in height of the water is from the halfway point (y = 2 meters) to the top of the tank (y = 4 meters).
Δy = 4 m - 2 m
Δy = 2 m
Work required (W) to pump the water to the top:
The work done is equal to the change in potential energy.
Potential energy (PE) = Mass * Gravitational acceleration * Change in height
PE = m_water * g * Δy
PE = 4000 kg * 9.8 m/s² * 2 m
PE = 78,400 kg·m²/s²
Since 1 Joule (J) is equal to 1 kg·m²/s², the work required (W) is 78,400 Joules.
Rounded to one decimal place, the work required to pump the water to the top of the tank is approximately 78,400.0 Joules.