Calculate Work Homework: Hemispherical Tank Full of Water

[norcal]

Homework Statement

The hemispherical tank shown is full of water. Given that water weighs 62.5 lb/ft3, find the work required to pump the water out of the tank. (the water will be pumped out of a spigot at the top)

Homework Equations

INTEGRAL from 0 to 5 of: pi*(5-h)(25-h^2)dh

The Attempt at a Solution

INTEGRAL from 0 to 5 of: pi*(h^3-5h^2-25h+125)dh

pi*((h^4)/4 - (5h^3)/3 - 12.5h^2 + 125h)

= (3125/12)*pi ft-lb.

This answer is incorrect though...anyone know what I am doing wrong? Thanks.

 

[Dick]

You didn't use the density of water data. How would you incorporate this?

 

[m2003]

As a scientist, it is important to always double check your calculations and equations to ensure accuracy. In this case, it seems that you may have made a mistake in your integration or in setting up the equation. It is always helpful to show your work and steps so that others can help identify where the error may have occurred.

Additionally, it is important to note the units in your final answer. The given weight of water is in pounds per cubic foot, so the work should also be in units of foot-pounds. It may be helpful to convert the units before solving the problem.

Another approach to solving this problem could be using the work-energy principle, which states that the work done by an external force is equal to the change in kinetic energy of the system. In this case, the initial kinetic energy of the system (water in the tank) is zero, and the final kinetic energy is also zero since the water is being pumped out of the tank at a constant rate. Therefore, the work done by the pump is equal to the change in potential energy of the system, which is equal to the weight of the water (62.5 lb/ft3) multiplied by the height of the tank (5 ft). This would give a final answer of 312.5 ft-lb, which is the same as your attempted solution.

Overall, it is important to carefully check your equations, units, and approach when solving a problem like this. If you are still unsure, it is always helpful to seek assistance from a peer or instructor.