Christopher B. answered • 03/09/22
Tutor
5.0
(308)
Mechanical Engineering Major and Former Physics Teacher
Hey Shannen,
You could actually do this without calculus, since the cross-sectional area of the tank remains constant in a cylinder. For this approach, here are some steps:
- Determine the volume of water in the tank in m3 based on the dimensions given (cross-sectional area times height).
- Find the mass of this water using the general equation for density.
- Find the weight of that water using F = mg
- Now, the work done is equal to the force times the distance. Since the weight is evenly distributed vertically (ie: the pool does not change width at all), we can take the average distance of each drop of water.
- The shortest distance is .5 m (for part a) and the longest is 4.0m, making the average distance 2.25 m.
- Multiply the weight of the water by the average distance to find the total work to pump out the water.
Calculus method: (You will need this for problems when the cross-sectional area is changing)
- The main difference is that, instead of the total volume of water, you want to find the volume of an infinitesimally thin horizontal sliver of water. We take a horizontal sliver because each "droplet" of that sliver requires the same amount of work to pump out, since it moves the same vertical distance.
- dV = 25 π dh
- Use the same steps as before to determine the weight of this sliver, in terms of dh.
- Now the question is what height does this sliver have to move vertically to get out of the tank.
- You can write this distance as 4-h, since it has to go from the height that it is now up to 4.0 m.
- Now you can set up your integral: W = ∫ weight * d
- Careful with the bounds - they will be from 0 to 3.5 m.
- You can check with the non-calculus method to make sure it's the same.
The only difference for part b is that your expression for distance changes slightly.
