A tank is full of water. Find the work W required to pump the water out of the spout. (Use 9.8 m/s2 for g. Use 1000 kg/m3 as the weight density of water.

Question

A water tank shaped like a triangular prism is given. The base of the prism is at the top and a spout is located at the top of the tank. The dimensions of the tank are 9 meters long, 4 meters wide, and 4 meters deep. The spout is 3 meters long.

Answer 1

1. **Volume element for water (differential slice):**

  A small horizontal slice of water at a depth y meters below the top of the tank has volume dV.   The cross-sectional area of the tank is rectangular (9 meters by 4 meters), so the volume of a small slice of thickness dy is:

  

  dV = 9 × 4 × dy = 36 dy

2. **Mass of the water slice:**

  The mass of a small slice is:

  

  dm = ρ × dV = 1000 × 36 dy = 36000 dy

3. **Work to pump each slice out:**

  The force to lift this slice is F = dm × g:

  

  F = 36000 × 9.8 dy = 352800 dy

  

  The distance the water must be lifted depends on the depth y. The distance to lift a slice from depth y is (4 - y).

  Therefore, the work to lift this slice is:

  

  dW = 352800 × (4 - y) dy

4. **Total Work:**

  To find the total work, integrate from y = 0 to y = 4 (the depth of the tank):

  

  W = ∫ 0^4 352800 × (4 - y) dy

  

  Performing the integration:

  

  W = 352800 × [4y - (y² / 2)] from 0 to 4

  W = 352800 × [16 - (16 / 2)] = 352800 × 8 = 2,822,400 Joules