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.) The tank is an inverted triangular prism with a length of 18 m, a base of 4 m,a height of 4 m, and a spout that is 4 m tall on one end.

To find the work \( W \) required to pump the water out of the spout from the tank, we can use the concept of work in terms of lifting a mass. The work done to move an object against gravity is given by:

\[
W = \int F \, dy
\]

Where \( F \) is the force required to lift a small slice of water over a distance \( y \), and \( y \) is the height from the bottom of the tank. In this case, we need to consider the geometry of the tank and how the volume of water changes with height.

### Step 1: Analyze the geometry of the tank
The tank is an inverted triangular prism. The cross-section of the tank is a triangle with a base of 4 m and a height of 4 m. The length of the tank is 18 m. 

The area of a triangular slice of the tank at height \( y \) is proportional to the square of the height because the triangle is similar at all heights.

Let’s define a slice at a height \( y \) above the bottom of the tank. The width of the tank at this height is given by:

\[
\text{Width at height } y = \frac{4}{4} \times y = y
\]

The area of the triangular slice at height \( y \) is:

\[
A(y) = \frac{1}{2} \times \text{base at height } y \times \text{height at height } y = \frac{1}{2} \times y \times 4 = 2y
\]

The volume of the small slice of water is the area of the triangular slice times the thickness \( dy \), and the mass of this slice is the density of water times the volume.

### Step 2: Calculate the force
The density of water is \( 1000 \, \text{kg/m}^3 \), so the weight of a slice of water is:

\[
dF = \text{density} \times \text{volume of the slice} \times g
\]

The volume of the slice at height \( y \) is:

\[
\text{Volume} = A(y) \times \text{thickness} = 2y \times 18 \, dy = 36y \, dy
\]

Thus, the force required to lift this slice is:

\[
dF = 1000 \times 36y \, dy \times 9.8 = 352800 y \, dy
\]

### Step 3: Set up the work integral
The slice at height \( y \) needs to be lifted to the top of the spout, which is located 4 meters above the tank. Therefore, the distance each slice must be lifted is \( (8 - y) \), since the total height of the system is 8 meters (4 meters for the spout and 4 meters for the tank).

The work required to lift a slice is:

\[
dW = dF \times (8 - y) = 352800 y \, dy \times (8 - y)
\]

To find the total work, we integrate from \( y = 0 \) to \( y = 4 \):

\[
W = \int_0^4 352800 y (8 - y) \, dy
\]

### Step 4: Solve the integral
Let's expand the integrand:

\[
W = 352800 \int_0^4 (8y - y^2) \, dy
\]

Now, solve the integral:

\[
W = 352800 \left[ \frac{8y^2}{2} - \frac{y^3}{3} \right]_0^4
\]

Evaluating the integral at the limits \( y = 0 \) and \( y = 4 \):

\[
W = 352800 \left[ \frac{8(4^2)}{2} - \frac{4^3}{3} \right]
\]

\[
W = 352800 \left[ \frac{8 \times 16}{2} - \frac{64}{3} \right]
\]

\[
W = 352800 \left[ 64 - \frac{64}{3} \right]
\]

\[
W = 352800 \times \left( \frac{192}{3} - \frac{64}{3} \right)
\]

\[
W = 352800 \times \frac{128}{3}
\]

\[
W = \frac{352800 \times 128}{3}
\]

\[
W = 15084800 \, \text{J}
\]

Thus, the total work required to pump the water out of the spout is approximately \( W = 15,084,800 \, \text{Joules} \).

---

Let me know if you'd like any further clarification!

### Relative Questions:
1. How would the work change if the spout height increased by 2 meters?
2. What happens to the work if the tank's base changes to 5 meters instead of 4 meters?
3. How does the shape of the tank impact the amount of work required?
4. Can you explain how the density of water affects the total work?
5. How would the problem differ if the tank were cylindrical instead of a triangular prism?

### Tip:
Remember that the work required to pump liquid depends not just on the volume but also on how far the liquid must be lifted!

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.) The tank is an inverted triangular prism with a length of 18 m, a base of 4 m, a height of 4 m, and a spout that is 4 m tall on one end.

Learn how to calculate the work required to pump water out of an inverted triangular prism tank. Using calculus, integration, and the work-energy theorem, we solve the problem step-by-step, considering the tank's geometry and fluid properties. Suitable for college-level students.

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