Question

The tank shown is full of water.

(a) Find the work required to pump the water out of the spout.

(b) Suppose that the pump breaks down after\(4.7 \times {10^5}\;{\rm{J}}\)of work has been done. What is the depth of the water remaining in the tank?

Short answer

(a) The work done to pump water out from the spout is \(1.06 \times {10^6}J\).

(b) After the pump breaks down the level of water that remains in the tank is approximately \(2\) meters.

Step-by-step solution

Definition of work done.

Work done: The term "work done" refers to both the force given to the body and the displacement of the body.

Given

\(4.7 \times {10^5}\;{\rm{J}}\) work has been done

Calculating the total work done.

(a)

Let's consider a rectangular slice of water (the blue slice).

First we need to determine its volume: \(dV = 2xdy \times 8\) using similar triangles we have:

\(\begin{aligned}{}\frac{x}{y} &= \frac{{1.5}}{3} \to x\\ = \frac{1}{2}y \to 2x\\ =& y\end{aligned}\)

We conclude that:

\(\begin{aligned}{}dV &= 2xdy \times 8\\ =& ydy \times 8\end{aligned}\)

Let's determine its mass and weight\({m_i} = \)density\( \times \)volume

\(\begin{aligned}{} &= 1000ydy \times 8\\ =& 8 \cdot {10^3}ydy\end{aligned}\)

The weight of this slice is:

\(\begin{aligned}{}{F_i} &= {m_i}g\\ =& 9.8 \times 8 \times {10^3}ydy\end{aligned}\)

The work \({W_i}\) to raise this slice to the top is:

\(\begin{aligned}{}{W_i} &= {F_i}(5 - y)\\ =& 9.8 \times 8 \times {10^3}y(5 - y)dy\end{aligned}\)

To find the total work done in emptying the entire tank, we calculate the integral :\(\begin{aligned}{}W &= \int_0^3 9 .8 \times 8 \times {10^3}y(5 - y)dyW - \int_0^3 7 8400y(5 - y)dyW\\ &= 78400\int_0^3 {\left( {5y - {y^2}} \right)} dyW\\ &= \left. {78400\left( {\frac{{5{y^2}}}{2} - \frac{{{y^3}}}{3}} \right)} \right|_0^3W\\ &= 78400\left( {\frac{{45}}{2} - 9} \right)\\ =& 1058400J \approx 1.06 \times {10^6}J\end{aligned}\)

Therefore, the total work done is \(2\).

Calculating to get the solution of the cubic equation.

(b)

From part (a) we know that the work done to pump water from the spout is:

\(W - \int_0^3 7 8400\)

We are interested in compute the depth of the water at the moment in which the pump breaks down after\(4.7 \cdot {10^5}\;{\rm{J}}.\)

Then we put\(W = 4.7 \cdot {10^5}\;{\rm{J}}\)in (\(1\)) with the lower limit of integration as\(d\), where\(d\)is the claimed depth.

Hence:\(\int_d^3 7 8400y(5 - y)dy = 4.7 \cdot {10^5}\)

Integrating the above equation:

\(\begin{aligned}{}4.7 \cdot {10^5} &= 78400\int_d^3 y (5 - y)dy\\ = 78400\int {\left( {5y - {y^2}} \right)} dy\\ = 78400\left( {\int_d^3 5 ydy - \int_d^3 {{y^2}} dy} \right)\\ = 78400\left( {5\int_d^3 y dy - \int_d^3 {{y^2}} dy} \right)\\ = \left. {78400\left( {\frac{5}{2}{y^2} - \frac{1}{3}{y^3}} \right)} \right|_d^3\\ = 78400\left( {\frac{{5{{(3)}^2}}}{2} - \frac{{{{(3)}^3}}}{3} - \frac{5}{2}{d^2} + \frac{1}{3}{d^3}} \right)\\ =& 78400\left( {\frac{{45}}{2} - 9 - \frac{5}{2}{d^2} + \frac{1}{3}{d^3}} \right)\\ \approx 1.06 \cdot {10^6} - 196000{d^2} + 26133,33{d^3}n\\ \approx 1.06 \cdot {10^6} - 1.96 \cdot {10^5}{d^2} + 2.61 \cdot {10^4}{d^3}\end{aligned}\)

Then, we can write the above result as:

\(4.7 \cdot {10^5} \approx 1.06 \cdot {10^6} - 1.96 \cdot {10^5}{d^2} + 2.61 \cdot {10^4}{d^3}\)

or:

\(0.26{d^3} - 1.96{d^2} + 5.9 = 0\)

The above equation is cubic, and can be solved graphically. Below we show the graph of the function:

\(f(d) = 0.26{d^3} - 1.96{d^2} + 5.9\)

As we can see from the graph above, the solution of the cubic equation is:

\(d = 2.03 \approx 2\)meters.

Therefore, after the pump breaks down the level of water that remains in the tank is approximately \(2\) meters.