Work — Question 5

Question 5 – Solution

1. Choose horizontal slices.

Let $y$ be height above the tank bottom, with $0\le y\le4$. A slice at height $y$ has thickness $dy$.

2. Find the slice volume and weight.

The horizontal circular area is $A=\pi(1)^2=\pi\ \mathrm{m^2}$. Hence$$dV=\pi\,dy,\qquad dF=9800\,dV=9800\pi\,dy.$$The density is already a weight density.

3. Find the distance to the outlet.

The outlet is at the top, $y=4$, so the slice must rise $4-y$ meters:$$dW=9800\pi(4-y)\,dy.$$

4. Set up and evaluate the work integral.

$$W=9800\pi\int_0^4(4-y)\,dy=9800\pi\left[4y-\frac{y^2}{2}\right]_0^4.$$$$W=9800\pi(16-8)=78400\pi.$$

5. State and check the work.

$$\boxed{W=78400\pi\ \mathrm{J}}.$$The total water weight is $9800(4\pi)=39200\pi$ N. Its average lifting distance is $2$ m, giving $78400\pi$ J.