Answer:
The work W required to pump the oir out of the spout is \(W=62344.9 J\)
Step-by-step explanation:
We are given the density of the oil, and we can calculate the volume of the oil as
\(V=\pi r^2\frac{h}{2}\)
because the tank is half full.
The we can calculate the oil mass as
\(m=\rho V=\rho \pi r^2\frac{h}{2}\)
Now, to calculate the work W necessary to pump the oil out, we have that
\(W=\int\limits^{h_{f}} _{h_{i}} {mg} \, dh\)
where m is already calculated, g is given in the problem, and h goes from 0m to 1m.
Finally we can calculate the work W:
\(W=\int\limits^{1m} _{0m} {\rho g \pi r^2 \frac{h}{2}} \, dh=\frac{\rho}{2} g \pi r^2\int\limits^{1m} _{0m} {h} \, dh\)
Therefore
\(W=\frac{\rho}{4} g \pi r^2 m^2=\frac{900}{4}\frac{kg}{m^3}*9.8\frac{m}{s^2}*9\pi m^2*m^2=19845\pi Nm=62344.9 J\)
is the work required to pump the oil out of the spout.
