A tank in the shape of an inverted right circular cone (the ice cream goes on top) has height 7 meters and radius 3 meters. It is filled with 3 meters of hot chocolate. Find the work required to empty the tank by pumping the hot chocolate over the top of the tank. The density of hot chocolate is 1050 kg/m^3.

  • 1
    $\begingroup$ Start with the definition of work. $\endgroup$
    – shade4159
    Commented Dec 5, 2013 at 6:27
  • $\begingroup$ Is that "filled to a depth of 3 metres" or "filled with 3 cubic metres"? $\endgroup$
    – apt1002
    Commented Dec 5, 2013 at 6:46
  • $\begingroup$ Filled to a depth of 3 meters. Here is what I did and I don't know where I went wrong: y=7/3x so x=3/7y V= pi(3/7)^2*1050 kg/m^3* 9.8m/s^2*(7-y) Then I took the integral of that from 0 to 3 $\endgroup$
    – shmoo
    Commented Dec 5, 2013 at 7:03

2 Answers 2


The force due to gravity is $F=mg$.

The work required to lift a body through a height $h$ is $W=Fh=mgh$.

An infinitesimal slice of chocolate $\Delta y$ at height $y$ has a radius of $\frac{3}{7}y$, a volume of $\pi \frac{9}{49} y^2 \Delta y$ and mass of $\pi \frac{9}{49} y^2 \Delta y \rho$ and needs to be raised a height of $7-y$ which requires work of $\pi \frac{9}{49} y^2 \Delta y \rho g(7-y)$. To empty the tank requires:

$$\begin{align} W&=\int_0^3 \pi \frac{9}{49} \rho g (7y^2-y^3)dy\\ &=\pi \frac{9}{49} \rho g \left(7\frac{3^3}{3}-\frac{3^4}{4}\right)\\ &=\pi \frac{9}{49}\times 1050 \times 9.8 \times\frac{171}{4}\\ &\approx 254\text{kJ}\\ \end{align}$$


The radius of the cone formed by liquid chocolate is $\frac37\times 3=\frac97m$. The volume of this chocolate cone is $\frac13\pi \times (\frac97)^2 \times 3=\frac{81}{49}\pi m^3$. Its mass is $1050\times \frac{81}{49}\pi=\frac{12150}{7}\pi kg$.

The centre of mass of the chocolate in the cone is $\frac34 \times 3=\frac94m$ from the vertex, which is itself $7=\frac{28}{4}m$ below the base, so the CM is $\frac{19}{4}m$ below the base.

To pump the chocolate out of the cone its CM must be raised up to the height of the base, ie by a distance of $\frac{19}{4}m$. The work required to do this is $\frac{12150}{7}\pi\times 9.8 \times \frac{19}{4} \approx 254 kJ$.


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