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.
asked Dec 5, 2013 at 6:11 21 1 1 gold badge 1 1 silver badge 3 3 bronze badges $\begingroup$ Start with the definition of work. $\endgroup$ Commented Dec 5, 2013 at 6:27 $\begingroup$ Is that "filled to a depth of 3 metres" or "filled with 3 cubic metres"? $\endgroup$ 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 $\endgroup$
Commented Dec 5, 2013 at 7:03The 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 $\fracy$, a volume of $\pi \frac y^2 \Delta y$ and mass of $\pi \frac y^2 \Delta y \rho$ and needs to be raised a height of $7-y$ which requires work of $\pi \frac y^2 \Delta y \rho g(7-y)$. To empty the tank requires:
$$\begin W&=\int_0^3 \pi \frac \rho g (7y^2-y^3)dy\\ &=\pi \frac \rho g \left(7\frac-\frac\right)\\ &=\pi \frac\times 1050 \times 9.8 \times\frac\\ &\approx 254\text\\ \end$$
answered Dec 10, 2013 at 0:44 2,979 1 1 gold badge 13 13 silver badges 23 23 bronze badges $\begingroup$The radius of the cone formed by liquid chocolate is $\frac37\times 3=\frac97m$ . The volume of this chocolate cone is $\frac13\pi \times (\frac97)^2 \times 3=\frac\pi m^3$ . Its mass is $1050\times \frac\pi=\frac\pi kg$ .
The centre of mass of the chocolate in the cone is $\frac34 \times 3=\frac94m$ from the vertex, which is itself $7=\fracm$ below the base, so the CM is $\fracm$ 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 $\fracm$ . The work required to do this is $\frac\pi\times 9.8 \times \frac \approx 254 kJ$ .
answered Apr 9, 2019 at 16:44 sammy gerbil sammy gerbil 910 7 7 silver badges 12 12 bronze badgesTo subscribe to this RSS feed, copy and paste this URL into your RSS reader.
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