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A cylindrical glass contains a liquid of density rho to a height h0. The volume of the liquid is V0. A straw of uniform cross-section A is inserted and a person drinks with constant speed u by maintaining the top pressure below atmospheric pressure. The total work done by the person to consume all the liquid is (where h is the height of the person's mouth above the base of the glass):
- W = rho*V0 * [g*(h - h0/2) + (1/2)*u²]
- W = rho*V0 * [g*(h - h0) + (1/2)*u²]
- W = (1/2)*rho*V0*u² + rho*V0*g*h
- W = (1/2)*rho*V0*u² + rho*g*h0
Correct answer: W = rho*V0 * [g*(h - h0/2) + (1/2)*u²]
Solution
The person does work equal to the energy needed to raise all liquid from its average initial height (h0/2) to the mouth height h, plus the kinetic energy imparted as liquid exits at speed u. This gives W = rho*V0*[g*(h-h0/2) + (1/2)*u²].
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