Question

A baloon filled with helium rises against gravity increasing its potential energy. The speed of the baloon also increases as it rises. How do you reconcile this with the law of conservation of mechanical energy? You can neglect viscous drag of air and assume that density of air is constant.

Answer 1

Let m , V ρ_He denote respectively the mass, volume and density of helium baloon and ρ_air be density of air

Volume V of baloon displaces volume V of air.

If the baloon rises to a height h, from s = ut + (1/2)at², we, geAt h = (1/2)at² 

So, as the baloon goes up, an equal volume of air comes down, increase in PE and KE of the baloon is at the cost of PE of air [which comes down].