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.

Explanation

Let `V=` volume of helium (baloon), `m=` mass of baloon
`rho_(He)=` density of helium, `rho _(a i r)=` density of air
Volume `V` of baloon displaces volume `V` of air.
`:.` The baloon experience as upthrust `=V(rho_(air)-rho_(He))g=ma=m(dupsilon)/(dt)` ...(i)
Integrating both sides w.r.t. t, we get
`v(rho_(air)_rho_(He))g t=m upsilon` ...(ii)
From (i), `a=(V(rho_(air)-rho_(He))g)/(m)`
If the baloon rises to a height `h` , then from
`s=ut=(1)/(2)at^(2)`
`h=0+(1)/(2)(V(rho_(air)-rho_(He))g t^(2))/(m)` ...(iii)
`K.E.` of baloon `=(1)/(2)m upsilon^(2)=((m upsilon)^(2))/(2m)=((rho_(air)-rho_(He))^(2)V^(2)g^(2)t^(2))/(2m)` ...using (ii)
`=(1)/(2) m upsilon^(2)=V(rho_(air)-rho_(He))gxx[((rho_(air)-rho_(He))V)/(2m)g t^(2)]`
using (iii) , we get
`(1)/(2) m upsilon^(2)=V(rho_(air)-rho_(He))g h`
`(1)/(2)m upsilon^(2)+Vrho_(He)gh=Vrho_(air)gh, i.e., ` K.E. of baloon `+` P.E. of baloon `=` change in P.E. of air.
Hence we conclude that as the Helium baloon goes up, an equal volume of air comes down. Increase in P.E. and K.E. of baloon is at the cost of P.E. of air (that comes down).